PHYS 101 - ELEMENTS OF PHYSICS
- Moment of Inertia
Question Bank - Set 4
Liberty University
Question 1
Question
Find the moment of inertia of a thin rod of mass Mand length Labout an axis
perpendicular to the rod and passing through one end.
Solution
Let’s consider a thin rod of mass Mand length Lwith the rotation axis passing
through one end. We will calculate the moment of inertia about this axis.
Step 1: Divide the rod into small segments ∆mand express the mass of
each segment in terms of its length.
Let xbe the distance of a segment from the end where the rotation axis
passes through. Therefore, the length of the segment is ∆xand the mass of the
segment is ∆m=M
L∆x.
Step 2: Calculate the moment of inertia of each small segment about the
axis.
The moment of inertia of a small segment ∆Iabout the axis is given by
∆I= ∆m·x2.
Substitute the expression for ∆mto get ∆I=M
L∆xx2.
Step 3: Sum up the contributions of all small segments to find the total
moment of inertia.
Integrate ∆Iover the length of the rod:
I=ZL
0
M
Lx2dx
Step 4: Solve the integral to find the moment of inertia.
I=M
LZL
0
x2dx
I=M
Lx3
3L
0
I=M
LL3
3−0
I=M
L·L3
3
I=ML2
3
Therefore, the moment of inertia of the thin rod about the axis perpendicular
to the rod and passing through one end is ML2
3.
Question 2
Question
Find the moment of inertia of a solid sphere of radius Rand mass Mabout a
diameter (axis passing through the center and perpendicular to the surface).
Solution
To find the moment of inertia of the solid sphere about a diameter, we can
integrate the moment of inertia of thin circular disks along the diameter axis.
Step 1: Consider a thin disk within the sphere at a distance xfrom the
center. The mass of this thin disk is dm.
Step 2: The moment of inertia of this thin disk about the axis passing
through the center is dI =1
2r2dm, where ris the distance of the thin disk from
the axis.
Step 3: Since the solid sphere has uniform mass distribution, the mass dm
of the thin disk is M
VdV , where Vis the volume of the sphere and dV is the
volume of the thin disk.
Step 4: Using the equation for the volume of a sphere V=4
3πR3and the
volume of a disk dV =πx2dx, we can express dm in terms of dx.
Step 5: Substitute dm in the expression for dI and integrate from x= 0 to
x=Rto find the total moment of inertia Iof the solid sphere.
Step 6: After integrating, we find that the moment of inertia of the solid
sphere about a diameter is I=2
5MR2.
2
Question 3
Question
Find the moment of inertia of a solid cone with radius Rand height Habout
its central axis. Assume the density of the cone is uniform.
Solution
Given: Radius of the cone, RHeight of the cone, H
The moment of inertia Iof a solid cone about its central axis can be calcu-
lated using the formula:
I=3
10MR2+3
20MH2
Where Mis the mass of the cone. We can find Musing the volume formula for
a cone:
V=1
3πR2H
Step 1: Find the mass Mof the cone Using the density formula ρ=M
V,
we can rearrange to find M:
M=ρ·V=ρ·1
3πR2H
Step 2: Substitute Minto the formula for moment of inertia Now
that we have M, we can substitute it into the formula for moment of inertia:
I=3
10 ρ·1
3πR2HR2+3
20 ρ·1
3πR2HH2
Step 3: Simplify the equation
I=1
10ρπR4H+1
20ρπR2H3
Hence, the moment of inertia of the solid cone about its central axis is
1
10 ρπR4H+1
20 ρπR2H3.
Question 4
Question
Calculate the moment of inertia of a uniform solid cylinder with mass M, radius
R, and height Habout an axis passing through its center and perpendicular to
its circular base.
3
Solution
Let’s break down the solution into steps:
Step 1: Determine the moment of inertia of a thin disk with respect to the
axis passing through its center and perpendicular to its circular base.
The moment of inertia of a thin disk of mass dm and radius rabout its center
and perpendicular to its circular base is given by the formula dI =1
2r2dm. To
find the total moment of inertia of the entire disk, we need to integrate this
expression over the entire disk.
Idisk =ZdI =ZR
0
1
2r2M
πR2dr =1
2M
πR2ZR
0
r2dr
Step 2: Evaluate the integral to find the moment of inertia of the thin disk.
Idisk =1
2M
πR2r3
3R
0
=1
6M
πR2R3=1
6MR2
Step 3: Use the moment of inertia of the thin disk to find the moment of
inertia of the solid cylinder.
Since the solid cylinder can be considered as a collection of such disks stacked
along the height H, the moment of inertia of the entire cylinder can be found
by summing the moments of inertia of all the disks.
Icylinder =ZdIdisk =ZH
0
1
6MR2dr
R=1
6MR2ZH
0
dr =1
6MR2H
Therefore, the moment of inertia of the uniform solid cylinder about the
given axis is 1
6MR2H.
Question 5
Question
Find the moment of inertia of a uniform solid cylinder of mass Mand radius
Rabout an axis passing through its center and perpendicular to the cylinder’s
axis. The density of the cylinder is ρ.
Solution
Let’s consider a small element of the cylinder located at a distance rfrom the
axis of rotation. The mass of this element can be given by δm =ρ·2πr ·δr,
where δr is the thickness of the element.
Step 1: Express moment of inertia for the whole cylinder.
Let’s integrate over the whole cylinder to find the moment of inertia.
dI=δm ·r2=ρ·2πr ·dr·r2
4
I=ZR
0
2πρr3dr=2
4πρR4
So, the moment of inertia of the cylinder about the given axis is 1
2MR2.
Question 6
Question
Find the moment of inertia of a uniform circular disk of radius Rabout an axis
perpendicular to the disk and passing through its center.
Solution
Step 1: Consider an elemental ring of radius rand width dr on the disk. Step
2: The mass of this elemental ring can be given by dm =ρ·dA, where ρis the
density of the disk material and dA = 2πr ·dr is the area of the elemental ring.
Step 3: The moment of inertia dI of this elemental ring about the axis is given
by dI =r2·dm. Step 4: Substitute the expressions for dm and dA into dI to
get dI =ρ·2πr3·dr. Step 5: Integrate dI from 0 to Rto find the total moment
of inertia Iof the disk:
I=ZR
0
ρ·2πr3·dr
Step 6: Simplify the expression and solve the integral to find I.
Question 7
Question
A thin uniform rod of length Land mass Mis rotated about an axis passing
through one end perpendicular to the rod. Find the moment of inertia of the
rod about this axis.
Solution
Let’s consider the rod to have mass per unit length λ=M
L.
Step 1: Divide the rod into small elements ∆m=λ∆x.
Step 2: The moment of inertia of each small element ∆Iabout the axis is
given by ∆I= (∆m)(x2).
Step 3: Sum up the moments of inertia of all small elements by integrating
from 0 to L:
I=ZL
0
λx2dx
5
Step 4: Solve the integral:
I=λZL
0
x2dx =λx3
3L
0
=λL3
3
Step 5: Substitute back the expression for λto find the moment of inertia
I:
I=ML2
3
Therefore, the moment of inertia of the rod about the given axis is ML2
3.
Question 8
Question
A uniform thin rod of length Land mass Mis rotating about an axis perpen-
dicular to the rod and passing through one end of the rod. Find the moment of
inertia of the rod about this axis.
Solution
Step 1: The moment of inertia of a point mass about an axis is given by I=mr2,
where mis the mass of the point mass and ris the perpendicular distance from
the point mass to the axis of rotation.
Let’s consider a small element δm of the rod at a distance xfrom the axis
of rotation. The mass of this element can be expressed as δm =M
Lδx, where
δx is a small length element of the rod.
Step 2: The perpendicular distance of this element from the axis of rotation
is x. Therefore, the moment of inertia of this element about the axis is dI =
M
Lx2δx.
Step 3: To find the total moment of inertia of the rod, we need to integrate
this expression over the entire length of the rod:
I=ZdI =ZL
0
M
Lx2dx
Step 4: Solving the integral gives:
I=M
LZL
0
x2dx
I=M
L1
3x3L
0
I=M
L1
3L3−1
3(0)3
6
I=M
L·1
3L3
Step 5: Simplifying the expression further:
I=1
3ML2
Therefore, the moment of inertia of the rod about the given axis is 1
3ML2.
Question 9
Question
A thin-walled hollow cylinder has an outer radius of 6 cm, an inner radius of
4 cm, and a height of 10 cm. Calculate the moment of inertia of the cylinder
about its central axis.
Solution
To find the moment of inertia of the hollow cylinder, we can use the formula for
the moment of inertia of a thin-walled cylinder:
I=1
2×M×(r2
outer +r2
inner)
Step 1: Calculate the mass of the cylinder The mass of the cylinder
can be found using the formula:
M= Volume ×Density
The volume of the hollow cylinder is:
Volume = π×h×(r2
outer −r2
inner)
where his the height of the cylinder.
Substitute the given values into the formula:
Volume = π×10 ×((6)2−(4)2) = 80πcm3
Given that the density of the material is ρ= 1 g/cm3, we can find the mass:
M= 80π×1 = 80πg
Step 2: Calculate the moment of inertia Now, plug the values of M,
router and rinner into the formula:
I=1
2×80π×((6)2+ (4)2) = 1
2×80π×52 = 2080πcm4
Therefore, the moment of inertia of the hollow cylinder about its central axis
is 2080πcm4.
