PHYS 101 - ELEMENTS OF PHYSICS
- Moment of Inertia
Question Bank - Set 1
Liberty University
Question 1
Question
Calculate the moment of inertia of a thin uniform rod of length Land mass M
about an axis perpendicular to the rod and passing through one end.
Solution
Let’s consider the thin uniform rod of length Lto be along the x-axis with one
end at the origin. We will calculate the moment of inertia Iabout the axis
perpendicular to the rod and passing through the origin.
Step 1: Determine the mass per unit length of the rod. The mass per unit
length λof the rod is given by:
λ=M
L
Step 2: Express the differential mass element dm in terms of dx. To express
the differential mass element dm, consider a small element of length dx at a
distance xfrom the origin. The mass of this element dm is given by:
dm =λ dx
Step 3: Express the moment of inertia dI of the differential mass element.
The moment of inertia dI of the differential mass element about the axis passing
through the origin is given by:
dI = (x2)dm = (x2)·(λ dx)
Step 4: Integrate to find the total moment of inertia I. Integrating the
expression for dI over the length of the rod from 0 to Lgives the total moment
of inertia I:
I=ZL
0
(x2)·(λ dx)
I=ZL
0
(x2)·M
Ldx
I=M
LZL
0
x2dx
Step 5: Evaluate the integral to find the moment of inertia I.
I=M
L1
3x3L
0
I=M
L1
3L3−0
I=1
3ML2
Therefore, the moment of inertia of the thin uniform rod about an axis
perpendicular to the rod and passing through one end is 1
3ML2.
Question 2
Question
A thin uniform rod of length Land mass Mis rotating about an axis passing
perpendicular to the rod at its midpoint with an angular velocity ω. Determine
the moment of inertia of the rod about this axis.
Solution
Let’s denote the moment of inertia of the rod about its center of mass (which
is the axis of rotation) as Icm. We can use the parallel axis theorem to find the
moment of inertia of the rod about the axis passing perpendicular to the rod at
its midpoint.
Step 1: Determine the moment of inertia of the rod about its center of mass.
The moment of inertia of a thin rod rotating about an axis passing through its
center of mass and perpendicular to the length of the rod is given by the formula:
Icm =1
12ML2
Step 2: Apply the parallel axis theorem. The parallel axis theorem states
that the moment of inertia about any axis parallel to the axis passing through
the center of mass can be found by adding the mass of the object times the
square of the distance between the two axes:
I=Icm +Md2
where dis the distance between the center of mass axis and the new axis of
rotation (in this case, the midpoint of the rod).
2
Step 3: Calculate the distance between the two axes. Since the axis of
rotation passes through the midpoint of the rod, the distance dis equal to half
the length of the rod:
d=L
2
Step 4: Substitute the values and calculate the moment of inertia about the
new axis. Plugging in the values for Icm and dinto the parallel axis theorem
formula, we have:
I=1
12ML2+ML
22
Simplifying the expression:
I=1
12ML2+1
4ML2=1
3ML2
Therefore, the moment of inertia of the rod about the axis passing perpen-
dicular to the rod at its midpoint is 1
3ML2.
Question 3
Question
A solid cylinder of mass Mand radius Rrolls without slipping down an inclined
plane of angle θwith the horizontal. Calculate the moment of inertia of the
cylinder about an axis through its center parallel to the inclined plane.
Solution
Step 1: Find the acceleration of the cylinder down the plane. The net torque
about the center of mass of the cylinder is due to the force of gravity and the
frictional force. Since the cylinder is rolling without slipping, the frictional force
provides the torque necessary for rolling motion and does no work. Hence, the
net torque is provided by the component of the force of gravity perpendicular
to the plane.
Net Torque = Iα =Rsin(θ)Mg
where Iis the moment of inertia of the cylinder about its center of mass. Since
the cylinder is rolling without slipping, we also know that the linear acceleration
down the plane is related to the angular acceleration by α=a/R, where ais
the linear acceleration. Thus,
Ia
R=Rsin(θ)Mg
a=Rsin(θ)g
Step 2: Calculate the moment of inertia about an axis parallel to the inclined
plane through the center of the cylinder. The parallel axis theorem states that
3
the moment of inertia about an axis parallel to the axis through the center of
mass is equal to the moment of inertia about the center of mass plus Md2,
where dis the perpendicular distance between the two axes. In this case, the
distance dis equal to Rcos(θ). So, the moment of inertia about an axis through
the center parallel to the inclined plane would be:
I′=I+Md2=1
2MR2+M(Rcos(θ))2
I′=1
2MR2+MR2cos2(θ)
I′=3
2MR2cos2(θ)
Therefore, the moment of inertia of the cylinder about an axis through its
center parallel to the inclined plane is 3
2MR2cos2(θ).
Question 4
Question
Find the moment of inertia of a thin uniform rod of length Land mass Mabout
an axis perpendicular to the rod and passing through one end.
Solution
To find the moment of inertia of the rod, we need to integrate the infinitesimal
mass elements along the length of the rod.
Step 1: Consider the infinitesimal mass element δm of the rod at a distance
xfrom the end where the axis passes through. The mass of the rod is uniformly
distributed, so we can express δm in terms of x.
δm =M
Ldx
Step 2: We need to find the moment of inertia of this infinitesimal mass
element δI. The moment of inertia of a point mass mat a distance rfrom an
axis is I=mr2.
For the infinitesimal mass element at position x, the distance from the axis
is r=x. Therefore, the moment of inertia δI for this element is:
δI = (δm) (x2) = M
Ldx(x2)
Step 3: Now, we can find the total moment of inertia Iof the entire rod by
integrating δI over the length of the rod from 0 to L.
I=ZL
0
δI =ZL
0
M
Lx2dx
4
Step 4: Solving the integral:
I=M
LZL
0
x2dx =M
Lx3
3L
0
=M
3LL3−03
I=M
3L2
Step 5: Therefore, the moment of inertia of the thin uniform rod about an
axis perpendicular to the rod and passing through one end is 1
3ML2.
Question 5
Question
Find the moment of inertia of a thin spherical shell of radius Rand mass M
about a diameter of the sphere.
Solution
Let’s denote the moment of inertia of the spherical shell as I. We can split the
spherical shell into thin concentric shells of radius rand thickness dr. The mass
of each thin shell is dm.
Step 1: Find an expression for dm. The mass of the thin shell can be
expressed as dm =ρ·dV , where ρis the mass density. The volume of the thin
shell is dV = 4πr2·dr. Hence, dm =ρ·4πr2·dr.
Step 2: Find an expression for I. The moment of inertia of a thin shell about
an axis passing through its center is dI =r2·dm. Substitute dm =ρ·4πr2·dr
into the expression for dI:
dI =r2·(ρ·4πr2·dr).
Step 3: Integrate to find I. Integrate dI over the entire mass of the spherical
shell:
I=ZdI =Zr2·(ρ·4πr2·dr).
Step 4: Determine the limits of the integral. The radius of the thin shell
varies from 0 to R, so the integration limits are from 0 to R:
I=ZR
0
4πρr4dr.
Step 5: Evaluate the integral.
I= 4πρ ZR
0
r4dr = 4πρ r5
5R
0
.
I=4
5πρR5.
5
Step 6: Substitute mass density. Since the mass Mis distributed uniformly
over the spherical shell, ρ=M
V, where V=4
3πR3is the volume of the spherical
shell. Thus, ρ=3M
4πR3. Substitute ρback into the expression for I:
I=4
5π·3M
4πR3R5=3
5MR2.
Therefore, the moment of inertia of the thin spherical shell of radius Rand
mass Mabout a diameter of the sphere is 3
5MR2.
Question 6
Question
A thin circular ring of radius Rand mass Mis rotating about an axis passing
through its center and perpendicular to its plane with an angular velocity ω.
Calculate the moment of inertia of the ring about an axis passing through a
point on its edge and perpendicular to its plane.
Solution
Let’s denote the moment of inertia of the ring about its center (axis of rotation)
as Icenter. We know that the moment of inertia of a ring about an axis passing
through its center and perpendicular to its plane is Icenter =MR2.
Step 1: Determine the additional moment of inertia when the axis of ro-
tation is shifted to the edge of the ring. This can be calculated by using the
parallel axis theorem, which states that Iedge =Icenter +M d2, where dis the
distance between the two axes. In this case, d=R. Thus,
Iedge =MR2+MR2= 2MR2
Therefore, the moment of inertia of the ring about an axis passing through
a point on its edge and perpendicular to its plane is 2MR2.
Question 7
Question
Calculate the moment of inertia of a thin rod of length Land mass Mabout
an axis perpendicular to the rod and passing through one end.
Solution
Let’s consider the rod to be along the x-axis with one end at the origin and the
other end at (L, 0,0). The linear density of the rod is given by λ=M
L.
Step 1: Assume a small element dm of the rod at position xwith width dx.
Thus, dm =λ dx. The distance of this element from the axis of rotation is x.
6
Step 2: The moment of inertia dI of this element about the axis of rotation
is given by dI =x2dm =λx2dx.
Step 3: Integrate dI from 0 to Lto find the total moment of inertia I.
I=ZL
0
λx2dx
=ZL
0
M
Lx2dx
=M
Lx3
3L
0
=M
LL3
3−0
=M
3L2
Thus, the moment of inertia of the thin rod about an axis perpendicular to
the rod and passing through one end is M
3L2.
Question 8
Question
Determine the moment of inertia of a solid cylinder of mass Mand radius R
about an axis passing through its center perpendicular to the symmetry axis.
Solution
Let’s consider an elemental disc of the cylinder with radius rand thickness dr
at a distance xfrom the center. The mass of this elemental disc is given by
dm =ρ·2πr ·dx, where ρis the density of the cylinder.
Step 1: Express the moment of inertia of the elemental disc about the axis
of rotation.
The moment of inertia of the elemental disc with respect to the axis is
dI =r2·dm. Thus, dI =r2·ρ·2πr ·dx.
Step 2: Integrate to find the total moment of inertia.
