PHYS 101 - ELEMENTS OF PHYSICS
- Moment of Inertia
Question Bank - Set 3
Liberty University
Question 1
Question
A thin uniform rod of length Land mass Mis rotating about an axis passing
through one end and perpendicular to the rod. Calculate the moment of inertia
of the rod about this axis.
Solution
Step 1: To find the moment of inertia of the thin uniform rod, we will consider it
to be composed of infinitesimally small particles along its length. The moment
of inertia of each particle dm about the rotation axis is r2dm, where ris the
distance of the particle from the axis of rotation.
Step 2: We can express rin terms of the position xalong the rod. The
distance rfrom the axis of rotation to a particle at position xis L−x, where
0≤x≤L.
Step 3: We can express dm in terms of Mand dx (infinitesimal length
along the rod). Since the rod is uniform, the mass per unit length is M/L, so
dm =M
Ldx.
Step 4: Now, we can express the total moment of inertia Iof the entire rod
by integrating the contribution from each infinitesimal particle along the length
of the rod:
I=ZL
0
(L−x)2·M
Ldx
Step 5: Simplifying the expression and solving the integral gives:
I=ZL
0
(L2−2Lx +x2)·M
Ldx
=ML2x
L−2Lx2
2L+x3
3L
L
0
=ML−L+L
3
=1
3ML2
Therefore, the moment of inertia of the thin uniform rod about the given
axis of rotation is 1
3ML2.
Question 2
Question
Find the moment of inertia of a thin uniform rod of length Land mass Mabout
an axis passing through one end perpendicular to the rod.
Solution
To find the moment of inertia of the thin uniform rod about an axis passing
through one end perpendicular to the rod, we can consider the rod as a collection
of infinitesimally small masses along its length. We can then sum up the moment
of inertia of each of these masses to find the total moment of inertia of the rod.
Step 1: Consider an infinitesimal mass dm at a distance xfrom the end
where the axis passes.
The mass dm can be expressed as:
dm =M
Ldx
Step 2: Find the moment of inertia dI of this infinitesimal mass dm.
The moment of inertia of an infinitesimal mass dm about the given axis is
given by:
dI =dm ·x2
Substitute the expression for dm:
dI =M
Ldx ·x2
Step 3: Integrate to find the total moment of inertia Iof the rod.
2
Integrating dI over the entire length of the rod gives the total moment of
inertia I:
I=ZdI =ZM
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
passing through one end perpendicular to the rod is M L2
3.
Question 3
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 of its ends.
Solution
Step 1: Choose an element of mass δm at a distance xfrom the axis passing
through one end of the rod.
Step 2: The mass δm of the element is given by δm =M
Ldx.
Step 3: The moment of inertia of this element about the axis is dI =δm·x2=
M
L·x2dx.
Step 4: To find the total moment of inertia, we need to integrate the expres-
sion for dI over the length of the rod.
I=ZL
0
M
Lx2dx
Step 5: Solving the integral gives:
I=M
Lx3
3L
0
=M
LL3
3−0
Step 6: Therefore, the moment of inertia of the rod about the axis passing
through one of its ends is:
I=1
3ML2
3
Question 4
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 end.
Solution
To find the moment of inertia of the rod, we will integrate the element of mass
over the entire length of the rod.
Step 1: Consider a small element of length dx at distance xfrom the end
of the rod. The mass of this element is dm =m
Ldx.
Step 2: The moment of inertia of this element about the axis is dI =
dm ·x2=m
Ldx·x2.
Step 3: Integrate dI over the entire length of the rod from 0 to L.
I=ZL
0
dI =ZL
0m
Ldx·x2
Step 4: Solve the integral.
I=m
LZL
0
x2dx =m
Lx3
3L
0
=m
L·L3
3=mL2
3
Step 5: Therefore, the moment of inertia of the thin uniform rod about the
given axis is mL2
3.
Question 5
Question
A thin rod of length Land mass Mis pivoted at one end. Find the moment of
inertia of the rod about an axis perpendicular to the rod and passing through
the free end.
Solution
To find the moment of inertia of the thin rod about an axis perpendicular to
the rod and passing through the free end, we can use the parallel axis theorem.