7
Question 10
Question
A thin rod of length Land mass Mis rotating about an axis passing through
one end perpendicular to the rod with an angular velocity ω. Calculate the
moment of inertia of the rod about the axis of rotation.
Solution
Step 1: The moment of inertia of a thin rod rotating about an axis passing
through one end perpendicular to the rod is given by the formula I=1
3ML2.
Step 2: This formula can be derived by integrating the expression for the
moment of inertia of a small mass element dm along the length of the rod. The
moment of inertia of a small mass element dm at a distance rfrom the axis of
rotation is dI =r2dm.
Step 3: By expressing dm in terms of the linear mass density λ=M
L, we
have dm =λdx where dx is a small length element along the rod.
Step 4: Substituting dm into the expression for dI gives dI =r2λdx.
Step 5: Integrating dI along the length of the rod from 0 to Lgives the total
moment of inertia I=RL
0r2λdx.
Step 6: Evaluating the integral gives I=λRL
0r2dx.
Step 7: Since rranges from 0 to L, we have I=λRL
0r2dx =λhr3
3iL
0.
Step 8: Simplifying the expression, we get I=1
3λL3.
Step 9: Substituting λ=M
Linto the equation gives I=1
3ML2. Therefore,
the moment of inertia of the thin rod about the axis of rotation is 1
3ML2.
Question 11
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 mass elements
of the rod along its length.
Step 1: Consider a small mass element dm of length dx at a distance xfrom
the end of the rod. The mass of this element can be expressed as dm =M
Ldx.
Step 2: The moment of inertia dI of this small mass element about the axis
perpendicular to the rod and passing through one end is given by dI = (x)2dm.
8
Step 3: Substitute dm =M
Ldx into the expression for dI and integrate from
x= 0 to x=Lto find the total moment of inertia I.
I=ZL
0
x2M
Ldx
=M
LZL
0
x2dx
=M
L1
3x3L
0
=M
L1
3L3−0
=1
3ML2
Step 4: 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 12
Question
A thin uniform rod of length Land mass Mis rotated about an axis passing
through one end perpendicular to the rod. Find the moment of inertia of the
rod about this axis.
Solution
Let’s consider the rod to be along the x-axis with the origin at the end where
the axis of rotation passes. The density of the rod is M
L.
Step 1: Divide the rod into small elements δm along its length, each at a
distance xfrom the origin.
Mass of each element: dm =M
Ldx
The moment of inertia of one small element about the axis of rotation:
dI = (x)2dm = (x)2M
Ldx
Step 2: To find the total moment of inertia of the rod, we integrate over
the entire length:
9
I=ZdI
=ZL
0
x2M
Ldx
=M
LZL
0
x2dx
=M
Lx3
3L
0
=M
LL3
3−03
=M
L·L3
3
=1
3ML2
Therefore, the moment of inertia of the rod about the given axis is 1
3ML2.
Question 13
Question
A thin uniform rod of mass mand length Lis rotating about an axis passing
through one end and perpendicular to the rod at a constant angular velocity ω.
Calculate 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 one
end perpendicular to the rod can be calculated using the formula:
I=Zr2dm
where ris the distance of an element of mass dm from the axis of rotation.
Step 2: Consider a small element dx of the rod at a distance xfrom the end
where the axis of rotation passes. The mass dm of this element is given by:
dm =m
Ldx
Step 3: The distance rof the element dm from the axis of rotation is simply
x. Therefore, r=x.
Step 4: Substituting dm and rinto the moment of inertia formula gives:
I=ZL
0
x2m
Ldx
10
Step 5: Simplifying the integral, we get:
I=m
LZL
0
x2dx
Step 6: Solving the integral, we have:
I=m
Lx3
3L
0
I=m
LL3
3−0
I=mL2
3
Step 7: Therefore, the moment of inertia of the rod about the axis passing
through one end and perpendicular to the rod is mL2
3.
Question 14
Question
Find 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 divide the rod into small elements of mass dm at a distance xfrom the
end through which the axis passes. The mass of each element is dm =M
Ldx.
Step 1: Calculate the moment of inertia dI of each small element about the
given axis. The moment of inertia of a small element of mass dm at a distance
xfrom the axis is given by dI =dm ·x2. So, dI =M
Ldx·x2=M
Lx2dx.
Step 2: Determine the total moment of inertia Iof the entire rod by inte-
grating dI over the length of the rod.
I=ZL
0
M
Lx2dx
Step 3: Solve the integral to find I.
I=M
LZL
0
x2dx
I=M
Lx3
3L
0
I=M
LL3
3−0
11
I=M
3L2
Therefore, the moment of inertia of the rod about the given axis is M
3L2.
Question 15
Question
A thin uniform rod of length Land mass Mis rotating about an axis perpendic-
ular to the rod and passing through one end. Calculate the moment of inertia
of the rod about this axis.
Solution
Step 1: Divide the rod into small elements ∆m=M
L∆x. Step 2: The distance
of each element from the axis of rotation is x. The moment of inertia of each
element about the axis is dI = ∆m·x2. Step 3: The total moment of inertia is
obtained by summing up all the elements:
I=ZdI
Step 4: Substitute ∆m=M
L∆xand dI =M
L∆x·x2into the integral. Step 5:
Integrate over the length of the rod from 0 to L.
I=ZL
0
M
L·x2dx
Step 6: Simplify the integral and solve for I.
I=M
L·x3
3L
0
=M
L·L3
3=1
3ML2
Step 7: Therefore, the moment of inertia of the rod about the given axis is
1
3ML2.
Question 16
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.
12
Solution
Step 1: Divide the rod into small elements δm of length δx.
Step 2: Express δm in terms of linear density λand δx:δm =λδx.
Step 3: Find the mass dm of a small element of length dx at a distance x
from the axis of rotation: dm =λdx.
Step 4: The moment of inertia dI of this small element about the axis of
rotation is given by dI =r2dm, where ris the distance of the element from the
axis.
Step 5: Express rin terms of x:r=L−x.
Step 6: Rewrite dI in terms of x:dI = (L−x)2λdx.
Step 7: Integrate dI from x= 0 to x=Lto find the total moment of inertia
I:
I=ZL
0
(L−x)2λdx
Step 8: Solve the integral:
I=ZL
0
(L2−2Lx +x2)λdx
I=λL3
3−2LL2
2+L3
3
I=λL3
3−L3
3=1
3λL3
Step 9: Substitute λ=M
Lto find the moment of inertia I:
I=1
3M
LL3=1
3ML2
Therefore, the moment of inertia of the thin uniform rod about the given
axis is 1
3ML2.
Question 17
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 its midpoint.
Solution
To find the moment of inertia of the rod, we will consider it to be made up of
infinitesimally small elements, each with mass dm and located at a distance x
from the axis of rotation. Then, we will integrate over all these elements to find
the total moment of inertia of the rod.
13
Step 1: Let’s consider an element of length dx at a distance xfrom the
center of the rod. The mass of this element dm can be expressed in terms of dx
and Msince the rod is uniform. The density λof the rod is given by λ=M
L.
So, dm =λ·dx =M
Ldx.
Step 2: The moment of inertia dI of this element about the axis can be
calculated using the formula dI =dm ·x2. Substituting for dm,dI =M
Ldx·
x2=Mx2
Ldx.
Step 3: Now, we can find the total moment of inertia Iby integrating dI
over the entire length of the rod. I=RdI =RL/2
−L/2
Mx2
Ldx.
Step 4: Solving the integral, I=M
LRL/2
−L/2x2dx =M
Lhx3
3iL/2
−L/2.I=
M
Lh(L
2)3
3−(−L
2)3
3i.I=M
LhL3
24 +L3
24 .I = M L2
12 .
Therefore, the moment of inertia of the thin uniform rod about an axis
perpendicular to the rod and passing through its midpoint is ML2
12 .
Question 18
Question
Find the moment of inertia of a thin rod of mass Mand length Labout an axis
perpendicular to the rod and passing through one of its ends.
Solution
To find the moment of inertia of the rod, we will consider it as made up of
infinitesimally small elements.
Step 1: Consider an element of length dx at a distance xfrom the end of
the rod, where 0 ≤x≤L.
Step 2: The mass dm of this element can be given as M
Ldx.
Step 3: The moment of inertia dI of this element about the axis perpen-
dicular to the rod and passing through one end is given by dI =dm ·x2.
Step 4: Substituting the expression for mass dm into the above equation,
we get dI =M
L·x2dx.
Step 5: The total moment of inertia Iof the rod can be found by integrating
dI from 0 to L:
I=ZL
0
dI =ZL
0
M
L·x2dx
Step 6: Solving the integral, we get:
I=M
L·x3
3L
0
=M
L·L3
3−0=ML2
3
Therefore, the moment of inertia of the thin rod of mass Mand length L
about an axis perpendicular to the rod and passing through one of its ends is
ML2
3.
14
Question 19
Question
A thin rod of mass mand length Lis rotated about an axis perpendicular to
the rod and passing through one end of the rod. Find the moment of inertia of
the rod about this axis.
Solution
Step 1: The moment of inertia of a point mass mat distance rfrom an axis of
rotation is I=mr2.
Step 2: The rod can be thought of as a collection of point masses, each
at a different distance from the axis of rotation. We can divide the rod into
infinitesimally small masses dm and write the moment of inertia contribution
from each as dI =dm ·r2.