Integrating dI over the entire cylinder, the moment of inertia Iis:
I=ZR
0
r2·ρ·2πr dr = 2πρ ZR
0
r3dr
Step 3: Solve the integral.
I= 2πρ r4
4R
0
=1
2πρR4
Hence, the moment of inertia of the solid cylinder about the given axis is
1
2πρR4.
7
Question 9
Question
A thin uniform rod of mass Mand length Lis bent at its center to form an
angle of 90◦. Determine the moment of inertia of this bent rod about an axis
passing through its center of mass and perpendicular to the plane of the rod.
Solution
Step 1: We will first find the moment of inertia of each half of the bent rod
about the center of mass. Let Icbe the moment of inertia of each half about
its center of mass. The moment of inertia of each half of the bent rod about its
own center of mass can be calculated as:
Ic=ML2
12
Step 2: We will use the parallel axis theorem to find the moment of inertia
of each half about the center of mass. The distance between the center of each
half and the center of mass can be calculated as L/4. Therefore, the moment
of inertia of each half about the axis passing through the center of mass is:
I=Ic+ML
42
Step 3: Now, let’s substitute Ic=ML2
12 and simplify:
I=ML2
12 +ML
42
I=ML2
12 +ML2
16
I=7ML2
48
Therefore, the moment of inertia of the bent rod about an axis passing
through its center of mass and perpendicular to the plane of the rod is 7M L2
48 .
Question 10
Question
Find the moment of inertia of a thin uniform rod of length Land mass Mabout
an axis perpendicular to the rod passing through one end.
8
Solution
Let’s denote the moment of inertia as I.
Step 1: Determine the mass per unit length λof the rod. The mass per
unit length λis given by λ=M
L.
Step 2: Find the moment of inertia of a small mass segment δm at a
distance xfrom the axis. The moment of inertia of a small mass segment δm
at a distance xfrom the axis is given by dI =δm ·x2.
Step 3: Express δm in terms of λand δx.δm =λ·δx
Step 4: Express xin terms of δx. From geometry, x=L
2−δx
Step 5: Substitute δm and xinto the formula for dI.dI = (λ·δx)·(L
2−δx)2
Step 6: Integrate dI from 0 to L
2to find the total moment of inertia.
I=ZL
2
0
(λ·δx)·(L
2−δx)2
Step 7: Simplify the integral and solve for I.
I=ZL
2
0
λ·δx ·(L2
4−Lδx +δx2)
I=λ·(L2
4δx −L
2δx2+1
3δx3)
L
2
0
I=λL3
24
Step 8: Substitute λ=M
Linto the equation.
I=M·L2
24
Thus, the moment of inertia of the thin uniform rod about an axis perpen-
dicular to the rod passing through one end is M·L2
24 .
Question 11
Question
Find the moment of inertia of a solid cone of radius Rand height hwith respect
to its central axis.
Solution
Let’s consider an elemental disk of radius rand thickness dr at a distance x
from the base of the cone.
Step 1: The mass of the elemental disk can be expressed in terms of its
volume and density. Using similar triangles, we can determine that the radius
of the disk at a distance xfrom the base is r=R
hx.
9
Step 2: The volume of the disk is the area of the circular base times the
thickness.
dV =πr2dr =πR
hx2
dr =πR2
h2x2dr
Step 3: The mass of the disk is the density multiplied by the volume.
dm =ρdV =ρ·πR2
h2x2dr
Step 4: The moment of inertia of the disk can be calculated as dI =r2·dm.
dI =R
hx2
·ρ·πR2
h2x2dr =ρ·πR4
h4x4dr
Step 5: Integrating dI over the entire cone gives the moment of inertia I.
I=Zh
0
dI =Zh
0
ρ·πR4
h4x4dx
Step 6: Solving the integral, we obtain the moment of inertia Iof the cone.
I=ρ·πR4
h4·1
5x5
h
0
=ρ·πR4
h4·1
5h5=3
10ρπR4h
Question 12
Question
Find the moment of inertia of a thin uniform rod of mass mand length Labout
an axis passing through one end of the rod and perpendicular to the rod.
Solution
Let’s consider the thin uniform rod to be along the x-axis with one end at the
origin and the other end at the point (L, 0,0). The rod is of mass m.
Step 1: Determine the mass per unit length of the rod. Since the rod is of
mass mand length L, the mass per unit length λof the rod is given by:
λ=m
L
Step 2: Calculate the moment of inertia of an infinitesimal mass δm at a
distance xfrom the axis. The moment of inertia of an infinitesimal mass δm
located at a distance xfrom the axis of rotation is given by dI =δm ·x2.
Step 3: Express δm in terms of the linear mass density λ. Since the mass
per unit length λis m
L, the mass δm of an infinitesimal length dx at distance x
along the rod is δm =λ·dx =m
L·dx.
10
Step 4: Integrate to find the total moment of inertia I. The total moment
of inertia Iabout the axis passing through one end of the rod can be obtained
by integrating the infinitesimal moment of inertia dI over the entire length of
the rod:
I=ZL
0
dI
Step 5: Substitute and solve the integral. Substituting the expression for
dI from Step 2 and the expression for δm from Step 3 into the integral yields:
I=ZL
0m
L·dx ·x2
I=m
LZL
0
x2dx
I=m
Lx3
3L
0
I=m
LL3
3−03
3
I=mL2
3
Therefore, the moment of inertia of the thin uniform rod about the axis
passing through one end of the rod and perpendicular to the rod is mL2
3.
Question 13
Question
Find the moment of inertia of a solid cone of radius Rand height Habout its
axis of symmetry.
Solution
Step 1: Consider an elemental disk-shaped slice of the cone at a distance x
from its base, with thickness dx. The radius of this disk can be expressed as a
function of x.
Step 2: The radius of the disk at a distance xfrom the base of the cone can
be given by the expression r(x) = R1−x
H.
Step 3: The differential mass dm of this elemental disk can be expressed in
terms of dx and the volume density ρof the cone. As the density is constant,
dm =ρ·π·[r(x)]2·dx.
Step 4: The moment of inertia dI of the elemental disk about its axis of
symmetry can be given by the expression dI =r2
dm ·dm, where rdm is the
distance of the elemental mass dm from the axis of rotation.
11
Step 5: Substituting the expressions for dm and r(x) into the formula for
dI, we have dI =ρ·π·[R1−x
H]2·x·dx.
Step 6: Integrate dI with limits from 0 to Hto find the total moment of
inertia Iof the cone about its axis of symmetry.
I=ZH
0
dI
Step 7: Substitute the expression for dI and integrate with respect to x.
Step 8: Calculate the final result to find the moment of inertia of the solid
cone about its axis of symmetry.
I=3
10 ·M·R2
Therefore, the moment of inertia of a solid cone of radius Rand height H
about its axis of symmetry is 3
10 ·M·R2, where Mis the total mass of the cone.
Question 14
Question
Determine the moment of inertia of a solid cone of mass Mand radius Rabout
its central axis.
Solution
Let’s consider an elemental disk of radius rand thickness dr at a distance h
from the base of the cone. The mass of this elemental disk can be expressed as
dm =ρ·dV , where ρis the density of the cone. The volume of the elemental
disk can be written as dV =πr2·dh, and the density ρcan be expressed as
M
Vcone , where Vcone is the volume of the cone.
Step 1: Find an expression for the mass dm of the elemental disk.
dV =πr2·dh
dm =ρ·dV
=M
Vcone ·πr2·dh
=M
1
3πR2H·πr2·dh
=3M
R2H·r2·dh
Step 2: Find an expression for the moment of inertia dI of the elemental
12
disk with respect to the central axis.
dI =r2·dm
=r2·3M
R2H·r2·dh
=3M
R2H·r4·dh
Step 3: Integrate the expression for dI from r= 0 to r=Rand h= 0 to
h=Hto find the total moment of inertia I.
I=ZH
0ZR
0
3M
R2H·r4·dr ·dh
=3M
R2HZH
0r5
5R
0
dh
=3M
R2HZH
0
R5
5dh
=3MR5
5R2HZH
0
dh
=3MR5
5R2[h]H
0
=3MH(R2)
5
Thus, the moment of inertia of the solid cone about its central axis is
3
5MR2H.
Question 15
Question
A thin uniform rod with mass Mand length Lis rotated about an axis passing
through one end perpendicular to the rod. Determine the moment of inertia of
the rod about this axis.
Solution
Step 1: Consider a small element of length dx at a distance xfrom the axis of
rotation. Step 2: The mass of this element is dm =M
Ldx. Step 3: The moment
of inertia of this element about the axis of rotation is dI =dm ·x2. Step 4:
Substituting dm and simplifying, we get dI =M
Lx2dx. Step 5: To find the total
moment of inertia, integrate dI from x= 0 to x=L. Step 6: I=RL
0
M
Lx2dx.
Step 7: Solving the integral, I=M
L1
3x3L
0. Step 8: Calculating, I=M
L·1
3L3.
Step 9: Thus, the moment of inertia of the rod about the given axis is I=1
3ML2.
13
Question 16
Question
Calculate the moment of inertia of a solid cone of height hand base radius R
rotating about its central axis. Assume the density of the cone is constant.
Solution
Step 1: The moment of inertia of a solid cone can be found by integrating the
differential mass elements over the volume of the cone:
I=Zr2dm
Step 2: Using the formula for the volume of a cone, we have:
V=1
3πR2h
Step 3: The mass can be found by multiplying the volume by the density, ρ:
m=ρV =1
3πρR2h
Step 4: The differential mass element, dm, can be expressed in terms of the
density and volume element, dV :
dm =ρdV
Step 5: Substituting dm into the moment of inertia formula and expressing
rin terms of hand R,
r=h
HR
where His the height of the cone.
Step 6: Now, we can express dm in terms of dh:
dm =ρ1
3πR2dh
Step 7: Substitute dm back into the formula for moment of inertia:
I=Zr2dm =Zh
0h
HR2
ρ1
3πR2dh
Step 8: Simplify the integrand and integrate over the limits to find the
moment of inertia of the cone rotating about its central axis.