Step 1: Find the moment of inertia of the rod about its center
of mass The moment of inertia of the rod about its center of mass can be
calculated using the formula for a thin rod rotated about an axis perpendicular
to the rod and passing through its center:
ICM =1
12ML2
4
Step 2: Apply the parallel axis theorem The parallel axis theorem
states that the moment of inertia about any axis parallel to and at a distance d
from the axis passing through the center of mass is given by:
I=ICM +M d2
Step 3: Calculate the moment of inertia about the free end In this
case, the distance dis equal to L/2 (half the length of the rod). Substituting
into the parallel axis theorem formula:
I=1
12ML2+ML
22
I=1
12ML2+1
4ML2
I=1
3ML2
Therefore, the moment of inertia of the thin rod about an axis perpendicular
to the rod and passing through the free end is 1
3ML2.
Question 6
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 end.
Solution
Step 1: Divide the rod into infinitesimally small elements of length dx at a
distance xfrom the end where the axis passes through.
Step 2: The mass of each element dm is given by dm =M
Ldx.
Step 3: The moment of inertia of an elemental mass dm about the axis is
dI =dm ·x2.
Step 4: Substitute dm and xto obtain dI =M
Lx2dx.
Step 5: The total moment of inertia of the rod is obtained by integrating
over the entire length:
I=ZdI =ZL
0
M
Lx2dx
Step 6: Solve the integral:
I=M
Lx3
3L
0
Step 7: Substitute the limits of integration and simplify:
I=M
LL3
3−03
3
5
Step 8: Further simplify to obtain the moment of inertia:
I=M
L·L3
3=M
3L2
Therefore, the moment of inertia of the thin uniform rod of mass Mand
length Labout an axis perpendicular to the rod and passing through one end
is M
3L2.
Question 7
Question
Find the moment of inertia of the region bounded by the curve y=x2and the
y-axis, rotated about the line y=−2.
Solution
To find the moment of inertia, we will first need to determine the area of the
region of interest and its centroid.
Step 1: Find the area of the region
1. We need to find the points of intersection between the curve y=x2and
the y-axis.
2. Setting x= 0 in y=x2, we find that the parabola intersects the y-axis at
the point (0,0).
3. The area of the region can be expressed as an integral:
A=Z1
0
x2dx
4. Solving the integral, we have:
A=1
3x3
1
0
=1
3
Step 2: Find the x-coordinate of the centroid
1. The x-coordinate of the centroid is given by:
¯x=1
AZ1
0
x·x2dx
2. Simplifying the expression, we get:
¯x=R1
0x3dx
1
3
=1
4
6
Step 3: Find the moment of inertia
1. The moment of inertia about the line y=−2 is given by:
I=Z1
0
(x2)·(x−¯x)2dx
2. Substituting the value of ¯x, we get:
I=Z1
0
x2(x−1
4)2dx
3. Solving the integral will give us the moment of inertia.
Question 8
Question
A thin uniform rod of length Land mass Mis rotating about an axis passing
through one end and perpendicular to the rod. Find the moment of inertia of
the rod with respect to this axis.
Solution
Step 1: Consider an elemental mass δm at a distance xfrom the axis of rotation.
Step 2: The moment of inertia of this elemental mass about the axis of
rotation is δI = (δm)x2.
Step 3: We need to express δm in terms of xin order to integrate. Since the
rod is uniform, the linear mass density λ=M
L.
Step 4: The elemental mass δm =λdx.
Step 5: Substituting δm =λdx into δI = (δm)x2gives δI = (λdx)x2=
λx2dx.
Step 6: The total moment of inertia Iof the rod about the given axis is
obtained by integrating δI from 0 to L.
Step 7: Therefore, I=RL
0λx2dx.
Step 8: Evaluating the integral, we get I=hλx3
3iL
0.
Step 9: Plugging in the values, we get I=λL3
3.
Step 10: Finally, substituting λ=M
L, we find I=ML2
3.
Step 11: So, the moment of inertia of the rod with respect to the given axis
is ML2
3.
7
Question 9
Question
A thin uniform rod of mass Mand length Lis rotating about an axis passing
through one end and perpendicular to the rod. What is the moment of inertia
of the rod about this axis?
Solution
To find the moment of inertia of the rod about an axis passing through one
end and perpendicular to the rod, we can treat the rod as a collection of in-
finitesimally small particles, each with mass dm, at a distance rfrom the axis
of rotation.