Step 3: The linear mass density λof the rod is given by λ=m
L. Therefore,
the infinitesimal mass dm is dm =λ·dx.
Step 4: The distance of an infinitesimal mass dm from the axis of rotation
is r=x.
Step 5: Expressing min terms of λgives m=λ·L.
Step 6: Substituting the expressions for dm and minto dI =dm ·r2gives
dI = (λ·dx)·(x2) = λx2dx
.
Step 7: Integrating over the length of the rod gives the total moment of
inertia:
I=ZL
0
λx2dx =ZL
0
m
Lx2dx
.
Step 8: Solving the integral gives
I=m
Lx3
3L
0
=m
L·L3
3=mL2
3
.
Therefore, the moment of inertia of the rod about the given axis is mL2
3.
Question 20
Question
A hollow sphere of radius Rand mass Mis rotating about an axis passing
through its center with an angular velocity ω. Find the moment of inertia of
the sphere about an axis passing through a point on its surface.
15
Solution
Step 1: The moment of inertia of a hollow sphere of radius Rand mass Mabout
an axis passing through its center is given by the formula:
I=2
3MR2
Step 2: Let’s find the moment of inertia of the sphere about an axis passing
through a point on its surface. According to the parallel axis theorem, the
moment of inertia about this axis is equal to the sum of the moment of inertia
about an axis passing through the center and the product of its mass and the
square of the distance between the two axes.
Step 3: Let the distance between the two axes be l. Using the parallel axis
theorem, the moment of inertia about the new axis is:
I′=I+Ml2
Step 4: Substituting the given values, we have:
I′=2
3MR2+Ml2
Step 5: To find l, we can use the Pythagorean theorem with the two sides
being Rand land the hypotenuse being R. Therefore:
R2=R2+l2
l=√R2=R
Step 6: Substituting l=Rinto the equation for I′, we get:
I′=2
3MR2+MR2
I′=5
3MR2
Therefore, the moment of inertia of the sphere about an axis passing through
a point on its surface is 5
3MR2.
Question 21
Question
A thin uniform rod of mass mand length Lis rotated about an axis perpen-
dicular to the rod and passing through its midpoint. Determine the moment of
inertia of the rod about this axis.
16
Solution
Step 1: The moment of inertia of a rod about an axis perpendicular to the rod
and passing through one end is 1
3mL2. We can use the Parallel Axis Theorem to
find the moment of inertia of the rod about an axis passing through its midpoint.
Step 2: The distance from the midpoint to the end of the rod is L
2. To use
the Parallel Axis Theorem, we will need to add the moment of inertia about the
midpoint to the moment of inertia about an axis through the endpoint.
Step 3: The moment of inertia about the endpoint is 1
3mL2, and the distance
between the two axes is L
2. Applying the Parallel Axis Theorem: Imidpoint =
Iend +md2.
Step 4: Substituting the values, Imidpoint =1
3mL2+mL
22.
Step 5: Simplifying further, Imidpoint =1
3mL2+1
4mL2.
Step 6: Combining the terms, Imidpoint =7
12 mL2.
Therefore, the moment of inertia of the rod about an axis perpendicular to
the rod and passing through its midpoint is 7
12 mL2.
Question 22
Question
A thin uniform 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
Step 1: Choose an element of length dx at a distance xfrom the axis of rotation.
Step 2: The mass of this element dm is given by
dm =M
Ldx
Step 3: The moment of inertia of this element about the axis of rotation is
dI = (dm)x2
Step 4: Substitute dm from Step 2 into the above equation to get
dI =M
Ldxx2
dI =M
Lx2dx
Step 5: Integrate dI from x= 0 to x=Lto find the total moment of inertia I.
I=ZL
0
M
Lx2dx
17
Step 6: Solve the integral
I=M
Lx3
3L
0
I=M
LL3
3−03
3
I=M
L·L3
3
I=ML2
3
Therefore, the moment of inertia of the rod about the given axis is ML2
3.
Question 23
Question
A thin uniform rectangular plate of width wand height his rotated about an
axis through its center perpendicular to its plane. Find the moment of inertia
of the plate with respect to this axis.
Solution
Step 1: The moment of inertia of the rectangular plate about an axis through
its center perpendicular to its plane can be calculated using the formula for the
moment of inertia of a rectangular plate rotated about its edge (I=1
3Mh2,
where Mis the mass of the plate).
Step 2: To find the mass Mof the rectangular plate, we need to find its area
first. The area Aof the plate is given by A=w·h.
Step 3: The mass Mcan be calculated using the density ρof the plate and
the area A. The mass Mis given by M=ρ·A.
Step 4: Substituting the expressions for Aand Minto the formula for mo-
ment of inertia, we get
I=1
3(Mh2) = 1
3(ρ·A·h2)
Step 5: Simplifying further, we get
I=1
3(ρ·w·h·h2) = 1
3ρwh3
Therefore, the moment of inertia of the rectangular plate about an axis
through its center perpendicular to its plane is 1
3ρwh3.
18
Question 24
Question
A thin uniform rod of mass mand length Lis rotating about an axis that is
perpendicular to the rod and passing through one end. Calculate the moment
of inertia of the rod about this axis.
Solution
Step 1: Consider the moment of inertia of an infinitesimally small mass element
dm at a distance xfrom the axis of rotation. The moment of inertia of this
mass element is given by dI =x2dm.
Step 2: Express dm in terms of dx. The linear mass density λof the rod is
λ=m
L. Therefore, dm =λ dx.
Step 3: Substituting dm =λ dx into dI =x2dm. We get dI =x2λ dx.
Step 4: Integrate both sides of the equation to find the total moment of
inertia I.RdI =Rx2λ dx
Step 5: Determine the limits of integration. For a rod of length L, the limits
of integration are from 0 to L.
Step 6: Integrate the right side of the equation. I=RL
0x2λ dx =RL
0x2m
Ldx
Step 7: Simplify the integral. I=m
LRL
0x2dx =m
Lhx3
3iL
0
Step 8: Evaluate the integral. I=m
LL3
3−03
3=m
L·L3
3
Step 9: Simplify the expression. I=mL2
3
Therefore, the moment of inertia of the rod about the given axis is mL2
3.
Question 25
Question
Determine the moment of inertia of a thin rod of mass Mand length Labout
an axis perpendicular to the rod passing through one end.
Solution
Let’s consider the thin rod as a 1D system.
Step 1: Determine the linear mass density of the rod. The linear mass
density λis defined as mass per unit length.
λ=M
L
Step 2: Express the moment of inertia in terms of an integral. The moment
of inertia Iof the rod about an axis perpendicular to the rod and passing through
19
one end can be expressed as an integral of r2dm, where ris the distance from
the axis and dm is a small mass element on the rod.
I=Zr2dm
Step 3: Express dm in terms of linear mass density. Since λ=M
L, we have
dm =λdx, where dx is a small length element along the rod.
Step 4: Express rin terms of x. The distance rfrom the axis to a small
mass element at a distance xfrom the end is given by r=x.
Step 5: Substitute dm and rinto the expression for I.
I=ZL
0
x2λdx
Step 6: Substitute λ=M
Linto the integral.
I=ZL
0
x2M
Ldx
Step 7: Integrate to find the moment of inertia I.
I=M
LZL
0
x2dx
I=M
Lx3
3L
0
I=M
LL3
3
I=ML2
3
Therefore, the moment of inertia of the thin rod about an axis perpendicular
to the rod passing through one end is ML2
3.
Question 26
Question
A thin rod of mass Mand length Lis rotating about an axis passing through
its center and perpendicular to its length. Determine the moment of inertia of
the rod about this axis.
20
Solution
Step 1: We know that the moment of inertia of an object rotating about an axis
is given by the formula I=Rr2dm, where ris the distance from the axis of
rotation, and dm is an infinitesimally small mass element of the object.
Step 2: Let’s consider an infinitesimally small mass element dm of the rod
at a distance xfrom the center. The mass of this element can be expressed as
dm =M
Ldx.
Step 3: The distance rof this mass element from the axis of rotation is given
by r=L
2−x.
Step 4: Substituting dm and rinto the formula I=Rr2dm, we get:
I=ZL/2
−L/2
(L
2−x)2M
Ldx
Step 5: Simplifying the integrand, we have:
I=M
LZL/2
−L/2
(L2
4−Lx +x2)dx
Step 6: Integrating term by term, we get:
I=M
L[L3
4x−L2
2x2+1
3x3]L/2
−L/2
Step 7: Evaluating the integral at the limits of integration, we obtain:
I=M
L[L3
4(L
2)−L2
2(L2
4) + 1
3(L3
8)]
Step 8: Simplifying further, we get:
I=ML2
12
Step 9: Therefore, the moment of inertia of the rod about the given axis is
ML2
12 .
Question 27
Question
A thin rod of mass Mand length Lis rotating around its one end with an
angular velocity ω. Determine the moment of inertia of the rod about its far
end.
21
Solution
1. The moment of inertia of the rod about its center can be calculated using
the formula for a rod rotating about an axis perpendicular to its length:
Icm =1
12ML2
2. To find the moment of inertia about the far end, we can use the parallel
axis theorem, which states:
I=Icm +Md2
where dis the distance between the two axes of rotation.
3. In this case, the distance dis equal to half the length of the rod:
d=L
2
4. Substituting the values into the parallel axis theorem:
I=1
12ML2+ML
22
I=1
12ML2+1
4ML2
I=1
12ML2+3
12ML2
I=1
12ML2+3
12ML2
I=4
12ML2=1
3ML2
5. Therefore, the moment of inertia of the rod about its far end is 1
3ML2.