14
Question 17
Question
A thin uniform rod of mass Mand length Lis rotated about an axis perpendic-
ular to the rod and passing through one end. Determine the moment of inertia
of the rod about this axis.
Solution
Let’s consider the rod to be along the x-axis with one end at the origin and
the other end at the point (L, 0,0). We can use the formula for the moment of
inertia of a continuous mass distribution:
I=Zr2dm
Step 1: Find the expression for dm.
Since the rod is thin and uniform, the linear mass density λis constant.
Therefore, the mass dm of an infinitesimally small element of length dx at a
distance xfrom the origin is:
dm =λ dx
Step 2: Find the expression for r.
The distance rbetween the infinitesimal mass dm and the axis of rotation
(which passes through the end at (L, 0,0)) is given by:
r=pL2+x2
Step 3: Express the moment of inertia using the above expressions.
Substitute the expressions for dm and rinto the formula for the moment of
inertia:
I=Zr2dm =ZL
0
(pL2+x2)2λ dx
Step 4: Perform the integration.
I=ZL
0
(L2+x2)λ dx
I=λZL
0
(L2+x2)dx =λL3+1
3L3
I=λ×4
3L3=4
3λL3
Step 5: Substitute the given mass for the rod.
Since M=λL, substitute Mfor λL in the expression for I:
I=4
3·M
L·L3=4
3ML2
Therefore, the moment of inertia of the rod about the specified axis is 4
3ML2.
15
Question 18
Question
A thin rod of length Lis rotating about an axis passing through its one end
perpendicular to its length. The mass per unit length of the rod is λ. Calculate
the moment of inertia of the rod about the axis of rotation.
Solution
Step 1: Consider an elemental mass δm of the rod at a distance xfrom the axis
of rotation.
Step 2: The mass per unit length is λ, so the mass of the elemental length
dx is λ dx.
Step 3: The moment of inertia of the elemental mass dm about the axis is
given by dI =δm ·x2.
Step 4: Substituting δm =λ dx and integrating from x= 0 to x=L, we
have
I=ZL
0
λx2dx.
Step 5: Solve the integral to get the moment of inertia:
I=λZL
0
x2dx =λx3
3L
0
=λL3
3.
Therefore, the moment of inertia of the rod about the given axis is λL3
3.
Question 19
Question
A thin rod of length Lis bent at its midpoint so that the two halves form a right
angle with each other. Calculate the moment of inertia of this bent rod about
an axis passing through the intersection of the two halves and perpendicular to
the plane containing them.
Solution
Step 1: Consider the moment of inertia of each half of the rod individually.
Let mbe the mass of the rod and ρbe the mass per unit length. The mass
of the entire rod is 2mand the total length is 2L. So, the linear density λis
defined as λ=m
2L.
For a thin rod rotating about an axis perpendicular to the rod at one end,
the moment of inertia is given by I=1
3mL2.
For each half of the bent rod, the moment of inertia is 1
3m
2L
22=
1
3m
8L2=mL2
24 .
16
Step 2: Use the parallel axis theorem to find the moment of inertia of the
bent rod.
The distance between the original axis of rotation (center of mass) and the
new axis passing through the intersection of the two halves is L
2.
The moment of inertia of each half about the new axis is mL2
24 .
By the parallel axis theorem, the moment of inertia of the bent rod about
the new axis is 2 mL2
24 + 2 mL2
24 =mL2
6.
Therefore, the moment of inertia of the bent rod about an axis passing
through the intersection of the two halves and perpendicular to the plane con-
taining them is mL2
6.
Question 20
Question
A thin uniform rod of mass Mand length Lis rotated about an axis passing
through one of its ends and perpendicular to the rod. Determine the moment
of inertia of the rod about this axis.
Solution
Step 1: We can consider the rod to be made up of infinitesimally thin discs
along its length. The moment of inertia of each disc can be given as I=1
2mr2,
where mis the mass of the infinitesimal disc and ris the distance of the disc
from the axis of rotation.
Step 2: Let’s express min terms of Mand rin terms of x(the distance of
the disc from the end of the rod). The mass of the disc is proportional to its
length, so dm =M
Ldx and r=L−x.
Step 3: Now, let’s substitute dm and rin the moment of inertia equation.
We get dI =1
2M
Ldx(L−x)2=M
2L(L2−2Lx +x2)dx.
Step 4: To find the total moment of inertia of the rod, we integrate dI from
x= 0 to x=L. This gives I=RL
0
M
2L(L2−2Lx +x2)dx.
Step 5: Simplifying the integral, we get I=M
2L1
3L3−L2x+1
3x3
L
0
.
Step 6: Evaluating the expression gives I=M
2L1
3L3−L3+1
3L3=1
3ML2.
Therefore, the moment of inertia of the rod about the axis passing through
one of its ends and perpendicular to the rod is 1
3ML2.
Question 21
Question
A thin uniform rod of length Land mass Mis rotated about an axis perpen-
dicular to the rod and passing through one end. Find the moment of inertia of
17
the rod about this axis.
Solution
Step 1: Determine the moment of inertia of an infinitesimally small element of
length dx at a distance xfrom the axis of rotation. The mass of the element is
dm =M
Ldx.
Step 2: Use the formula for moment of inertia of a point mass about an
axis:
dI =r2dm
Where ris the perpendicular distance from the axis to the element. In this case,
r=x. Therefore,
dI =x2M
Ldx
Step 3: Integrate to find the total moment of inertia I:
I=ZdI =Zx2M
Ldx
Step 4: Perform the integration:
I=M
LZL
0
x2dx
Step 5: Solve the integral:
I=M
Lx3
3L
0
I=M
LL3
3−0
I=ML2
3
Step 6: Therefore, the moment of inertia of the rod about the axis perpen-
dicular to the rod and passing through one end is M L2
3.
Question 22
Question
Find the moment of inertia of a thin uniform rod of length Land mass Mabout
an axis passing through one end of the rod and perpendicular to it.
18
Solution
Step 1: We can find the moment of inertia of the rod by considering it as a
collection of infinitesimally small mass elements δm along the length of the rod.
Step 2: Let us consider a small mass element δm at a distance xfrom the
end of the rod. The mass of this element is proportional to the length of the
rod, so δm =M
Ldx.
Step 3: The moment of inertia of this small mass element about the axis
passing through one end of the rod is given by dI =δm ·x2. Substituting the
value of δm gives dI =M
Lx2dx.
Step 4: To find the total moment of inertia of the rod, we integrate the
expression dI =M
Lx2dx from 0 to L(the length of the rod):
I=ZL
0
M
Lx2dx
Step 5: Simplifying the integral gives:
I=M
LZL
0
x2dx =M
Lx3
3L
0
Step 6: Evaluating the integral gives:
I=M
LL3
3−03
3=M
L·L3
3=M
3L2
Step 7: Therefore, the moment of inertia of the rod about an axis passing
through one end of the rod and perpendicular to it is M
3L2.
Question 23
Question
A thin uniform rod of length Land mass Mis rotating about an axis passing
through one end perpendicular to the rod with an angular velocity ω. Determine
the moment of inertia of the rod about this axis.
Solution
Step 1: The moment of inertia of the rod about an axis passing through its
center perpendicular to the rod is 1
12 ML2.
Step 2: The parallel axis theorem states that the moment of inertia about
an axis parallel to and a distance daway from an axis through the center of
mass is given by I=Icm +Md2, where Icm is the moment of inertia about the
center of mass.
Step 3: In this case, the distance dis L/2 since the axis is passing through
one end of the rod.
19
Step 4: Substituting Icm =1
12 ML2and d=L/2 into the parallel axis
theorem, we have I=1
12 ML2+ML
22.
Step 5: Simplifying the expression, we get I=1
12 ML2+1
4ML2.
Step 6: Combining the terms, we find I=1
3ML2.
Therefore, the moment of inertia of the rod about an axis passing through
one end perpendicular to the rod is 1
3ML2.
Question 24
Question
Find the moment of inertia of a rectangular plate of mass Mand dimensions
a×babout an axis passing through one corner of the plate and perpendicular
to the plate.
Solution
To find the moment of inertia of the rectangular plate about the given axis, we
can use the parallel axis theorem. First, we find the moment of inertia of the
plate about its center of mass and then apply the parallel axis theorem to shift
the axis to the desired location.
Step 1: Find the moment of inertia of the rectangular plate about its center
of mass.
The moment of inertia of a rectangular plate about an axis passing through
its center and perpendicular to the plate is given by the expression:
Icenter =M
12 (a2+b2)
where Mis the mass of the plate, ais the length of the plate, and bis the
width of the plate.
Step 2: Apply the parallel axis theorem to find the moment of inertia about
the desired axis.
The parallel axis theorem states that the moment of inertia about a parallel
axis is equal to the moment of inertia about the center of mass plus the product
of the mass and the square of the distance between the two axes.
In this case, the distance between the center of mass and the desired axis is
a
2+b
2=a+b
2. Therefore, the moment of inertia about the desired axis is:
I=Icenter +Ma+b
22
Substitute Icenter from step 1 into the equation and simplify to get the final
expression for the moment of inertia about the given axis.
20
Question 25
Question
A thin uniform rod of mass Mand length Lis rotating about an axis perpen-
dicular to the rod and passing through its midpoint with an angular velocity ω.
Determine the moment of inertia of the rod about this axis.
Solution
Step 1: The moment of inertia of a thin rod rotating about an axis perpendicular
to the rod and passing through its midpoint can be calculated using the formula
I=1
12 mL2, where mis the mass of the rod and Lis its length.
Step 2: First, we need to express the mass Mof the rod in terms of its linear
density λ=M
L.
Step 3: The mass mof an infinitesimally small segment of length dx located
at a distance xfrom the midpoint can be expressed as dm =λdx.
Step 4: The moment of inertia dI of this segment about the axis can be
calculated as dI = (dm)x2.
Step 5: Substituting the expression for dm into the equation for dI gives
dI =λx2dx.
Step 6: To find the total moment of inertia Iof the entire rod, we integrate
dI with respect to xover the length of the rod. Hence, I=RL/2
−L/2λx2dx.