Step 1: Let’s express dm in terms of the linear density λof the rod and
the length element dx. The mass of an infinitesimal length dx of the rod is
dm =λ dx.
Step 2: Next, we express rin terms of x, where xis the distance from
the end of the rod to the mass element dm. Since the rod is rotating along its
length, r=L−x.
Step 3: Now, we can express I(moment of inertia) as the sum of the
moments of inertia of all the mass elements dm along the length of the rod.
I=Zr2dm
Step 4: Substitute for dm and r:
I=ZL
0
(L−x)2λ dx
Step 5: Expand and solve the integral:
I=λZL
0
(L2−2Lx +x2)dx
=λL3
3−L2x
2+x3
3
L
0
=λL3
3−L3
2+L3
3
=1
3λL3
Step 6: Finally, substitute the linear density λ=M
Linto the expression for
I:
I=1
3M
LL3=1
3ML2
Therefore, the moment of inertia of the rod about an axis passing through
one end and perpendicular to the rod is 1
3ML2.
8
Question 10
Question
A solid cylinder of mass Mand radius Ris rotated about an axis passing through
its center and perpendicular to its length. Find the moment of inertia of the
cylinder with respect to this axis.
Solution
Let’s denote the moment of inertia of the cylinder with respect to the given axis
as I.
Step 1: Determine the moment of inertia of a small mass element.
Consider a small mass element of the cylinder at a distance rfrom the axis
of rotation. The moment of inertia of this mass element dm with respect to the
axis is r2dm.
Step 2: Express dm in terms of rand dr.
The mass of the small element dm can be expressed in terms of the linear
mass density λand the length dr as dm =λdr. Since the cylinder has a uniform
density, we have λ=M
πR2.
Step 3: Express the moment of inertia Iin terms of rand dr.
Integrating the moment of inertia of the small mass elements from r= 0 to
r=R, we get:
I=ZR
0
r2dm
Substitute dm =λdr into the expression:
I=ZR
0
r2λdr
Step 4: Integrate to find the moment of inertia I.
Now, substitute λ=M
πR2into the integral:
I=ZR
0
r2M
πR2dr
I=M
πR2ZR
0
r2dr
I=M
πR2r3
3R
0
I=M
πR2R3
3−0
I=MR2
3
Therefore, the moment of inertia of the cylinder with respect to the given
axis is M R2
3.
9
Question 11
Question
Find the moment of inertia of a uniform thin rod of mass Mand length Labout
an axis perpendicular to the rod and passing through its midpoint.
Solution
Step 1: Determine the mass per unit length λof the rod.
λ=M
L
Step 2: Calculate the moment of inertia of an infinitesimal element of the
rod at a distance xfrom the midpoint.
dI =λ dx ·x2
Step 3: Integrate the expression for dI from −L/2 to L/2 to find the total
moment of inertia I.
I=ZL/2
−L/2
λ·x2dx
I=ZL/2
−L/2
M
L·x2dx
I=M
LZL/2
−L/2
x2dx
I=M
Lx3
3L/2
−L/2
I=M
L(L/2)3
3−(−L/2)3
3
I=M
LL3
24 +L3
24
I=M
L·L3
12
I=1
12ML2
Therefore, the moment of inertia of the uniform thin rod of mass Mand
length Labout an axis perpendicular to the rod and passing through its mid-
point is 1
12 ML2.
10
Question 12
Question
A thin hoop of radius Rand mass Mis rotating about its central axis with
an angular velocity ω. Find the moment of inertia of the hoop about an axis
passing through its edge and perpendicular to its plane.
Solution
Step 1: The moment of inertia of the hoop rotating about its central axis is
Ic=MR2.
Step 2: Consider the parallel axis theorem, which states that the moment
of inertia about an axis parallel to and a distance daway from the center of
rotation is given by I=Ic+Md2. In this case, the distance from the central
axis to the edge is R, so the moment of inertia about an axis passing through
its edge and perpendicular to its plane is given by
I=Ic+MR2=MR2+MR2= 2M R2.
Question 13
Question
A thin rod of length Land mass Mis rotating about an axis perpendicular to
its length passing through one of its ends. Find the moment of inertia of the
rod about this axis.