Question 28
Question
A thin rod of length Land mass Mis rotating about an axis passing through
one end perpendicular to the rod with an angular velocity ω. Calculate the
moment of inertia of the rod about the axis of rotation.
Solution
Step 1: The moment of inertia for a point mass rotating about an axis is given
by I=mr2, where mis the mass of the point mass and ris the distance of the
point mass from the axis of rotation.
22
Step 2: To find the moment of inertia of the entire rod, we need to consider
infinitesimally small point masses that make up the rod. Let’s consider a small
elemental mass dm at a distance xfrom the pivot point.
Step 3: The mass of the elemental length dx is given by dm =M
Ldx.
Step 4: The moment of inertia of this elemental mass about the axis of
rotation is dI =dm ·x2.
Step 5: Substituting the expression for dm into the moment of inertia for-
mula, we get dI =M
Ldxx2.
Step 6: The total moment of inertia of the rod can be obtained by integrating
dI over the entire length of the rod: I=RdI =RM
Ldxx2from 0 to L.
Step 7: Solving the integral, we get I=RL
0
M
Lx2dx.
Step 8: Simplifying the expression, we find I=M
LRL
0x2dx.
Step 9: By evaluating the integral, we get I=M
Lhx3
3iL
0.
Step 10: Therefore, the moment of inertia of the rod about the axis of
rotation is I=M
LL3
3=1
3ML2.
Question 29
Question
Determine the moment of inertia of a solid cylinder of mass Mand radius R
about an axis passing through its center perpendicular to its height.
Solution
Let’s consider a solid cylinder with mass Mand radius Ras shown below.
solid_cylinder.png
Step 1: To find the moment of inertia, we need to consider the cylinder as
a collection of infinitesimally thin disks. The moment of inertia of each disk is
I=1
2mr2, where mis the mass of the disk and ris the distance from the axis
of rotation.
Step 2: First, express the mass of each disk in terms of its radius rand the
cylinder’s total mass M. The mass dm of a disk with radius rand thickness dr
is given by dm =M
πR2·2πr ·dr.
Step 3: Now, substitute dm into the formula for the moment of inertia
I=1
2mr2, yielding dI =1
2M
πR2·2πr ·drr2.
Step 4: Integrate dI over the whole cylinder from 0 to Rto find the total
moment of inertia I:
I=ZR
0
1
2M
πR2·2πr ·drr2
23
Step 5: Simplify and solve the integral to find the moment of inertia I:
I=ZR
0
1
2·M
πR2·2πr3dr =M
2R2ZR
0
r3dr
Step 6: Evaluate the integral:
I=M
2R2r4
4R
0
=M
2R2·R4
4=1
8MR2
Therefore, the moment of inertia of the solid cylinder about an axis passing
through its center perpendicular to its height is 1
8MR2.
Question 30
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
To calculate the moment of inertia of the rod, we need to consider it as a
continuous mass distribution along its length. We can express the mass element
dm of the rod in terms of its linear mass density λ.
Step 1: Express the linear mass density λin terms of the total mass Mand
length Lof the rod.
λ=M
L
Step 2: Express dm (small mass element) in terms of λand dx (small length
element).
dm =λ dx =M
Ldx
Step 3: Express the moment of inertia Iof the rod by integrating the
contribution of each element of mass along the entire length of the rod.
I=Zr2dm
where ris the perpendicular distance of the mass element dm from the axis
of rotation.
Step 4: Express rin terms of x, the distance from the end of the rod.
r=L−x
24
Step 5: Substitute dm and rinto the equation for Iand perform the inte-
gration.
I=ZL
0
(L−x)2M
Ldx
=M
LZL
0
(L2−2Lx +x2)dx
=M
LL3x−L2x2+1
3x3
L
0
=M
L1
3L3
=1
3ML2
Step 6: Therefore, the moment of inertia of the thin rod of length Land
mass Mabout the axis perpendicular to the rod and passing through one end
is 1
3ML2.
Question 31
Question
A thin rod of mass mand length Lis rotating in a horizontal plane about one
end with an angular velocity ω. Determine the moment of inertia of the rod
about an axis perpendicular to the rod passing through its midpoint.
Solution
Step 1: To find the moment of inertia of the rod about an axis perpendicular
to the rod passing through its midpoint, we can use the parallel-axis theorem.
Let’s first find the moment of inertia of the rod about its center of mass.
Step 2: The moment of inertia of a rod rotating about an axis perpendicular
to the rod and passing through its center is given by the formula Icm =1
12 mL2.
Step 3: Now, using the parallel-axis theorem, the moment of inertia Iof
the rod about an axis perpendicular to the rod passing through its midpoint is
given by I=Icm +md2, where dis the distance between the midpoint and the
center of mass.
Step 4: The distance dbetween the midpoint and the center of mass of the
rod is d=L
2.
Step 5: Substituting Icm =1
12 mL2and d=L
2into the formula I=Icm +
md2, we get I=1
12 mL2+m(L
2)2.
Step 6: Simplifying the expression, we find I=1
12 mL2+1
4mL2=1
3mL2.
Therefore, the moment of inertia of the rod about an axis perpendicular to
the rod passing through its midpoint is 1
3mL2.
25
Question 32
Question
Find the moment of inertia of a uniform thin rod of length Land mass Mabout
an axis perpendicular to the rod and passing through one of its ends.
Solution
Step 1: The moment of inertia of an object about an axis can be calculated
using the formula: I=Rr2dm, where ris the distance from the axis of rotation
to the mass element dm. In this case, we will consider an elemental mass dm
at a distance xfrom the end of the rod.
Step 2: The mass dm of the rod can be expressed as dm =M
Ldx, where dx
is an elemental length of the rod.
Step 3: The distance rof the mass element dm from the axis of rotation is
r=x.
Step 4: Substituting dm and rinto the formula for moment of inertia, we
have: I=RL
0x2·M
Ldx.
Step 5: Solving the integral, we get: I=M
LRL
0x2dx.
Step 6: Integrating x2from 0 to Lgives: I=M
Lhx3
3iL
0.
Step 7: Simplifying further, we get: I=M
LL3
3−0.
Step 8: Finally, the moment of inertia of the rod about an axis perpendicular
to the rod and passing through one of its ends is: I=ML2
3.
Question 33
Question
Find the moment of inertia of a solid cylinder of radius Rand mass Mabout
an axis passing through its center perpendicular to the cylinder’s height.
Solution
Step 1: We can calculate the moment of inertia of the solid cylinder by inte-
grating over its volume. The moment of inertia Ican be expressed as:
I=Zr2dm
where ris the distance from the axis of rotation to the differential mass element
dm.
Step 2: To find dm, we need to express it in terms of the cylinder’s density.
The volume element dV of a cylinder in cylindrical coordinates is given by:
dV =Rdθdrdz
26
Step 3: Since the cylinder is of uniform density ρ, the mass located in a
differential volume dV is:
dm =ρdV =ρRdθdrdz
Step 4: Now we substitute dm into the equation for moment of inertia:
I=Zr2ρRdθdrdz
Step 5: To simplify the integral, we need to express rin terms of r,θ, and z.
For a point at coordinates (r, θ, z) in cylindrical coordinates, rcan be calculated
as:
r=pR2+z2
Step 6: Now we substitute rinto the equation for moment of inertia:
I=Z(pR2+z2)2ρRdθdrdz
Step 7: We can simplify the integral and limits since the cylinder is sym-
metric about the z-axis. Therefore, we integrate over zfrom −h/2 to h/2 and
over θfrom 0 to 2π.
Step 8: After performing the integrals, we find the moment of inertia of the
cylinder:
I=1
2πρR2h3
Thus, the moment of inertia of the solid cylinder about the given axis is
1
2πρR2h3.
Question 34
Question
Calculate the moment of inertia of a uniform thin rod of length Land mass M
about an axis perpendicular to the rod and passing through one of its ends.
Solution
Let’s denote the rod as a 1-dimensional object with mass uniformly distributed
along its length. We will use the formula for the moment of inertia of a contin-
uous mass distribution:
I=Zr2dm
Step 1: Determine the mass element dm in terms of dx,M, and L. The
linear density λof the rod is given by:
λ=M
L
27
The mass element dm can be expressed in terms of the linear density λand
a differential length dx along the rod:
dm =λ dx =M
Ldx
Step 2: Express rin terms of x. The distance rof the mass element dm
from the axis of rotation is the position of the mass element along the rod. Since
the axis of rotation is at one end of the rod, rcan be expressed in terms of x:
r=x
Step 3: Substitute dm and rinto the expression for I. The moment of
inertia Ican be calculated by integrating r2dm along the length of the rod:
I=Zr2dm =Zx2M
Ldx
Step 4: Integrate to find the moment of inertia.
I=M
LZL
0
x2dx =M
Lx3
3L
0
I=M
L·L3
3=ML2
3
Therefore, the moment of inertia of the rod about the axis passing through
one of its ends is ML2
3.
Question 35
Question
A solid cylinder with radius R, mass M, and height His rotated about an
axis passing through its center and perpendicular to its height. Determine the
moment of inertia of the cylinder about this axis.