Step 7: Simplifying the integration, we get I=λhx3
3iL/2
−L/2.
Step 8: Evaluating the limits of integration, we find I=λ
3hL
23−−L
23i.
Step 9: Simplifying further, we get I=λ
3hL3
8+L3
8i.
Step 10: Thus, the moment of inertia Iof the rod about the given axis is
I=2
3λL3=2
3M
LL3=1
3ML2.
Question 26
Question
Find the moment of inertia of a thin uniform rod of mass mand length Labout
an axis perpendicular to the rod and passing through one of its ends.
Solution
Let’s divide the rod into small elements of mass dm at a distance xfrom the end
of the rod. The mass dm of a small element of length dx is given by dm =m
Ldx.
Step 1: Express the moment of inertia Ias the sum of moments of inertia
of all the small elements.
21
The moment of inertia of a small element dm about the perpendicular axis
passing through one end is dI =x2dm. Therefore, the moment of inertia of the
entire rod is:
I=ZL
0
x2dm
Step 2: Express dm in terms of dx.
Given that dm =m
Ldx, substitute this into the expression for I:
I=ZL
0
x2m
Ldx
Step 3: Simplify the expression and solve for I.
I=ZL
0
m
Lx2dx
I=m
LZL
0
x2dx
I=m
Lx3
3L
0
I=m
LL3
3−0
I=mL2
3
Therefore, the moment of inertia of the thin uniform rod about an axis
perpendicular to the rod and passing through one of its ends is mL2
3.
Question 27
Question
A solid uniform cylinder of radius Rand mass Mis rolling without slipping
along a horizontal surface with angular speed ω. Calculate the moment of
inertia of the cylinder about an axis perpendicular to its symmetry axis and
passing through its center.
Solution
Step 1: The moment of inertia of the cylinder about its symmetry axis passing
through its center is 1
2MR2.
Step 2: To find the moment of inertia of the cylinder about an axis per-
pendicular to its symmetry axis and passing through its center, we can use the
parallel axis theorem.
22
Step 3: The distance between the two axes is R. The parallel axis theorem
states that I=Icm +Md2, where Iis the moment of inertia about the new
axis, Icm is the moment of inertia about the center of mass axis, Mis the mass
of the object, and dis the distance between the two axes.
Step 4: Substituting the values, we get I=1
2MR2+MR2=3
2MR2.
Therefore, the moment of inertia of the cylinder about an axis perpendicular
to its symmetry axis and passing through its center is 3
2MR2.
Question 28
Question
A thin rod of length Land mass Mis rotated about an axis that is perpendicular
to the rod and passes through one end. Find the moment of inertia of the rod
about this axis.
Solution
1. Consider a small element of the rod of length dx at a distance xfrom the
end of the rod. The mass of this element can be approximated as M
Ldx.
2. The moment of inertia dI of this element about the axis is given by
dI =M
Lx2dx.
3. To find the total moment of inertia Iof the entire rod, we need to integrate
dI over the length of the rod.
I=ZL
0
M
Lx2dx
4. Solving the integral, we get
I=M
Lx3
3L
0
=M
LL3
3−0
5. Therefore, the moment of inertia of the rod about the axis passing through
one end is
I=M
3L2
Question 29
Question
Find the moment of inertia of a thin-walled semi-circular cylinder of radius R
and uniform mass density ρ, rotated about an axis passing through its diameter.
23
Solution
Let’s consider the thin-walled semi-circular cylinder as a collection of thin rings,
each at a distance rfrom the axis of rotation. We will sum up the moments of
inertia of all these thin rings to find the total moment of inertia of the semi-
circular cylinder.
Step 1: Consider a thin ring of radius rand thickness dr at a distance r
from the axis of rotation.
The mass of this thin ring can be calculated by considering the area of a
ring:
dM = 2πrdrρ
The moment of inertia of this thin ring about the axis of rotation is given
by:
dI =r2dM = 2πr3drρ
Step 2: Integrate to find the total moment of inertia of the semi-circular
cylinder.
We can now integrate dI over the entire semi-circular cylinder:
I=ZdI =ZR
0
2πr3drρ
I= 2πρ ZR
0
r3dr
I= 2πρ r4
4R
0
I= 2πρ R4
4−04
4
I= 2πρ R4
4
I=1
2πR4ρ
Thus, the moment of inertia of the thin-walled semi-circular cylinder of radius R
and uniform mass density ρ, rotated about an axis passing through its diameter
is 1
2πR4ρ.
Question 30
Question
A thin uniform rod of length Land mass Mis rotating about an axis perpendic-
ular to the rod and passing through one end with an angular velocity ω. Find
the moment of inertia of the rod about this axis.
24
Solution
Step 1: Divide the rod into small elements ∆mieach at a distance rifrom the
axis of rotation.
Step 2: Calculate the moment of inertia of each element Ii= (∆mi)·(ri)2.
Step 3: Sum up all the moments of inertia of each element using the formula
for moment of inertia: I=PIi.
Step 4: Express ∆miin terms of the total mass M, and riin terms of the
distance xialong the rod.
Step 5: Integrate I=RIidm over the length of the rod to find the total
moment of inertia I.
Step 6: Finally, substitute the result of the integral and simplify to express
the moment of inertia of the rod in terms of its length L, mass M, and angular
velocity ω.
Question 31
Question
Find the moment of inertia of a uniform solid cylinder of mass Mand radius
R, rotating about an axis that is tangential to the cylinder’s surface.
Solution
Let’s denote the moment of inertia as I. To find the moment of inertia of the
solid cylinder, we need to integrate over the mass distribution of the cylinder.
Step 1: Determine the mass distribution function. We can express the mass
distribution function as ρ(r), where ris the distance from the axis of rotation.
For a solid cylinder of uniform density, ρ(r) is constant.
Step 2: Define an element of mass. Consider an elemental disk of radius
rand thickness dr at distance rfrom the axis of rotation. The mass of this
elemental disk can be expressed as dm =ρ(r)·dV , where dV is the volume of
the elemental disk.
Step 3: Express dV in terms of rand dr. The volume of the elemental disk
can be expressed as dV =πr2·dr.
Step 4: Calculate dm in terms of rand dr. Substitute dV =πr2·dr into
dm =ρ(r)·dV to get dm =πρ(r)r2·dr.
Step 5: Find the moment of inertia dI of the elemental disk. The moment
of inertia of the elemental disk about the axis of rotation is dI =r2·dm.
Substitute dm =πρ(r)r2·dr into this expression to get dI =πρ(r)r4·dr.
Step 6: Integrate to find the total moment of inertia I. Integrate dI =
πρ(r)r4·dr over the entire volume of the cylinder to find the total moment of
inertia I.
Step 7: Substitute values and evaluate the integral. Substitute ρ(r) with
the density of the cylinder, and integrate over the limits of rfrom 0 to Rwhere
25
Ris the radius of the cylinder. Evaluate the integral to find the moment of
inertia I.
Question 32
Question
Calculate the moment of inertia about the x-axis of the region bounded by the
curves y=x2and y=x.
Solution
To find the moment of inertia about the x-axis, we will use the formula:
Ix=Zb
a
y2dx
where aand bare the x-values where the two curves intersect.
Step 1: Find the points of intersection. Setting x2=x, we find the
points of intersection to be x= 0 and x= 1.
Step 2: Set up the integral. For the given region, we have y=x2above
and y=xbelow. Thus, the integral becomes:
Ix=Z1
0
(x2−x)2dx
Expanding the integrand gives:
Ix=Z1
0
(x4−2x3+x2)dx
Step 3: Evaluate the integral. Integrating each term separately, we get:
Ix=1
5x5−1
2x4+1
3x31
0
Evaluating this from 0 to 1 gives:
Ix=1
5−1
2+1
3−(0 −0 + 0)
Ix=1
5−1
2+1
3
Ix=1
30
Therefore, the moment of inertia about the x-axis for the given region is 1
30 .
26
Question 33
Question
A thin rod of length Land mass Mis rotated about an axis that is perpendicular
to the rod and passes through one end. Determine the moment of inertia of the
rod about this axis.
Solution
To find the moment of inertia of the thin rod about an axis perpendicular to
the rod and passing through one end, we will use the formula for the moment
of inertia of a continuous mass distribution:
I=Zr2dm
where ris the distance from the axis of rotation to an infinitesimal mass
element dm.
Step 1: Choose a variable Let xbe the distance along the rod from the
end where the axis of rotation passes through. The infinitesimal mass dm can
be expressed as a function of x.
Step 2: Expressing elemental mass The mass per unit length of the rod
can be expressed as M
L. Therefore, the infinitesimal mass dm is dm =M
Ldx.
Step 3: Calculate the moment of inertia Substitute r=xand dm =
M
Ldx into the formula for moment of inertia:
I=ZL
0
x2M
Ldx
I=M
LZL
0
x2dx
I=M
Lx3
3L
0
I=M
LL3
3−0
I=ML2
3
Therefore, the moment of inertia of the thin rod about an axis perpendicular
to the rod and passing through one end is M L2
3.
27
Question 34
Question
A thin disc of radius Rand mass Mis rotating about an axis passing through
its center with angular velocity ω. Determine the moment of inertia of the disc
with respect to an axis perpendicular to the disc passing through its edge.
Solution
Step 1: The moment of inertia of a thin disc about an axis passing through its
center and perpendicular to the disc is given by the formula Icenter =1
2MR2.
Step 2: To determine the moment of inertia of the disc about an axis passing
through its edge (perpendicular to the disc), we can use the parallel axis theo-
rem. According to the theorem, Iedge =Icenter +M d2, where dis the distance
between the two axes.
Step 3: The distance dbetween the edge of the disc and the center is equal
to the radius R. Therefore, d=R.
Step 4: Substituting d=Rinto the formula Iedge =Icenter +Md2, we get
Iedge =1
2MR2+MR2.
Step 5: Simplifying, we find Iedge =3
2MR2. Hence, the moment of inertia
of the disc about an axis perpendicular to the disc passing through its edge is
3
2MR2.