Solution
Step 1: Consider a small element of the rod at a distance xfrom the axis of
rotation.
Step 2: The mass of this element can be considered as M
Ldx.
Step 3: The moment of inertia of this element about the axis of rotation is
dI =M
Lx2dx (using I=Rr2dm).
Step 4: To find the total moment of inertia of the rod, we need to integrate
dI from 0 to L.
Step 5: So, I=RL
0
M
Lx2dx.
Step 6: Solving the integral, we get I=M
Lhx3
3iL
0.
Step 7: Therefore, I=M
3LL3=1
3ML2.
Step 8: Hence, the moment of inertia of the rod about the given axis is
1
3ML2.
11
Question 14
Question
A thin uniform rod of mass mand length Lis rotating about an axis passing
through one end of the rod perpendicular to the length of the rod, with an
angular speed ω. Calculate the moment of inertia of the rod about this axis.
Solution
Step 1: The moment of inertia of the rod can be calculated using the formula
I=Rr2dm, where ris the distance of an element of mass dm from the axis of
rotation.
Step 2: Consider an element of mass dm located at a distance xfrom the
axis of rotation. The mass of this element can be expressed in terms of dx as
dm =m
Ldx.
Step 3: The distance rof this element dm from the axis of rotation is
r=L−x.
Step 4: Substitute dm and rinto the formula for moment of inertia: I=
Rr2dm =RL
0(L−x)2m
Ldx.
Step 5: Expand and simplify the integrand: I=mRL
0(L2−2Lx +x2)dx.
Step 6: Integrate term by term: I=mhL2x−Lx2+x3
3iL
0.
Step 7: Evaluate the definite integral: I=mhL3−L·L2+L3
3i.
Step 8: Simplify the expression: I=mL3−L3+L3
3.
Step 9: Final result: I=1
3mL2.
Therefore, the moment of inertia of the rod about the given axis is 1
3mL2.
Question 15
Question
A thin uniform rod of length Land mass Mis rotating about an axis passing
perpendicular to one end and parallel to the rod. Find the moment of inertia
of the rod about this axis.
Solution
To find the moment of inertia of the rod about the given axis, we need to use
the formula for the moment of inertia of a uniform rod rotating about an axis
perpendicular to one end:
I=1
3ML2
12
Step 1: Determine the moment of inertia of the rod about its center. First,
we consider the moment of inertia of the rod about the center (where the axis
passes). Let’s denote this as Icenter.
Icenter =1
12ML2
Step 2: Use the parallel axis theorem. We now need to use the parallel axis
theorem to find the moment of inertia of the rod about the given axis passing
perpendicular to one end. The parallel axis theorem states that the moment
of inertia about any axis parallel to an axis through the center of mass can be
found by:
I=Icenter +Md2
where dis the perpendicular distance between the two axes, which in this case
is L/2 since the axis of rotation is perpendicular to one end and parallel to the
rod.
Step 3: Apply the formula. Substitute Icenter and dinto the formula to find
the moment of inertia of the rod about the given axis:
I=1
12ML2+ML
22
I=1
12ML2+1
4ML2
I=1
12ML2+3
12ML2
I=1
4ML2
Therefore, the moment of inertia of the thin uniform rod about the given
axis is 1
4ML2.
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 of its ends.
Solution
Let’s consider the rod as being made up of infinitesimally small mass elements
along its length. Let δm be the mass of an infinitesimal mass element at a
distance xfrom the axis passing through one end of the rod. The mass of this
element is δm =M
Ldx.
The moment of inertia of this infinitesimal mass element about the given
axis is dI = (δm)(x2) = M
Ldx(x2).
13
To find the total moment of inertia Iof the entire rod, we integrate dI from
0 to L:
I=ZL
0M
Ldx(x2)
Step 1: Simplify the integrand
I=ZL
0
Mx2
Ldx
Step 2: Integrate with respect to x
I=Mx3
3LL
0
I=ML2
3
Step 3: Final result Therefore, the moment of inertia of the thin uniform
rod of length Land mass Mabout an axis perpendicular to the rod and passing
through one of its ends is M L2
3.
Question 17
Question
A thin rod of length Land mass Mis rotated about an axis perpendicular to
the rod and passing through its center. Find 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 center
of the rod. The mass of this element can be approximated as dm =M
Ldx.