Solution
Step 1: We can calculate the moment of inertia of the cylinder by integrating
the square of the distance from each element of mass to the rotation axis. Let’s
consider a thin ring of radius rand thickness dr at a distance xfrom the center
of the cylinder.
Step 2: The mass dm of the ring can be expressed in terms of dr and dx:
dm =M
πR2H·2πr ·dr =2M r
RH ·dr
28
I=M
LZL
0
x2dx
I=M
Lx3
3L
0
I=M
LL3
3−0
I=M
L·L3
3
I=ML2
3
Therefore, the moment of inertia of the thin rod about the axis perpendicular
to the rod and passing through one end is ML2
3.
Question 2
Question
Find the moment of inertia of a solid sphere of radius Rand mass Mabout a
diameter (axis passing through the center and perpendicular to the surface).
Solution
To find the moment of inertia of the solid sphere about a diameter, we can
integrate the moment of inertia of thin circular disks along the diameter axis.
Step 1: Consider a thin disk within the sphere at a distance xfrom the
center. The mass of this thin disk is dm.
Step 2: The moment of inertia of this thin disk about the axis passing
through the center is dI =1
2r2dm, where ris the distance of the thin disk from
the axis.
Step 3: Since the solid sphere has uniform mass distribution, the mass dm
of the thin disk is M
VdV , where Vis the volume of the sphere and dV is the
volume of the thin disk.
Step 4: Using the equation for the volume of a sphere V=4
3πR3and the
volume of a disk dV =πx2dx, we can express dm in terms of dx.
Step 5: Substitute dm in the expression for dI and integrate from x= 0 to
x=Rto find the total moment of inertia Iof the solid sphere.
Step 6: After integrating, we find that the moment of inertia of the solid
sphere about a diameter is I=2
5MR2.
2
Question 3
Question
Find the moment of inertia of a solid cone with radius Rand height Habout
its central axis. Assume the density of the cone is uniform.
Solution
Given: Radius of the cone, RHeight of the cone, H
The moment of inertia Iof a solid cone about its central axis can be calcu-
lated using the formula:
I=3
10MR2+3
20MH2
Where Mis the mass of the cone. We can find Musing the volume formula for
a cone:
V=1
3πR2H
Step 1: Find the mass Mof the cone Using the density formula ρ=M
V,
we can rearrange to find M:
M=ρ·V=ρ·1
3πR2H
Step 2: Substitute Minto the formula for moment of inertia Now
that we have M, we can substitute it into the formula for moment of inertia:
I=3
10 ρ·1
3πR2HR2+3
20 ρ·1
3πR2HH2
Step 3: Simplify the equation
I=1
10ρπR4H+1
20ρπR2H3
Hence, the moment of inertia of the solid cone about its central axis is
1
10 ρπR4H+1
20 ρπR2H3.
Question 4
Question
Calculate the moment of inertia of a uniform solid cylinder with mass M, radius
R, and height Habout an axis passing through its center and perpendicular to
its circular base.
3
Solution
Let’s break down the solution into steps:
Step 1: Determine the moment of inertia of a thin disk with respect to the
axis passing through its center and perpendicular to its circular base.
The moment of inertia of a thin disk of mass dm and radius rabout its center
and perpendicular to its circular base is given by the formula dI =1
2r2dm. To
find the total moment of inertia of the entire disk, we need to integrate this
expression over the entire disk.
Idisk =ZdI =ZR
0
1
2r2M
πR2dr =1
2M
πR2ZR
0
r2dr
Step 2: Evaluate the integral to find the moment of inertia of the thin disk.
Idisk =1
2M
πR2r3
3R
0
=1
6M
πR2R3=1
6MR2
Step 3: Use the moment of inertia of the thin disk to find the moment of
inertia of the solid cylinder.
Since the solid cylinder can be considered as a collection of such disks stacked
along the height H, the moment of inertia of the entire cylinder can be found
by summing the moments of inertia of all the disks.
Icylinder =ZdIdisk =ZH
0
1
6MR2dr
R=1
6MR2ZH
0
dr =1
6MR2H
Therefore, the moment of inertia of the uniform solid cylinder about the
given axis is 1
6MR2H.
Question 5
Question
Find the moment of inertia of a uniform solid cylinder of mass Mand radius
Rabout an axis passing through its center and perpendicular to the cylinder’s
axis. The density of the cylinder is ρ.
Solution
Let’s consider a small element of the cylinder located at a distance rfrom the
axis of rotation. The mass of this element can be given by δm =ρ·2πr ·δr,
where δr is the thickness of the element.
Step 1: Express moment of inertia for the whole cylinder.
Let’s integrate over the whole cylinder to find the moment of inertia.
dI=δm ·r2=ρ·2πr ·dr·r2
4
I=ZR
0
2πρr3dr=2
4πρR4
So, the moment of inertia of the cylinder about the given axis is 1
2MR2.
Question 6
Question
Find the moment of inertia of a uniform circular disk of radius Rabout an axis
perpendicular to the disk and passing through its center.
Solution
Step 1: Consider an elemental ring of radius rand width dr on the disk. Step
2: The mass of this elemental ring can be given by dm =ρ·dA, where ρis the
density of the disk material and dA = 2πr ·dr is the area of the elemental ring.
Step 3: The moment of inertia dI of this elemental ring about the axis is given
by dI =r2·dm. Step 4: Substitute the expressions for dm and dA into dI to
get dI =ρ·2πr3·dr. Step 5: Integrate dI from 0 to Rto find the total moment
of inertia Iof the disk:
I=ZR
0
ρ·2πr3·dr
Step 6: Simplify the expression and solve the integral to find I.
Question 7
Question
A thin uniform rod of length Land mass Mis rotated about an axis passing
through one end perpendicular to the rod. Find the moment of inertia of the
rod about this axis.
Solution
Let’s consider the rod to have mass per unit length λ=M
L.
Step 1: Divide the rod into small elements ∆m=λ∆x.
Step 2: The moment of inertia of each small element ∆Iabout the axis is
given by ∆I= (∆m)(x2).
Step 3: Sum up the moments of inertia of all small elements by integrating
from 0 to L:
I=ZL
0
λx2dx
5
Step 4: Solve the integral:
I=λZL
0
x2dx =λx3
3L
0
=λL3
3
Step 5: Substitute back the expression for λto find the moment of inertia
I:
I=ML2
3
Therefore, the moment of inertia of the rod about the given axis is ML2
3.
Question 8
Question
A uniform thin rod of length Land mass Mis rotating about an axis perpen-
dicular to the rod and passing through one end of the rod. Find the moment of
inertia of the rod about this axis.
Solution
Step 1: The moment of inertia of a point mass about an axis is given by I=mr2,
where mis the mass of the point mass and ris the perpendicular distance from
the point mass to the axis of rotation.
Let’s consider a small element δm of the rod at a distance xfrom the axis
of rotation. The mass of this element can be expressed as δm =M
Lδx, where
δx is a small length element of the rod.
Step 2: The perpendicular distance of this element from the axis of rotation
is x. Therefore, the moment of inertia of this element about the axis is dI =
M
Lx2δx.
Step 3: To find the total moment of inertia of the rod, we need to integrate
this expression over the entire length of the rod:
I=ZdI =ZL
0
M
Lx2dx
Step 4: Solving the integral gives:
I=M
LZL
0
x2dx
I=M
L1
3x3L
0
I=M
L1
3L3−1
3(0)3
6
I=M
L·1
3L3
Step 5: Simplifying the expression further:
I=1
3ML2
Therefore, the moment of inertia of the rod about the given axis is 1
3ML2.
Question 9
Question
A thin-walled hollow cylinder has an outer radius of 6 cm, an inner radius of
4 cm, and a height of 10 cm. Calculate the moment of inertia of the cylinder
about its central axis.
Solution
To find the moment of inertia of the hollow cylinder, we can use the formula for
the moment of inertia of a thin-walled cylinder:
I=1
2×M×(r2
outer +r2
inner)
Step 1: Calculate the mass of the cylinder The mass of the cylinder
can be found using the formula:
M= Volume ×Density
The volume of the hollow cylinder is:
Volume = π×h×(r2
outer −r2
inner)
where his the height of the cylinder.
Substitute the given values into the formula:
Volume = π×10 ×((6)2−(4)2) = 80πcm3
Given that the density of the material is ρ= 1 g/cm3, we can find the mass:
M= 80π×1 = 80πg
Step 2: Calculate the moment of inertia Now, plug the values of M,
router and rinner into the formula:
I=1
2×80π×((6)2+ (4)2) = 1
2×80π×52 = 2080πcm4
Therefore, the moment of inertia of the hollow cylinder about its central axis
is 2080πcm4.
7
Question 10
Question
A thin rod of length Land mass Mis rotating about an axis passing through
one end perpendicular to the rod with an angular velocity ω. Calculate the
moment of inertia of the rod about the axis of rotation.
Solution
Step 1: The moment of inertia of a thin rod rotating about an axis passing
through one end perpendicular to the rod is given by the formula I=1
3ML2.
Step 2: This formula can be derived by integrating the expression for the
moment of inertia of a small mass element dm along the length of the rod. The
moment of inertia of a small mass element dm at a distance rfrom the axis of
rotation is dI =r2dm.
Step 3: By expressing dm in terms of the linear mass density λ=M
L, we
have dm =λdx where dx is a small length element along the rod.
Step 4: Substituting dm into the expression for dI gives dI =r2λdx.