Question 35
Question
A uniform thin rod of length Land mass Mis rotating about an axis perpen-
dicular to the rod and passing through one end. Find the moment of inertia of
the rod about this axis.
Solution
To find the moment of inertia (I) of the rod about the given axis, we can use the
formula for the moment of inertia of a thin rod rotating about one end, given
by I=1
3ML2.
Step 1: Write down the formula for moment of inertia of a thin rod rotating
about one end.
I=1
3ML2
Step 2: Substitute the given values Mand Linto the formula.
I=1
3·M·L2=1
3·M·L·L
28
I=ZL
0
(x2)·M
Ldx
I=M
LZL
0
x2dx
Step 5: Evaluate the integral to find the moment of inertia I.
I=M
L1
3x3L
0
I=M
L1
3L3−0
I=1
3ML2
Therefore, the moment of inertia of the thin uniform rod about an axis
perpendicular to the rod and passing through one end is 1
3ML2.
Question 2
Question
A thin uniform rod of length Land mass Mis rotating about an axis passing
perpendicular to the rod at its midpoint with an angular velocity ω. Determine
the moment of inertia of the rod about this axis.
Solution
Let’s denote the moment of inertia of the rod about its center of mass (which
is the axis of rotation) as Icm. We can use the parallel axis theorem to find the
moment of inertia of the rod about the axis passing perpendicular to the rod at
its midpoint.
Step 1: Determine the moment of inertia of the rod about its center of mass.
The moment of inertia of a thin rod rotating about an axis passing through its
center of mass and perpendicular to the length of the rod is given by the formula:
Icm =1
12ML2
Step 2: Apply the parallel axis theorem. The parallel axis theorem states
that the moment of inertia about any axis parallel to the axis passing through
the center of mass can be found by adding the mass of the object times the
square of the distance between the two axes:
I=Icm +Md2
where dis the distance between the center of mass axis and the new axis of
rotation (in this case, the midpoint of the rod).
2
Step 3: Calculate the distance between the two axes. Since the axis of
rotation passes through the midpoint of the rod, the distance dis equal to half
the length of the rod:
d=L
2
Step 4: Substitute the values and calculate the moment of inertia about the
new axis. Plugging in the values for Icm and dinto the parallel axis theorem
formula, we have:
I=1
12ML2+ML
22
Simplifying the expression:
I=1
12ML2+1
4ML2=1
3ML2
Therefore, the moment of inertia of the rod about the axis passing perpen-
dicular to the rod at its midpoint is 1
3ML2.
Question 3
Question
A solid cylinder of mass Mand radius Rrolls without slipping down an inclined
plane of angle θwith the horizontal. Calculate the moment of inertia of the
cylinder about an axis through its center parallel to the inclined plane.
Solution
Step 1: Find the acceleration of the cylinder down the plane. The net torque
about the center of mass of the cylinder is due to the force of gravity and the
frictional force. Since the cylinder is rolling without slipping, the frictional force
provides the torque necessary for rolling motion and does no work. Hence, the
net torque is provided by the component of the force of gravity perpendicular
to the plane.
Net Torque = Iα =Rsin(θ)Mg
where Iis the moment of inertia of the cylinder about its center of mass. Since
the cylinder is rolling without slipping, we also know that the linear acceleration
down the plane is related to the angular acceleration by α=a/R, where ais
the linear acceleration. Thus,
Ia
R=Rsin(θ)Mg
a=Rsin(θ)g
Step 2: Calculate the moment of inertia about an axis parallel to the inclined
plane through the center of the cylinder. The parallel axis theorem states that
3
the moment of inertia about an axis parallel to the axis through the center of
mass is equal to the moment of inertia about the center of mass plus Md2,
where dis the perpendicular distance between the two axes. In this case, the
distance dis equal to Rcos(θ). So, the moment of inertia about an axis through
the center parallel to the inclined plane would be:
I′=I+Md2=1
2MR2+M(Rcos(θ))2
I′=1
2MR2+MR2cos2(θ)
I′=3
2MR2cos2(θ)
Therefore, the moment of inertia of the cylinder about an axis through its
center parallel to the inclined plane is 3
2MR2cos2(θ).
Question 4
Question
Find the moment of inertia of a thin uniform rod of length Land mass Mabout
an axis perpendicular to the rod and passing through one end.
Solution
To find the moment of inertia of the rod, we need to integrate the infinitesimal
mass elements along the length of the rod.
Step 1: Consider the infinitesimal mass element δm of the rod at a distance
xfrom the end where the axis passes through. The mass of the rod is uniformly
distributed, so we can express δm in terms of x.
δm =M
Ldx
Step 2: We need to find the moment of inertia of this infinitesimal mass
element δI. The moment of inertia of a point mass mat a distance rfrom an
axis is I=mr2.
For the infinitesimal mass element at position x, the distance from the axis
is r=x. Therefore, the moment of inertia δI for this element is:
δI = (δm) (x2) = M
Ldx(x2)
Step 3: Now, we can find the total moment of inertia Iof the entire rod by
integrating δI over the length of the rod from 0 to L.
I=ZL
0
δI =ZL
0
M
Lx2dx
4
Step 4: Solving the integral:
I=M
LZL
0
x2dx =M
Lx3
3L
0
=M
3LL3−03
I=M
3L2
Step 5: Therefore, the moment of inertia of the thin uniform rod about an
axis perpendicular to the rod and passing through one end is 1
3ML2.
Question 5
Question
Find the moment of inertia of a thin spherical shell of radius Rand mass M
about a diameter of the sphere.
Solution
Let’s denote the moment of inertia of the spherical shell as I. We can split the
spherical shell into thin concentric shells of radius rand thickness dr. The mass
of each thin shell is dm.
Step 1: Find an expression for dm. The mass of the thin shell can be
expressed as dm =ρ·dV , where ρis the mass density. The volume of the thin
shell is dV = 4πr2·dr. Hence, dm =ρ·4πr2·dr.
Step 2: Find an expression for I. The moment of inertia of a thin shell about
an axis passing through its center is dI =r2·dm. Substitute dm =ρ·4πr2·dr
into the expression for dI:
dI =r2·(ρ·4πr2·dr).
Step 3: Integrate to find I. Integrate dI over the entire mass of the spherical
shell:
I=ZdI =Zr2·(ρ·4πr2·dr).
Step 4: Determine the limits of the integral. The radius of the thin shell
varies from 0 to R, so the integration limits are from 0 to R:
I=ZR
0
4πρr4dr.
Step 5: Evaluate the integral.
I= 4πρ ZR
0
r4dr = 4πρ r5
5R
0
.
I=4
5πρR5.
5
Step 6: Substitute mass density. Since the mass Mis distributed uniformly
over the spherical shell, ρ=M
V, where V=4
3πR3is the volume of the spherical
shell. Thus, ρ=3M
4πR3. Substitute ρback into the expression for I:
I=4
5π·3M
4πR3R5=3
5MR2.
Therefore, the moment of inertia of the thin spherical shell of radius Rand
mass Mabout a diameter of the sphere is 3
5MR2.
Question 6
Question
A thin circular ring of radius Rand mass Mis rotating about an axis passing
through its center and perpendicular to its plane with an angular velocity ω.
Calculate the moment of inertia of the ring about an axis passing through a
point on its edge and perpendicular to its plane.
Solution
Let’s denote the moment of inertia of the ring about its center (axis of rotation)
as Icenter. We know that the moment of inertia of a ring about an axis passing
through its center and perpendicular to its plane is Icenter =MR2.
Step 1: Determine the additional moment of inertia when the axis of ro-
tation is shifted to the edge of the ring. This can be calculated by using the
parallel axis theorem, which states that Iedge =Icenter +M d2, where dis the
distance between the two axes. In this case, d=R. Thus,
Iedge =MR2+MR2= 2MR2
Therefore, the moment of inertia of the ring about an axis passing through
a point on its edge and perpendicular to its plane is 2MR2.
Question 7
Question
Calculate the moment of inertia of a thin rod of length Land mass Mabout
an axis perpendicular to the rod and passing through one end.
Solution
Let’s consider the rod to be along the x-axis with one end at the origin and the
other end at (L, 0,0). The linear density of the rod is given by λ=M
L.
Step 1: Assume a small element dm of the rod at position xwith width dx.
Thus, dm =λ dx. The distance of this element from the axis of rotation is x.
6
Step 2: The moment of inertia dI of this element about the axis of rotation
is given by dI =x2dm =λx2dx.
Step 3: Integrate dI from 0 to Lto find the total moment of inertia I.
I=ZL
0
λx2dx
=ZL
0
M
Lx2dx
=M
Lx3
3L
0
=M
LL3
3−0
=M
3L2
Thus, the moment of inertia of the thin rod about an axis perpendicular to
the rod and passing through one end is M
3L2.
Question 8
Question
Determine the moment of inertia of a solid cylinder of mass Mand radius R
about an axis passing through its center perpendicular to the symmetry axis.
Solution
Let’s consider an elemental disc of the cylinder with radius rand thickness dr
at a distance xfrom the center. The mass of this elemental disc is given by
dm =ρ·2πr ·dx, where ρis the density of the cylinder.
Step 1: Express the moment of inertia of the elemental disc about the axis
of rotation.
The moment of inertia of the elemental disc with respect to the axis is
dI =r2·dm. Thus, dI =r2·ρ·2πr ·dx.
Step 2: Integrate to find the total moment of inertia.
Integrating dI over the entire cylinder, the moment of inertia Iis:
I=ZR
0
r2·ρ·2πr dr = 2πρ ZR
0
r3dr
Step 3: Solve the integral.
I= 2πρ r4
4R
0
=1
2πρR4
Hence, the moment of inertia of the solid cylinder about the given axis is
1
2πρR4.
7
Question 9
Question
A thin uniform rod of mass Mand length Lis bent at its center to form an
angle of 90◦. Determine the moment of inertia of this bent rod about an axis
passing through its center of mass and perpendicular to the plane of the rod.