Step 2: The moment of inertia of this element about the axis of rotation is
given by dI =r2dm, where ris the distance of the element from the axis of
rotation. In this case, r=x.
Step 3: Thus, the moment of inertia of the entire rod can be found by
summing up the moments of inertia of all such small elements:
I=ZdI =ZL/2
−L/2
x2M
Ldx
Step 4: Solving the integral gives:
I=M
LZL/2
−L/2
x2dx
14
Step 5: Evaluating the integral, we get:
I=M
Lx3
3L/2
−L/2
Step 6: Substituting the limits of integration and simplifying further yields:
I=1
3ML3
8−(−L)3
8
Step 7: Simplifying the expression gives:
I=1
12ML2
Thus, the moment of inertia of the thin rod about the axis passing through
its center is 1
12 ML2.
Question 18
Question
A thin spherical shell with radius Rand mass Mis rotating about an axis
passing through its center. Determine the moment of inertia of the shell with
respect to the axis of rotation.
Solution
We can calculate the moment of inertia of a thin spherical shell by considering
the shell as a collection of thin rings. The moment of inertia of each ring is
given by Iring = dm ·r2, where dm is the mass of the ring and ris the distance
of the ring from the axis of rotation.
Step 1: Divide the shell into thin rings. Since the spherical shell is thin, we
can consider it as a collection of infinitesimally thin rings. Let’s choose a ring
of radius rand thickness ∆r.
Step 2: Determine the mass of the ring. The mass of the ring can be
calculated using the surface density σof the shell:
dm = σ·2πr ·∆r
Given that the mass Mis distributed over the entire surface area 4πR2of the
shell, we have σ=M
4πR2.
Step 3: Calculate the moment of inertia of the ring. The moment of inertia
of the thin ring is:
Iring = dm ·r2=M
4πR2·2πr ·∆r·r2
15
Step 4: Integrate to find the total moment of inertia. To find the total
moment of inertia of the shell, we need to sum up the moments of inertia of all
the thin rings. Integrating from r= 0 to r=R:
I=ZR
0
Iring dr =ZR
0M
2Rr3dr
Step 5: Perform the integration.
I=M
2RZR
0
r3dr =M
2Rr4
4R
0
=1
8MR2
Therefore, the moment of inertia of the thin spherical shell with respect to
the axis of rotation is 1
8MR2.
Question 19
Question
A thin uniform rod of mass mand length Lis 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: To find the moment of inertia of the rod, we need to consider the
rotational inertia of each infinitesimal mass element along the rod and integrate
over the entire length. Let’s consider an infinitesimal mass element dm at a
distance xfrom the end of the rod.
Step 2: The rotational inertia of an infinitesimal mass element dm at a
distance xfrom the axis of rotation is given by dI =r2dm, where ris the
distance of dm from the axis of rotation.
Step 3: Using the geometry of the rod, the mass element dm is located at a
distance xfrom the end of the rod. Therefore, the distance ris equal to L−x.
Step 4: Substituting r=L−xand dm =m
Ldx into the expression dI =
r2dm, we have dI = (L−x)2m
Ldx.
Step 5: Now, we can integrate dI over the entire length of the rod. The total
moment of inertia Iof the rod is given by:
I=ZL
0
(L−x)2m
Ldx
Step 6: Simplifying the integral, we have:
I=mL3
3−L2
2x+L
3x2
L
0
16
Step 7: Evaluating the integral limits, we get:
I=mL3
3−L3
2+L3
3
Step 8: Simplifying further, we find:
I=mL2
3
Step 9: Therefore, the moment of inertia of the thin uniform rod rotating
about an axis perpendicular to the rod and passing through one end is mL2
3.
Question 20
Question
A thin uniform rod of length Land mass Mis pivoted at one end. Find the
moment of inertia of the rod about an axis perpendicular to the rod and passing
through the end of the rod where it is pivoted.
Solution
To find the moment of inertia Iof the rod about the given axis, we will consider
the rod to be made up of infinitesimally small mass elements.
Step 1: Identify the mass element Consider a small mass element dm
at a distance xfrom the pivot point. The mass of this element can be expressed
as:
dm =m
Ldx
where mis the total mass of the rod.