Step 5: Integrating dI along the length of the rod from 0 to Lgives the total
moment of inertia I=RL
0r2λdx.
Step 6: Evaluating the integral gives I=λRL
0r2dx.
Step 7: Since rranges from 0 to L, we have I=λRL
0r2dx =λhr3
3iL
0.
Step 8: Simplifying the expression, we get I=1
3λL3.
Step 9: Substituting λ=M
Linto the equation gives I=1
3ML2. Therefore,
the moment of inertia of the thin rod about the axis of rotation is 1
3ML2.
Question 11
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 mass elements
of the rod along its length.
Step 1: Consider a small mass element dm of length dx at a distance xfrom
the end of the rod. The mass of this element can be expressed as dm =M
Ldx.
Step 2: The moment of inertia dI of this small mass element about the axis
perpendicular to the rod and passing through one end is given by dI = (x)2dm.
8
Step 3: Substitute dm =M
Ldx into the expression for dI and integrate from
x= 0 to x=Lto find the total moment of inertia I.
I=ZL
0
x2M
Ldx
=M
LZL
0
x2dx
=M
L1
3x3L
0
=M
L1
3L3−0
=1
3ML2
Step 4: 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 12
Question
A thin uniform rod of length Land mass Mis rotated about an axis passing
through one end perpendicular to the rod. Find the moment of inertia of the
rod about this axis.
Solution
Let’s consider the rod to be along the x-axis with the origin at the end where
the axis of rotation passes. The density of the rod is M
L.
Step 1: Divide the rod into small elements δm along its length, each at a
distance xfrom the origin.
Mass of each element: dm =M
Ldx
The moment of inertia of one small element about the axis of rotation:
dI = (x)2dm = (x)2M
Ldx
Step 2: To find the total moment of inertia of the rod, we integrate over
the entire length:
9
I=ZdI
=ZL
0
x2M
Ldx
=M
LZL
0
x2dx
=M
Lx3
3L
0
=M
LL3
3−03
=M
L·L3
3
=1
3ML2
Therefore, the moment of inertia of the rod about the given axis is 1
3ML2.
Question 13
Question
A thin uniform rod of mass mand length Lis rotating about an axis passing
through one end and perpendicular to the rod at a constant angular velocity ω.
Calculate 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 one
end perpendicular to the rod can be calculated using the formula:
I=Zr2dm
where ris the distance of an element of mass dm from the axis of rotation.
Step 2: Consider a small element dx of the rod at a distance xfrom the end
where the axis of rotation passes. The mass dm of this element is given by:
dm =m
Ldx
Step 3: The distance rof the element dm from the axis of rotation is simply
x. Therefore, r=x.
Step 4: Substituting dm and rinto the moment of inertia formula gives:
I=ZL
0
x2m
Ldx
10
Step 5: Simplifying the integral, we get:
I=m
LZL
0
x2dx
Step 6: Solving the integral, we have:
I=m
Lx3
3L
0
I=m
LL3
3−0
I=mL2
3
Step 7: Therefore, the moment of inertia of the rod about the axis passing
through one end and perpendicular to the rod is mL2
3.
Question 14
Question
Find 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 divide the rod into small elements of mass dm at a distance xfrom the
end through which the axis passes. The mass of each element is dm =M
Ldx.
Step 1: Calculate the moment of inertia dI of each small element about the
given axis. The moment of inertia of a small element of mass dm at a distance
xfrom the axis is given by dI =dm ·x2. So, dI =M
Ldx·x2=M
Lx2dx.
Step 2: Determine the total moment of inertia Iof the entire rod by inte-
grating dI over the length of the rod.
I=ZL
0
M
Lx2dx
Step 3: Solve the integral to find I.
I=M
LZL
0
x2dx
I=M
Lx3
3L
0
I=M
LL3
3−0
11
I=M
3L2
Therefore, the moment of inertia of the rod about the given axis is M
3L2.
Question 15
Question
A thin uniform rod of length Land mass Mis rotating about an axis perpendic-
ular to the rod and passing through one end. Calculate the moment of inertia
of the rod about this axis.
Solution
Step 1: Divide the rod into small elements ∆m=M
L∆x. Step 2: The distance
of each element from the axis of rotation is x. The moment of inertia of each
element about the axis is dI = ∆m·x2. Step 3: The total moment of inertia is
obtained by summing up all the elements:
I=ZdI
Step 4: Substitute ∆m=M
L∆xand dI =M
L∆x·x2into the integral. Step 5:
Integrate over the length of the rod from 0 to L.
I=ZL
0
M
L·x2dx
Step 6: Simplify the integral and solve for I.
I=M
L·x3
3L
0
=M
L·L3
3=1
3ML2
Step 7: Therefore, the moment of inertia of the rod about the given axis is
1
3ML2.
Question 16
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.
12
Solution
Step 1: Divide the rod into small elements δm of length δx.
Step 2: Express δm in terms of linear density λand δx:δm =λδx.
Step 3: Find the mass dm of a small element of length dx at a distance x
from the axis of rotation: dm =λdx.
Step 4: The moment of inertia dI of this small element about the axis of
rotation is given by dI =r2dm, where ris the distance of the element from the
axis.
Step 5: Express rin terms of x:r=L−x.
Step 6: Rewrite dI in terms of x:dI = (L−x)2λdx.
Step 7: Integrate dI from x= 0 to x=Lto find the total moment of inertia
I:
I=ZL
0
(L−x)2λdx
Step 8: Solve the integral:
I=ZL
0
(L2−2Lx +x2)λdx
I=λL3
3−2LL2
2+L3
3
I=λL3
3−L3
3=1
3λL3
Step 9: Substitute λ=M
Lto find the moment of inertia I:
I=1
3M
LL3=1
3ML2
Therefore, the moment of inertia of the thin uniform rod about the given
axis is 1
3ML2.
Question 17
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 its midpoint.
Solution
To find the moment of inertia of the rod, we will consider it to be made up of
infinitesimally small elements, each with mass dm and located at a distance x
from the axis of rotation. Then, we will integrate over all these elements to find
the total moment of inertia of the rod.
13
Step 1: Let’s consider an element of length dx at a distance xfrom the
center of the rod. The mass of this element dm can be expressed in terms of dx
and Msince the rod is uniform. The density λof the rod is given by λ=M
L.
So, dm =λ·dx =M
Ldx.
Step 2: The moment of inertia dI of this element about the axis can be
calculated using the formula dI =dm ·x2. Substituting for dm,dI =M
Ldx·
x2=Mx2
Ldx.
Step 3: Now, we can find the total moment of inertia Iby integrating dI
over the entire length of the rod. I=RdI =RL/2
−L/2
Mx2
Ldx.
Step 4: Solving the integral, I=M
LRL/2
−L/2x2dx =M
Lhx3
3iL/2
−L/2.I=
M
Lh(L
2)3
3−(−L
2)3
3i.I=M
LhL3
24 +L3
24 .I = M L2
12 .
Therefore, the moment of inertia of the thin uniform rod about an axis
perpendicular to the rod and passing through its midpoint is ML2
12 .
Question 18
Question
Find the moment of inertia of a thin rod of mass Mand length Labout an axis
perpendicular to the rod and passing through one of its ends.
Solution
To find the moment of inertia of the rod, we will consider it as made up of
infinitesimally small elements.
Step 1: Consider an element of length dx at a distance xfrom the end of
the rod, where 0 ≤x≤L.
Step 2: The mass dm of this element can be given as M
Ldx.
Step 3: The moment of inertia dI of this element about the axis perpen-
dicular to the rod and passing through one end is given by dI =dm ·x2.
Step 4: Substituting the expression for mass dm into the above equation,
we get dI =M
L·x2dx.
Step 5: The total moment of inertia Iof the rod can be found by integrating
dI from 0 to L:
I=ZL
0
dI =ZL
0
M
L·x2dx
Step 6: Solving the integral, we get:
I=M
L·x3
3L
0
=M
L·L3
3−0=ML2
3
Therefore, the moment of inertia of the thin rod of mass Mand length L
about an axis perpendicular to the rod and passing through one of its ends is
ML2
3.
14
Question 19
Question
A thin rod of mass mand length Lis rotated about an axis perpendicular to
the rod and passing through one end of the rod. Find the moment of inertia of
the rod about this axis.
Solution
Step 1: The moment of inertia of a point mass mat distance rfrom an axis of
rotation is I=mr2.
Step 2: The rod can be thought of as a collection of point masses, each
at a different distance from the axis of rotation. We can divide the rod into
infinitesimally small masses dm and write the moment of inertia contribution
from each as dI =dm ·r2.
Step 3: The linear mass density λof the rod is given by λ=m
L. Therefore,
the infinitesimal mass dm is dm =λ·dx.
Step 4: The distance of an infinitesimal mass dm from the axis of rotation
is r=x.
Step 5: Expressing min terms of λgives m=λ·L.
Step 6: Substituting the expressions for dm and minto dI =dm ·r2gives
dI = (λ·dx)·(x2) = λx2dx
.
Step 7: Integrating over the length of the rod gives the total moment of
inertia:
I=ZL
0
λx2dx =ZL
0
m
Lx2dx
.
Step 8: Solving the integral gives
I=m
Lx3
3L
0
=m
L·L3
3=mL2
3
.
Therefore, the moment of inertia of the rod about the given axis is mL2
3.
Question 20
Question
A hollow sphere of radius Rand mass Mis rotating about an axis passing
through its center with an angular velocity ω. Find the moment of inertia of
the sphere about an axis passing through a point on its surface.