Solution
Step 1: We will first find the moment of inertia of each half of the bent rod
about the center of mass. Let Icbe the moment of inertia of each half about
its center of mass. The moment of inertia of each half of the bent rod about its
own center of mass can be calculated as:
Ic=ML2
12
Step 2: We will use the parallel axis theorem to find the moment of inertia
of each half about the center of mass. The distance between the center of each
half and the center of mass can be calculated as L/4. Therefore, the moment
of inertia of each half about the axis passing through the center of mass is:
I=Ic+ML
42
Step 3: Now, let’s substitute Ic=ML2
12 and simplify:
I=ML2
12 +ML
42
I=ML2
12 +ML2
16
I=7ML2
48
Therefore, the moment of inertia of the bent rod about an axis passing
through its center of mass and perpendicular to the plane of the rod is 7M L2
48 .
Question 10
Question
Find the moment of inertia of a thin uniform rod of length Land mass Mabout
an axis perpendicular to the rod passing through one end.
8
Solution
Let’s denote the moment of inertia as I.
Step 1: Determine the mass per unit length λof the rod. The mass per
unit length λis given by λ=M
L.
Step 2: Find the moment of inertia of a small mass segment δm at a
distance xfrom the axis. The moment of inertia of a small mass segment δm
at a distance xfrom the axis is given by dI =δm ·x2.
Step 3: Express δm in terms of λand δx.δm =λ·δx
Step 4: Express xin terms of δx. From geometry, x=L
2−δx
Step 5: Substitute δm and xinto the formula for dI.dI = (λ·δx)·(L
2−δx)2
Step 6: Integrate dI from 0 to L
2to find the total moment of inertia.
I=ZL
2
0
(λ·δx)·(L
2−δx)2
Step 7: Simplify the integral and solve for I.
I=ZL
2
0
λ·δx ·(L2
4−Lδx +δx2)
I=λ·(L2
4δx −L
2δx2+1
3δx3)
L
2
0
I=λL3
24
Step 8: Substitute λ=M
Linto the equation.
I=M·L2
24
Thus, the moment of inertia of the thin uniform rod about an axis perpen-
dicular to the rod passing through one end is M·L2
24 .
Question 11
Question
Find the moment of inertia of a solid cone of radius Rand height hwith respect
to its central axis.
Solution
Let’s consider an elemental disk of radius rand thickness dr at a distance x
from the base of the cone.
Step 1: The mass of the elemental disk can be expressed in terms of its
volume and density. Using similar triangles, we can determine that the radius
of the disk at a distance xfrom the base is r=R
hx.
9
Step 2: The volume of the disk is the area of the circular base times the
thickness.
dV =πr2dr =πR
hx2
dr =πR2
h2x2dr
Step 3: The mass of the disk is the density multiplied by the volume.
dm =ρdV =ρ·πR2
h2x2dr
Step 4: The moment of inertia of the disk can be calculated as dI =r2·dm.
dI =R
hx2
·ρ·πR2
h2x2dr =ρ·πR4
h4x4dr
Step 5: Integrating dI over the entire cone gives the moment of inertia I.
I=Zh
0
dI =Zh
0
ρ·πR4
h4x4dx
Step 6: Solving the integral, we obtain the moment of inertia Iof the cone.
I=ρ·πR4
h4·1
5x5
h
0
=ρ·πR4
h4·1
5h5=3
10ρπR4h
Question 12
Question
Find the moment of inertia of a thin uniform rod of mass mand length Labout
an axis passing through one end of the rod and perpendicular to the rod.
Solution
Let’s consider the thin uniform rod to be along the x-axis with one end at the
origin and the other end at the point (L, 0,0). The rod is of mass m.
Step 1: Determine the mass per unit length of the rod. Since the rod is of
mass mand length L, the mass per unit length λof the rod is given by:
λ=m
L
Step 2: Calculate the moment of inertia of an infinitesimal mass δm at a
distance xfrom the axis. The moment of inertia of an infinitesimal mass δm
located at a distance xfrom the axis of rotation is given by dI =δm ·x2.
Step 3: Express δm in terms of the linear mass density λ. Since the mass
per unit length λis m
L, the mass δm of an infinitesimal length dx at distance x
along the rod is δm =λ·dx =m
L·dx.
10
Step 4: Integrate to find the total moment of inertia I. The total moment
of inertia Iabout the axis passing through one end of the rod can be obtained
by integrating the infinitesimal moment of inertia dI over the entire length of
the rod:
I=ZL
0
dI
Step 5: Substitute and solve the integral. Substituting the expression for
dI from Step 2 and the expression for δm from Step 3 into the integral yields:
I=ZL
0m
L·dx ·x2
I=m
LZL
0
x2dx
I=m
Lx3
3L
0
I=m
LL3
3−03
3
I=mL2
3
Therefore, the moment of inertia of the thin uniform rod about the axis
passing through one end of the rod and perpendicular to the rod is mL2
3.
Question 13
Question
Find the moment of inertia of a solid cone of radius Rand height Habout its
axis of symmetry.
Solution
Step 1: Consider an elemental disk-shaped slice of the cone at a distance x
from its base, with thickness dx. The radius of this disk can be expressed as a
function of x.
Step 2: The radius of the disk at a distance xfrom the base of the cone can
be given by the expression r(x) = R1−x
H.
Step 3: The differential mass dm of this elemental disk can be expressed in
terms of dx and the volume density ρof the cone. As the density is constant,
dm =ρ·π·[r(x)]2·dx.
Step 4: The moment of inertia dI of the elemental disk about its axis of
symmetry can be given by the expression dI =r2
dm ·dm, where rdm is the
distance of the elemental mass dm from the axis of rotation.
11
Step 5: Substituting the expressions for dm and r(x) into the formula for
dI, we have dI =ρ·π·[R1−x
H]2·x·dx.
Step 6: Integrate dI with limits from 0 to Hto find the total moment of
inertia Iof the cone about its axis of symmetry.
I=ZH
0
dI
Step 7: Substitute the expression for dI and integrate with respect to x.
Step 8: Calculate the final result to find the moment of inertia of the solid
cone about its axis of symmetry.
I=3
10 ·M·R2
Therefore, the moment of inertia of a solid cone of radius Rand height H
about its axis of symmetry is 3
10 ·M·R2, where Mis the total mass of the cone.
Question 14
Question
Determine the moment of inertia of a solid cone of mass Mand radius Rabout
its central axis.
Solution
Let’s consider an elemental disk of radius rand thickness dr at a distance h
from the base of the cone. The mass of this elemental disk can be expressed as
dm =ρ·dV , where ρis the density of the cone. The volume of the elemental
disk can be written as dV =πr2·dh, and the density ρcan be expressed as
M
Vcone , where Vcone is the volume of the cone.
Step 1: Find an expression for the mass dm of the elemental disk.
dV =πr2·dh
dm =ρ·dV
=M
Vcone ·πr2·dh
=M
1
3πR2H·πr2·dh
=3M
R2H·r2·dh
Step 2: Find an expression for the moment of inertia dI of the elemental
12
disk with respect to the central axis.
dI =r2·dm
=r2·3M
R2H·r2·dh
=3M
R2H·r4·dh
Step 3: Integrate the expression for dI from r= 0 to r=Rand h= 0 to
h=Hto find the total moment of inertia I.
I=ZH
0ZR
0
3M
R2H·r4·dr ·dh
=3M
R2HZH
0r5
5R
0
dh
=3M
R2HZH
0
R5
5dh
=3MR5
5R2HZH
0
dh
=3MR5
5R2[h]H
0
=3MH(R2)
5
Thus, the moment of inertia of the solid cone about its central axis is
3
5MR2H.
Question 15
Question
A thin uniform rod with mass Mand length Lis rotated about an axis passing
through one end perpendicular to the rod. Determine the moment of inertia of
the rod about this axis.
Solution
Step 1: Consider a small element of length dx at a distance xfrom the axis of
rotation. Step 2: The mass of this element is dm =M
Ldx. Step 3: The moment
of inertia of this element about the axis of rotation is dI =dm ·x2. Step 4:
Substituting dm and simplifying, we get dI =M
Lx2dx. Step 5: To find the total
moment of inertia, integrate dI from x= 0 to x=L. Step 6: I=RL
0
M
Lx2dx.
Step 7: Solving the integral, I=M
L1
3x3L
0. Step 8: Calculating, I=M
L·1
3L3.
Step 9: Thus, the moment of inertia of the rod about the given axis is I=1
3ML2.
13
Question 16
Question
Calculate the moment of inertia of a solid cone of height hand base radius R
rotating about its central axis. Assume the density of the cone is constant.
Solution
Step 1: The moment of inertia of a solid cone can be found by integrating the
differential mass elements over the volume of the cone:
I=Zr2dm
Step 2: Using the formula for the volume of a cone, we have:
V=1
3πR2h
Step 3: The mass can be found by multiplying the volume by the density, ρ:
m=ρV =1
3πρR2h
Step 4: The differential mass element, dm, can be expressed in terms of the
density and volume element, dV :
dm =ρdV
Step 5: Substituting dm into the moment of inertia formula and expressing
rin terms of hand R,
r=h
HR
where His the height of the cone.
Step 6: Now, we can express dm in terms of dh:
dm =ρ1
3πR2dh
Step 7: Substitute dm back into the formula for moment of inertia:
I=Zr2dm =Zh
0h
HR2
ρ1
3πR2dh
Step 8: Simplify the integrand and integrate over the limits to find the
moment of inertia of the cone rotating about its central axis.
14
Question 17
Question
A thin uniform rod of mass Mand length Lis rotated about an axis perpendic-
ular to the rod and passing through one end. Determine the moment of inertia
of the rod about this axis.
Solution
Let’s consider the rod to be along the x-axis with one end at the origin and
the other end at the point (L, 0,0). We can use the formula for the moment of
inertia of a continuous mass distribution:
I=Zr2dm
Step 1: Find the expression for dm.
Since the rod is thin and uniform, the linear mass density λis constant.
Therefore, the mass dm of an infinitesimally small element of length dx at a
distance xfrom the origin is:
dm =λ dx
Step 2: Find the expression for r.
The distance rbetween the infinitesimal mass dm and the axis of rotation
(which passes through the end at (L, 0,0)) is given by:
r=pL2+x2
Step 3: Express the moment of inertia using the above expressions.