15
Solution
Step 1: The moment of inertia of a hollow sphere of radius Rand mass Mabout
an axis passing through its center is given by the formula:
I=2
3MR2
Step 2: Let’s find the moment of inertia of the sphere about an axis passing
through a point on its surface. According to the parallel axis theorem, the
moment of inertia about this axis is equal to the sum of the moment of inertia
about an axis passing through the center and the product of its mass and the
square of the distance between the two axes.
Step 3: Let the distance between the two axes be l. Using the parallel axis
theorem, the moment of inertia about the new axis is:
I′=I+Ml2
Step 4: Substituting the given values, we have:
I′=2
3MR2+Ml2
Step 5: To find l, we can use the Pythagorean theorem with the two sides
being Rand land the hypotenuse being R. Therefore:
R2=R2+l2
l=√R2=R
Step 6: Substituting l=Rinto the equation for I′, we get:
I′=2
3MR2+MR2
I′=5
3MR2
Therefore, the moment of inertia of the sphere about an axis passing through
a point on its surface is 5
3MR2.
Question 21
Question
A thin uniform rod of mass mand length Lis rotated about an axis perpen-
dicular to the rod and passing through its midpoint. Determine the moment of
inertia of the rod about this axis.
16
Solution
Step 1: The moment of inertia of a rod about an axis perpendicular to the rod
and passing through one end is 1
3mL2. We can use the Parallel Axis Theorem to
find the moment of inertia of the rod about an axis passing through its midpoint.
Step 2: The distance from the midpoint to the end of the rod is L
2. To use
the Parallel Axis Theorem, we will need to add the moment of inertia about the
midpoint to the moment of inertia about an axis through the endpoint.
Step 3: The moment of inertia about the endpoint is 1
3mL2, and the distance
between the two axes is L
2. Applying the Parallel Axis Theorem: Imidpoint =
Iend +md2.
Step 4: Substituting the values, Imidpoint =1
3mL2+mL
22.
Step 5: Simplifying further, Imidpoint =1
3mL2+1
4mL2.
Step 6: Combining the terms, Imidpoint =7
12 mL2.
Therefore, the moment of inertia of the rod about an axis perpendicular to
the rod and passing through its midpoint is 7
12 mL2.
Question 22
Question
A thin uniform 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
Step 1: Choose an element of length dx at a distance xfrom the axis of rotation.
Step 2: The mass of this element dm is given by
dm =M
Ldx
Step 3: The moment of inertia of this element about the axis of rotation is
dI = (dm)x2
Step 4: Substitute dm from Step 2 into the above equation to get
dI =M
Ldxx2
dI =M
Lx2dx
Step 5: Integrate dI from x= 0 to x=Lto find the total moment of inertia I.
I=ZL
0
M
Lx2dx
17
Step 6: Solve the integral
I=M
Lx3
3L
0
I=M
LL3
3−03
3
I=M
L·L3
3
I=ML2
3
Therefore, the moment of inertia of the rod about the given axis is ML2
3.
Question 23
Question
A thin uniform rectangular plate of width wand height his rotated about an
axis through its center perpendicular to its plane. Find the moment of inertia
of the plate with respect to this axis.
Solution
Step 1: The moment of inertia of the rectangular plate about an axis through
its center perpendicular to its plane can be calculated using the formula for the
moment of inertia of a rectangular plate rotated about its edge (I=1
3Mh2,
where Mis the mass of the plate).
Step 2: To find the mass Mof the rectangular plate, we need to find its area
first. The area Aof the plate is given by A=w·h.
Step 3: The mass Mcan be calculated using the density ρof the plate and
the area A. The mass Mis given by M=ρ·A.
Step 4: Substituting the expressions for Aand Minto the formula for mo-
ment of inertia, we get
I=1
3(Mh2) = 1
3(ρ·A·h2)
Step 5: Simplifying further, we get
I=1
3(ρ·w·h·h2) = 1
3ρwh3
Therefore, the moment of inertia of the rectangular plate about an axis
through its center perpendicular to its plane is 1
3ρwh3.
18
Question 24
Question
A thin uniform rod of mass mand length Lis rotating about an axis that is
perpendicular to the rod and passing through one end. Calculate the moment
of inertia of the rod about this axis.
Solution
Step 1: Consider the moment of inertia of an infinitesimally small mass element
dm at a distance xfrom the axis of rotation. The moment of inertia of this
mass element is given by dI =x2dm.
Step 2: Express dm in terms of dx. The linear mass density λof the rod is
λ=m
L. Therefore, dm =λ dx.
Step 3: Substituting dm =λ dx into dI =x2dm. We get dI =x2λ dx.
Step 4: Integrate both sides of the equation to find the total moment of
inertia I.RdI =Rx2λ dx
Step 5: Determine the limits of integration. For a rod of length L, the limits
of integration are from 0 to L.
Step 6: Integrate the right side of the equation. I=RL
0x2λ dx =RL
0x2m
Ldx
Step 7: Simplify the integral. I=m
LRL
0x2dx =m
Lhx3
3iL
0
Step 8: Evaluate the integral. I=m
LL3
3−03
3=m
L·L3
3
Step 9: Simplify the expression. I=mL2
3
Therefore, the moment of inertia of the rod about the given axis is mL2
3.
Question 25
Question
Determine the moment of inertia of a thin rod of mass Mand length Labout
an axis perpendicular to the rod passing through one end.
Solution
Let’s consider the thin rod as a 1D system.
Step 1: Determine the linear mass density of the rod. The linear mass
density λis defined as mass per unit length.
λ=M
L
Step 2: Express the moment of inertia in terms of an integral. The moment
of inertia Iof the rod about an axis perpendicular to the rod and passing through
19
one end can be expressed as an integral of r2dm, where ris the distance from
the axis and dm is a small mass element on the rod.
I=Zr2dm
Step 3: Express dm in terms of linear mass density. Since λ=M
L, we have
dm =λdx, where dx is a small length element along the rod.
Step 4: Express rin terms of x. The distance rfrom the axis to a small
mass element at a distance xfrom the end is given by r=x.
Step 5: Substitute dm and rinto the expression for I.
I=ZL
0
x2λdx
Step 6: Substitute λ=M
Linto the integral.
I=ZL
0
x2M
Ldx
Step 7: Integrate to find the moment of inertia I.
I=M
LZL
0
x2dx
I=M
Lx3
3L
0
I=M
LL3
3
I=ML2
3
Therefore, the moment of inertia of the thin rod about an axis perpendicular
to the rod passing through one end is ML2
3.
Question 26
Question
A thin rod of mass Mand length Lis rotating about an axis passing through
its center and perpendicular to its length. Determine the moment of inertia of
the rod about this axis.
20
Solution
Step 1: We know that the moment of inertia of an object rotating about an axis
is given by the formula I=Rr2dm, where ris the distance from the axis of
rotation, and dm is an infinitesimally small mass element of the object.
Step 2: Let’s consider an infinitesimally small mass element dm of the rod
at a distance xfrom the center. The mass of this element can be expressed as
dm =M
Ldx.
Step 3: The distance rof this mass element from the axis of rotation is given
by r=L
2−x.
Step 4: Substituting dm and rinto the formula I=Rr2dm, we get:
I=ZL/2
−L/2
(L
2−x)2M
Ldx
Step 5: Simplifying the integrand, we have:
I=M
LZL/2
−L/2
(L2
4−Lx +x2)dx
Step 6: Integrating term by term, we get:
I=M
L[L3
4x−L2
2x2+1
3x3]L/2
−L/2
Step 7: Evaluating the integral at the limits of integration, we obtain:
I=M
L[L3
4(L
2)−L2
2(L2
4) + 1
3(L3
8)]
Step 8: Simplifying further, we get:
I=ML2
12
Step 9: Therefore, the moment of inertia of the rod about the given axis is
ML2
12 .
Question 27
Question
A thin rod of mass Mand length Lis rotating around its one end with an
angular velocity ω. Determine the moment of inertia of the rod about its far
end.
21
Solution
1. The moment of inertia of the rod about its center can be calculated using
the formula for a rod rotating about an axis perpendicular to its length:
Icm =1
12ML2
2. To find the moment of inertia about the far end, we can use the parallel
axis theorem, which states:
I=Icm +Md2
where dis the distance between the two axes of rotation.
3. In this case, the distance dis equal to half the length of the rod:
d=L
2
4. Substituting the values into the parallel axis theorem:
I=1
12ML2+ML
22
I=1
12ML2+1
4ML2
I=1
12ML2+3
12ML2
I=1
12ML2+3
12ML2
I=4
12ML2=1
3ML2
5. Therefore, the moment of inertia of the rod about its far end is 1
3ML2.
Question 28
Question
A thin rod of length Land mass Mis rotating about an axis passing through
one end perpendicular to the rod with an angular velocity ω. Calculate the
moment of inertia of the rod about the axis of rotation.
Solution
Step 1: The moment of inertia for a point mass rotating about an axis is given
by I=mr2, where mis the mass of the point mass and ris the distance of the
point mass from the axis of rotation.
22
Step 2: To find the moment of inertia of the entire rod, we need to consider
infinitesimally small point masses that make up the rod. Let’s consider a small
elemental mass dm at a distance xfrom the pivot point.
Step 3: The mass of the elemental length dx is given by dm =M
Ldx.
Step 4: The moment of inertia of this elemental mass about the axis of
rotation is dI =dm ·x2.
Step 5: Substituting the expression for dm into the moment of inertia for-
mula, we get dI =M
Ldxx2.
Step 6: The total moment of inertia of the rod can be obtained by integrating
dI over the entire length of the rod: I=RdI =RM
Ldxx2from 0 to L.
Step 7: Solving the integral, we get I=RL
0
M
Lx2dx.
Step 8: Simplifying the expression, we find I=M
LRL
0x2dx.
Step 9: By evaluating the integral, we get I=M
Lhx3
3iL
0.
Step 10: Therefore, the moment of inertia of the rod about the axis of
rotation is I=M
LL3
3=1
3ML2.
Question 29
Question
Determine the moment of inertia of a solid cylinder of mass Mand radius R
about an axis passing through its center perpendicular to its height.
Solution
Let’s consider a solid cylinder with mass Mand radius Ras shown below.
solid_cylinder.png
Step 1: To find the moment of inertia, we need to consider the cylinder as
a collection of infinitesimally thin disks. The moment of inertia of each disk is
I=1
2mr2, where mis the mass of the disk and ris the distance from the axis
of rotation.
Step 2: First, express the mass of each disk in terms of its radius rand the
cylinder’s total mass M. The mass dm of a disk with radius rand thickness dr
is given by dm =M
πR2·2πr ·dr.
Step 3: Now, substitute dm into the formula for the moment of inertia
I=1
2mr2, yielding dI =1
2M
πR2·2πr ·drr2.
Step 4: Integrate dI over the whole cylinder from 0 to Rto find the total
moment of inertia I:
I=ZR
0
1
2M
πR2·2πr ·drr2
23
Step 5: Simplify and solve the integral to find the moment of inertia I:
I=ZR
0
1
2·M
πR2·2πr3dr =M
2R2ZR
0
r3dr
Step 6: Evaluate the integral:
I=M
2R2r4
4R
0
=M
2R2·R4
4=1
8MR2
Therefore, the moment of inertia of the solid cylinder about an axis passing
through its center perpendicular to its height is 1
8MR2.
Question 30
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
To calculate the moment of inertia of the rod, we need to consider it as a
continuous mass distribution along its length. We can express the mass element
dm of the rod in terms of its linear mass density λ.
Step 1: Express the linear mass density λin terms of the total mass Mand
length Lof the rod.
λ=M
L
Step 2: Express dm (small mass element) in terms of λand dx (small length
element).
dm =λ dx =M
Ldx
Step 3: Express the moment of inertia Iof the rod by integrating the
contribution of each element of mass along the entire length of the rod.
I=Zr2dm
where ris the perpendicular distance of the mass element dm from the axis
of rotation.
Step 4: Express rin terms of x, the distance from the end of the rod.
r=L−x
24
Step 5: Substitute dm and rinto the equation for Iand perform the inte-
gration.
I=ZL
0
(L−x)2M
Ldx
=M
LZL
0
(L2−2Lx +x2)dx
=M
LL3x−L2x2+1
3x3
L
0
=M
L1
3L3
=1
3ML2
Step 6: Therefore, the moment of inertia of the thin rod of length Land
mass Mabout the axis perpendicular to the rod and passing through one end
is 1
3ML2.
Question 31
Question
A thin rod of mass mand length Lis rotating in a horizontal plane about one
end with an angular velocity ω. Determine the moment of inertia of the rod
about an axis perpendicular to the rod passing through its midpoint.
Solution
Step 1: To find the moment of inertia of the rod about an axis perpendicular
to the rod passing through its midpoint, we can use the parallel-axis theorem.
Let’s first find the moment of inertia of the rod about its center of mass.
Step 2: The moment of inertia of a rod rotating about an axis perpendicular
to the rod and passing through its center is given by the formula Icm =1
12 mL2.
Step 3: Now, using the parallel-axis theorem, the moment of inertia Iof
the rod about an axis perpendicular to the rod passing through its midpoint is
given by I=Icm +md2, where dis the distance between the midpoint and the
center of mass.
Step 4: The distance dbetween the midpoint and the center of mass of the
rod is d=L
2.
Step 5: Substituting Icm =1
12 mL2and d=L
2into the formula I=Icm +
md2, we get I=1
12 mL2+m(L
2)2.
Step 6: Simplifying the expression, we find I=1
12 mL2+1
4mL2=1
3mL2.
Therefore, the moment of inertia of the rod about an axis perpendicular to
the rod passing through its midpoint is 1
3mL2.
25
Question 32
Question
Find the moment of inertia of a uniform thin rod of length Land mass Mabout
an axis perpendicular to the rod and passing through one of its ends.
Solution
Step 1: The moment of inertia of an object about an axis can be calculated
using the formula: I=Rr2dm, where ris the distance from the axis of rotation
to the mass element dm. In this case, we will consider an elemental mass dm
at a distance xfrom the end of the rod.
Step 2: The mass dm of the rod can be expressed as dm =M
Ldx, where dx
is an elemental length of the rod.
Step 3: The distance rof the mass element dm from the axis of rotation is
r=x.
Step 4: Substituting dm and rinto the formula for moment of inertia, we
have: I=RL
0x2·M
Ldx.
Step 5: Solving the integral, we get: I=M
LRL
0x2dx.
Step 6: Integrating x2from 0 to Lgives: I=M
Lhx3
3iL
0.
Step 7: Simplifying further, we get: I=M
LL3
3−0.
Step 8: Finally, the moment of inertia of the rod about an axis perpendicular
to the rod and passing through one of its ends is: I=ML2
3.
Question 33
Question
Find the moment of inertia of a solid cylinder of radius Rand mass Mabout
an axis passing through its center perpendicular to the cylinder’s height.
Solution
Step 1: We can calculate the moment of inertia of the solid cylinder by inte-
grating over its volume. The moment of inertia Ican be expressed as:
I=Zr2dm
where ris the distance from the axis of rotation to the differential mass element
dm.
Step 2: To find dm, we need to express it in terms of the cylinder’s density.
The volume element dV of a cylinder in cylindrical coordinates is given by:
dV =Rdθdrdz
26
Step 3: Since the cylinder is of uniform density ρ, the mass located in a
differential volume dV is:
dm =ρdV =ρRdθdrdz
Step 4: Now we substitute dm into the equation for moment of inertia:
I=Zr2ρRdθdrdz
Step 5: To simplify the integral, we need to express rin terms of r,θ, and z.
For a point at coordinates (r, θ, z) in cylindrical coordinates, rcan be calculated
as:
r=pR2+z2
Step 6: Now we substitute rinto the equation for moment of inertia:
I=Z(pR2+z2)2ρRdθdrdz
Step 7: We can simplify the integral and limits since the cylinder is sym-
metric about the z-axis. Therefore, we integrate over zfrom −h/2 to h/2 and
over θfrom 0 to 2π.
Step 8: After performing the integrals, we find the moment of inertia of the
cylinder:
I=1
2πρR2h3
Thus, the moment of inertia of the solid cylinder about the given axis is
1
2πρR2h3.
Question 34
Question
Calculate the moment of inertia of a uniform thin rod of length Land mass M
about an axis perpendicular to the rod and passing through one of its ends.
Solution
Let’s denote the rod as a 1-dimensional object with mass uniformly distributed
along its length. We will use the formula for the moment of inertia of a contin-
uous mass distribution:
I=Zr2dm
Step 1: Determine the mass element dm in terms of dx,M, and L. The
linear density λof the rod is given by:
λ=M
L
27
The mass element dm can be expressed in terms of the linear density λand
a differential length dx along the rod:
dm =λ dx =M
Ldx
Step 2: Express rin terms of x. The distance rof the mass element dm
from the axis of rotation is the position of the mass element along the rod. Since
the axis of rotation is at one end of the rod, rcan be expressed in terms of x:
r=x
Step 3: Substitute dm and rinto the expression for I. The moment of
inertia Ican be calculated by integrating r2dm along the length of the rod:
I=Zr2dm =Zx2M
Ldx
Step 4: Integrate to find the moment of inertia.
I=M
LZL
0
x2dx =M
Lx3
3L
0
I=M
L·L3
3=ML2
3
Therefore, the moment of inertia of the rod about the axis passing through
one of its ends is ML2
3.
Question 35
Question
A solid cylinder with radius R, mass M, and height His rotated about an
axis passing through its center and perpendicular to its height. Determine the
moment of inertia of the cylinder about this axis.
Solution
Step 1: We can calculate the moment of inertia of the cylinder by integrating
the square of the distance from each element of mass to the rotation axis. Let’s
consider a thin ring of radius rand thickness dr at a distance xfrom the center
of the cylinder.
Step 2: The mass dm of the ring can be expressed in terms of dr and dx:
dm =M
πR2H·2πr ·dr =2M r
RH ·dr
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Step 3: The moment of inertia dI of this ring about the given axis is given
by dI =r2·dm.
dI =r2·2Mr
RH ·dr=2M
H·r3·dr
Step 4: Next, we integrate dI from r= 0 to r=Rto find the total moment
of inertia Iof the cylinder.
I=ZR
0
2M
H·r3dr
Step 5: Evaluating the integral, we get:
I=1
2·2M
H·r4R
0
=1
2·2MR4
H=MR2
2
Step 6: Therefore, the moment of inertia of the solid cylinder about the
given axis is MR2
2.
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