Substitute the expressions for dm and rinto the formula for the moment of
inertia:
I=Zr2dm =ZL
0
(pL2+x2)2λ dx
Step 4: Perform the integration.
I=ZL
0
(L2+x2)λ dx
I=λZL
0
(L2+x2)dx =λL3+1
3L3
I=λ×4
3L3=4
3λL3
Step 5: Substitute the given mass for the rod.
Since M=λL, substitute Mfor λL in the expression for I:
I=4
3·M
L·L3=4
3ML2
Therefore, the moment of inertia of the rod about the specified axis is 4
3ML2.
15
Question 18
Question
A thin rod of length Lis rotating about an axis passing through its one end
perpendicular to its length. The mass per unit length of the rod is λ. Calculate
the moment of inertia of the rod about the axis of rotation.
Solution
Step 1: Consider an elemental mass δm of the rod at a distance xfrom the axis
of rotation.
Step 2: The mass per unit length is λ, so the mass of the elemental length
dx is λ dx.
Step 3: The moment of inertia of the elemental mass dm about the axis is
given by dI =δm ·x2.
Step 4: Substituting δm =λ dx and integrating from x= 0 to x=L, we
have
I=ZL
0
λx2dx.
Step 5: Solve the integral to get the moment of inertia:
I=λZL
0
x2dx =λx3
3L
0
=λL3
3.
Therefore, the moment of inertia of the rod about the given axis is λL3
3.
Question 19
Question
A thin rod of length Lis bent at its midpoint so that the two halves form a right
angle with each other. Calculate the moment of inertia of this bent rod about
an axis passing through the intersection of the two halves and perpendicular to
the plane containing them.
Solution
Step 1: Consider the moment of inertia of each half of the rod individually.
Let mbe the mass of the rod and ρbe the mass per unit length. The mass
of the entire rod is 2mand the total length is 2L. So, the linear density λis
defined as λ=m
2L.
For a thin rod rotating about an axis perpendicular to the rod at one end,
the moment of inertia is given by I=1
3mL2.
For each half of the bent rod, the moment of inertia is 1
3m
2L
22=
1
3m
8L2=mL2
24 .
16
Step 2: Use the parallel axis theorem to find the moment of inertia of the
bent rod.
The distance between the original axis of rotation (center of mass) and the
new axis passing through the intersection of the two halves is L
2.
The moment of inertia of each half about the new axis is mL2
24 .
By the parallel axis theorem, the moment of inertia of the bent rod about
the new axis is 2 mL2
24 + 2 mL2
24 =mL2
6.
Therefore, the moment of inertia of the bent rod about an axis passing
through the intersection of the two halves and perpendicular to the plane con-
taining them is mL2
6.
Question 20
Question
A thin uniform rod of mass Mand length Lis rotated about an axis passing
through one of its ends and perpendicular to the rod. Determine the moment
of inertia of the rod about this axis.
Solution
Step 1: We can consider the rod to be made up of infinitesimally thin discs
along its length. The moment of inertia of each disc can be given as I=1
2mr2,
where mis the mass of the infinitesimal disc and ris the distance of the disc
from the axis of rotation.
Step 2: Let’s express min terms of Mand rin terms of x(the distance of
the disc from the end of the rod). The mass of the disc is proportional to its
length, so dm =M
Ldx and r=L−x.
Step 3: Now, let’s substitute dm and rin the moment of inertia equation.
We get dI =1
2M
Ldx(L−x)2=M
2L(L2−2Lx +x2)dx.
Step 4: To find the total moment of inertia of the rod, we integrate dI from
x= 0 to x=L. This gives I=RL
0
M
2L(L2−2Lx +x2)dx.
Step 5: Simplifying the integral, we get I=M
2L1
3L3−L2x+1
3x3
L
0
.
Step 6: Evaluating the expression gives I=M
2L1
3L3−L3+1
3L3=1
3ML2.
Therefore, the moment of inertia of the rod about the axis passing through
one of its ends and perpendicular to the rod is 1
3ML2.
Question 21
Question
A thin uniform rod of length Land mass Mis rotated about an axis perpen-
dicular to the rod and passing through one end. Find the moment of inertia of
17
the rod about this axis.
Solution
Step 1: Determine the moment of inertia of an infinitesimally small element of
length dx at a distance xfrom the axis of rotation. The mass of the element is
dm =M
Ldx.
Step 2: Use the formula for moment of inertia of a point mass about an
axis:
dI =r2dm
Where ris the perpendicular distance from the axis to the element. In this case,
r=x. Therefore,
dI =x2M
Ldx
Step 3: Integrate to find the total moment of inertia I:
I=ZdI =Zx2M
Ldx
Step 4: Perform the integration:
I=M
LZL
0
x2dx
Step 5: Solve the integral:
I=M
Lx3
3L
0
I=M
LL3
3−0
I=ML2
3
Step 6: Therefore, the moment of inertia of the rod about the axis perpen-
dicular to the rod and passing through one end is M L2
3.
Question 22
Question
Find the moment of inertia of a thin uniform rod of length Land mass Mabout
an axis passing through one end of the rod and perpendicular to it.
18
Solution
Step 1: We can find the moment of inertia of the rod by considering it as a
collection of infinitesimally small mass elements δm along the length of the rod.
Step 2: Let us consider a small mass element δm at a distance xfrom the
end of the rod. The mass of this element is proportional to the length of the
rod, so δm =M
Ldx.
Step 3: The moment of inertia of this small mass element about the axis
passing through one end of the rod is given by dI =δm ·x2. Substituting the
value of δm gives dI =M
Lx2dx.
Step 4: To find the total moment of inertia of the rod, we integrate the
expression dI =M
Lx2dx from 0 to L(the length of the rod):
I=ZL
0
M
Lx2dx
Step 5: Simplifying the integral gives:
I=M
LZL
0
x2dx =M
Lx3
3L
0
Step 6: Evaluating the integral gives:
I=M
LL3
3−03
3=M
L·L3
3=M
3L2
Step 7: Therefore, the moment of inertia of the rod about an axis passing
through one end of the rod and perpendicular to it is M
3L2.
Question 23
Question
A thin uniform rod of length Land mass Mis rotating about an axis passing
through one end perpendicular to the rod with an angular velocity ω. Determine
the moment of inertia of the rod about this axis.
Solution
Step 1: The moment of inertia of the rod about an axis passing through its
center perpendicular to the rod is 1
12 ML2.
Step 2: The parallel axis theorem states that the moment of inertia about
an axis parallel to and a distance daway from an axis through the center of
mass is given by I=Icm +Md2, where Icm is the moment of inertia about the
center of mass.
Step 3: In this case, the distance dis L/2 since the axis is passing through
one end of the rod.
19
Step 4: Substituting Icm =1
12 ML2and d=L/2 into the parallel axis
theorem, we have I=1
12 ML2+ML
22.
Step 5: Simplifying the expression, we get I=1
12 ML2+1
4ML2.
Step 6: Combining the terms, we find I=1
3ML2.
Therefore, the moment of inertia of the rod about an axis passing through
one end perpendicular to the rod is 1
3ML2.
Question 24
Question
Find the moment of inertia of a rectangular plate of mass Mand dimensions
a×babout an axis passing through one corner of the plate and perpendicular
to the plate.
Solution
To find the moment of inertia of the rectangular plate about the given axis, we
can use the parallel axis theorem. First, we find the moment of inertia of the
plate about its center of mass and then apply the parallel axis theorem to shift
the axis to the desired location.
Step 1: Find the moment of inertia of the rectangular plate about its center
of mass.
The moment of inertia of a rectangular plate about an axis passing through
its center and perpendicular to the plate is given by the expression:
Icenter =M
12 (a2+b2)
where Mis the mass of the plate, ais the length of the plate, and bis the
width of the plate.
Step 2: Apply the parallel axis theorem to find the moment of inertia about
the desired axis.
The parallel axis theorem states that the moment of inertia about a parallel
axis is equal to the moment of inertia about the center of mass plus the product
of the mass and the square of the distance between the two axes.
In this case, the distance between the center of mass and the desired axis is
a
2+b
2=a+b
2. Therefore, the moment of inertia about the desired axis is:
I=Icenter +Ma+b
22
Substitute Icenter from step 1 into the equation and simplify to get the final
expression for the moment of inertia about the given axis.
20
Question 25
Question
A thin uniform rod of mass Mand length Lis rotating about an axis perpen-
dicular to the rod and passing through its midpoint with an angular velocity ω.
Determine the moment of inertia of the rod about this axis.
Solution
Step 1: The moment of inertia of a thin rod rotating about an axis perpendicular
to the rod and passing through its midpoint can be calculated using the formula
I=1
12 mL2, where mis the mass of the rod and Lis its length.
Step 2: First, we need to express the mass Mof the rod in terms of its linear
density λ=M
L.
Step 3: The mass mof an infinitesimally small segment of length dx located
at a distance xfrom the midpoint can be expressed as dm =λdx.
Step 4: The moment of inertia dI of this segment about the axis can be
calculated as dI = (dm)x2.
Step 5: Substituting the expression for dm into the equation for dI gives
dI =λx2dx.
Step 6: To find the total moment of inertia Iof the entire rod, we integrate
dI with respect to xover the length of the rod. Hence, I=RL/2
−L/2λx2dx.
Step 7: Simplifying the integration, we get I=λhx3
3iL/2
−L/2.
Step 8: Evaluating the limits of integration, we find I=λ
3hL
23−−L
23i.
Step 9: Simplifying further, we get I=λ
3hL3
8+L3
8i.
Step 10: Thus, the moment of inertia Iof the rod about the given axis is
I=2
3λL3=2
3M
LL3=1
3ML2.
Question 26
Question
Find the moment of inertia of a thin uniform rod of mass mand length Labout
an axis perpendicular to the rod and passing through one of its ends.
Solution
Let’s divide the rod into small elements of mass dm at a distance xfrom the end
of the rod. The mass dm of a small element of length dx is given by dm =m
Ldx.
Step 1: Express the moment of inertia Ias the sum of moments of inertia
of all the small elements.
21
The moment of inertia of a small element dm about the perpendicular axis
passing through one end is dI =x2dm. Therefore, the moment of inertia of the
entire rod is:
I=ZL
0
x2dm
Step 2: Express dm in terms of dx.
Given that dm =m
Ldx, substitute this into the expression for I:
I=ZL
0
x2m
Ldx
Step 3: Simplify the expression and solve for I.
I=ZL
0
m
Lx2dx
I=m
LZL
0
x2dx
I=m
Lx3
3L
0
I=m
LL3
3−0
I=mL2
3
Therefore, the moment of inertia of the thin uniform rod about an axis
perpendicular to the rod and passing through one of its ends is mL2
3.
Question 27
Question
A solid uniform cylinder of radius Rand mass Mis rolling without slipping
along a horizontal surface with angular speed ω. Calculate the moment of
inertia of the cylinder about an axis perpendicular to its symmetry axis and
passing through its center.
Solution
Step 1: The moment of inertia of the cylinder about its symmetry axis passing
through its center is 1
2MR2.
Step 2: To find the moment of inertia of the cylinder about an axis per-
pendicular to its symmetry axis and passing through its center, we can use the
parallel axis theorem.
22
Step 3: The distance between the two axes is R. The parallel axis theorem
states that I=Icm +Md2, where Iis the moment of inertia about the new
axis, Icm is the moment of inertia about the center of mass axis, Mis the mass
of the object, and dis the distance between the two axes.
Step 4: Substituting the values, we get I=1
2MR2+MR2=3
2MR2.
Therefore, the moment of inertia of the cylinder about an axis perpendicular
to its symmetry axis and passing through its center is 3
2MR2.
Question 28
Question
A thin rod of length Land mass Mis rotated about an axis that is perpendicular
to the rod and passes through one end. Find the moment of inertia of the rod
about this axis.
Solution
1. Consider a small element of the rod of length dx at a distance xfrom the
end of the rod. The mass of this element can be approximated as M
Ldx.
2. The moment of inertia dI of this element about the axis is given by
dI =M
Lx2dx.
3. To find the total moment of inertia Iof the entire rod, we need to integrate
dI over the length of the rod.
I=ZL
0
M
Lx2dx
4. Solving the integral, we get
I=M
Lx3
3L
0
=M
LL3
3−0
5. Therefore, the moment of inertia of the rod about the axis passing through
one end is
I=M
3L2
Question 29
Question
Find the moment of inertia of a thin-walled semi-circular cylinder of radius R
and uniform mass density ρ, rotated about an axis passing through its diameter.
23
Solution
Let’s consider the thin-walled semi-circular cylinder as a collection of thin rings,
each at a distance rfrom the axis of rotation. We will sum up the moments of
inertia of all these thin rings to find the total moment of inertia of the semi-
circular cylinder.
Step 1: Consider a thin ring of radius rand thickness dr at a distance r
from the axis of rotation.
The mass of this thin ring can be calculated by considering the area of a
ring:
dM = 2πrdrρ
The moment of inertia of this thin ring about the axis of rotation is given
by:
dI =r2dM = 2πr3drρ
Step 2: Integrate to find the total moment of inertia of the semi-circular
cylinder.
We can now integrate dI over the entire semi-circular cylinder:
I=ZdI =ZR
0
2πr3drρ
I= 2πρ ZR
0
r3dr
I= 2πρ r4
4R
0
I= 2πρ R4
4−04
4
I= 2πρ R4
4
I=1
2πR4ρ
Thus, the moment of inertia of the thin-walled semi-circular cylinder of radius R
and uniform mass density ρ, rotated about an axis passing through its diameter
is 1
2πR4ρ.
Question 30
Question
A thin uniform rod of length Land mass Mis rotating about an axis perpendic-
ular to the rod and passing through one end with an angular velocity ω. Find
the moment of inertia of the rod about this axis.
24
Solution
Step 1: Divide the rod into small elements ∆mieach at a distance rifrom the
axis of rotation.
Step 2: Calculate the moment of inertia of each element Ii= (∆mi)·(ri)2.
Step 3: Sum up all the moments of inertia of each element using the formula
for moment of inertia: I=PIi.
Step 4: Express ∆miin terms of the total mass M, and riin terms of the
distance xialong the rod.
Step 5: Integrate I=RIidm over the length of the rod to find the total
moment of inertia I.
Step 6: Finally, substitute the result of the integral and simplify to express
the moment of inertia of the rod in terms of its length L, mass M, and angular
velocity ω.
Question 31
Question
Find the moment of inertia of a uniform solid cylinder of mass Mand radius
R, rotating about an axis that is tangential to the cylinder’s surface.
Solution
Let’s denote the moment of inertia as I. To find the moment of inertia of the
solid cylinder, we need to integrate over the mass distribution of the cylinder.
Step 1: Determine the mass distribution function. We can express the mass
distribution function as ρ(r), where ris the distance from the axis of rotation.
For a solid cylinder of uniform density, ρ(r) is constant.
Step 2: Define an element of mass. Consider an elemental disk of radius
rand thickness dr at distance rfrom the axis of rotation. The mass of this
elemental disk can be expressed as dm =ρ(r)·dV , where dV is the volume of
the elemental disk.
Step 3: Express dV in terms of rand dr. The volume of the elemental disk
can be expressed as dV =πr2·dr.
Step 4: Calculate dm in terms of rand dr. Substitute dV =πr2·dr into
dm =ρ(r)·dV to get dm =πρ(r)r2·dr.
Step 5: Find the moment of inertia dI of the elemental disk. The moment
of inertia of the elemental disk about the axis of rotation is dI =r2·dm.
Substitute dm =πρ(r)r2·dr into this expression to get dI =πρ(r)r4·dr.
Step 6: Integrate to find the total moment of inertia I. Integrate dI =
πρ(r)r4·dr over the entire volume of the cylinder to find the total moment of
inertia I.
Step 7: Substitute values and evaluate the integral. Substitute ρ(r) with
the density of the cylinder, and integrate over the limits of rfrom 0 to Rwhere
25
Ris the radius of the cylinder. Evaluate the integral to find the moment of
inertia I.
Question 32
Question
Calculate the moment of inertia about the x-axis of the region bounded by the
curves y=x2and y=x.
Solution
To find the moment of inertia about the x-axis, we will use the formula:
Ix=Zb
a
y2dx
where aand bare the x-values where the two curves intersect.
Step 1: Find the points of intersection. Setting x2=x, we find the
points of intersection to be x= 0 and x= 1.
Step 2: Set up the integral. For the given region, we have y=x2above
and y=xbelow. Thus, the integral becomes:
Ix=Z1
0
(x2−x)2dx
Expanding the integrand gives:
Ix=Z1
0
(x4−2x3+x2)dx
Step 3: Evaluate the integral. Integrating each term separately, we get:
Ix=1
5x5−1
2x4+1
3x31
0
Evaluating this from 0 to 1 gives:
Ix=1
5−1
2+1
3−(0 −0 + 0)
Ix=1
5−1
2+1
3
Ix=1
30
Therefore, the moment of inertia about the x-axis for the given region is 1
30 .
26
Question 33
Question
A thin rod of length Land mass Mis rotated about an axis that is perpendicular
to the rod and passes through one end. Determine the moment of inertia of the
rod about this axis.
Solution
To find the moment of inertia of the thin rod about an axis perpendicular to
the rod and passing through one end, we will use the formula for the moment
of inertia of a continuous mass distribution:
I=Zr2dm
where ris the distance from the axis of rotation to an infinitesimal mass
element dm.
Step 1: Choose a variable Let xbe the distance along the rod from the
end where the axis of rotation passes through. The infinitesimal mass dm can
be expressed as a function of x.
Step 2: Expressing elemental mass The mass per unit length of the rod
can be expressed as M
L. Therefore, the infinitesimal mass dm is dm =M
Ldx.
Step 3: Calculate the moment of inertia Substitute r=xand dm =
M
Ldx into the formula for moment of inertia:
I=ZL
0
x2M
Ldx
I=M
LZL
0
x2dx
I=M
Lx3
3L
0
I=M
LL3
3−0
I=ML2
3
Therefore, the moment of inertia of the thin rod about an axis perpendicular
to the rod and passing through one end is M L2
3.
27
Question 34
Question
A thin disc of radius Rand mass Mis rotating about an axis passing through
its center with angular velocity ω. Determine the moment of inertia of the disc
with respect to an axis perpendicular to the disc passing through its edge.
Solution
Step 1: The moment of inertia of a thin disc about an axis passing through its
center and perpendicular to the disc is given by the formula Icenter =1
2MR2.
Step 2: To determine the moment of inertia of the disc about an axis passing
through its edge (perpendicular to the disc), we can use the parallel axis theo-
rem. According to the theorem, Iedge =Icenter +M d2, where dis the distance
between the two axes.
Step 3: The distance dbetween the edge of the disc and the center is equal
to the radius R. Therefore, d=R.
Step 4: Substituting d=Rinto the formula Iedge =Icenter +Md2, we get
Iedge =1
2MR2+MR2.
Step 5: Simplifying, we find Iedge =3
2MR2. Hence, the moment of inertia
of the disc about an axis perpendicular to the disc passing through its edge is
3
2MR2.
Question 35
Question
A uniform thin rod of length Land mass Mis rotating about an axis perpen-
dicular to the rod and passing through one end. Find the moment of inertia of
the rod about this axis.
Solution
To find the moment of inertia (I) of the rod about the given axis, we can use the
formula for the moment of inertia of a thin rod rotating about one end, given
by I=1
3ML2.
Step 1: Write down the formula for moment of inertia of a thin rod rotating
about one end.
I=1
3ML2
Step 2: Substitute the given values Mand Linto the formula.
I=1
3·M·L2=1
3·M·L·L
28
Step 3: Simplify the expression.
I=1
3·M·L·L=1
3·M·L2
Step 4: Therefore, the moment of inertia of the rod about the given axis is
1
3ML2.
29