Step 2: Calculate the moment of inertia of the mass element The
moment of inertia dI of the mass element dm about the pivot point is given by:
dI =r2dm =x2dm
where ris the distance of the mass element from the pivot point.
Substitute the expression for dm:
dI =x2m
Ldx=m
Lx2dx
Step 3: Integrate to find the total moment of inertia Integrate dI
from 0 to Lto find the total moment of inertia I:
I=ZL
0
m
Lx2dx
I=m
LZL
0
x2dx
17
I=m
Lx3
3L
0
I=m
LL3
3−0
I=mL2
3
Therefore, the moment of inertia of the rod about the given axis is mL2
3.
Question 21
Question
A uniform rod of mass Mand length Lis rotating about an axis passing through
one end perpendicular to the rod with angular speed ω. Calculate the moment
of inertia of the rod about this axis.
Solution
Step 1: Determine the moment of inertia of an infinitesimal element of the rod.
Let’s consider an infinitesimal element of length dx at distance xfrom the
axis of rotation. The mass of this element can be written as dm =M
Ldx. The
moment of inertia of this element about the given axis is dI =x2dm.
Step 2: Integrate to find the total moment of inertia.
Integrating from 0 to L:
I=ZL
0
x2M
Ldx
Step 3: Simplify and solve the integral.
I=M
LZL
0
x2dx =M
Lx3
3L
0
I=M
LL3
3−0=M
L·L3
3=M
3L2
Therefore, the moment of inertia of the rod about the given axis is 1
3ML2.
Question 22
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 end of the rod.
18
Solution
Step 1: Determine the mass per unit length of the rod. Step 2: Express the
moment of inertia of an infinitesimal element of the rod. Step 3: Integrate the
moment of inertia over the entire length of the rod to find the total moment of
inertia. Step 4: Simplify the expression for the moment of inertia and express
it in terms of the rod’s mass Mand length L. Step 5: Hence, the moment of
inertia of the thin uniform rod about the given axis is calculated.
Let’s solve it step by step.
Step 1: Determine the mass per unit length of the rod.
The mass per unit length, λ, of the rod is given by:
λ=M
L
Step 2: Express the moment of inertia of an infinitesimal element of the
rod.
Consider an infinitesimal element of length dx at a distance xfrom the end
of the rod. The mass of this element is dm =λdx.
The moment of inertia of this element about the axis perpendicular to the
rod and passing through one end is given by:
dI =dm ·x2=λdx ·x2
Step 3: Integrate the moment of inertia over the entire length of the rod to
find the total moment of inertia.
The total moment of inertia Iis obtained by integrating dI over the length
of the rod:
I=ZL
0
λx2dx
Step 4: Simplify the expression for the moment of inertia and express it in
terms of the rod’s mass Mand length L.
Substitute λ=M
Linto the integral:
I=ZL
0M
Lx2dx =M
L·x3
3L
0
I=M
L·L3
3=ML2
3
Step 5: Hence, the moment of inertia of the thin uniform rod about the
given axis is M L2
3.
Question 23
Question
A thin rod of length Land mass Mis rotating about an axis that passes through
one end of the rod perpendicular to its length. Find the moment of inertia of
the rod about this axis.
19
Solution
To find the moment of inertia of the rod about an axis passing through one end
perpendicular to its length, we can treat the rod as a collection of infinitesimally
small point masses. We will integrate the contributions to the moment of inertia
from each of these small masses.
Step 1: Choose a small mass element δm at a distance xfrom the axis of
rotation. The mass of the rod can be written as δm =M
Ldx.
Step 2: The moment of inertia dI of this mass element about the axis is
given by dI = (δm)x2.
Step 3: Substitute the expression for δm into the equation for dI:
dI =M
Ldxx2
Step 4: Integrate dI from 0 to Lto find the total moment of inertia Iof
the rod:
I=ZL
0M
Lx2dx
Step 5: Solve the integral:
I=M
LZL
0
x2dx
=M
Lx3
3L
0
=M
LL3
3−0
=1
3ML2
Therefore, the moment of inertia of the rod about the axis passing through
one end perpendicular to its length is 1
3ML2.
Question 24
Question
Find the moment of inertia of a solid cylinder of radius Rand mass M, about
an axis that passes through its center and is perpendicular to its symmetry axis.
Solution
The moment of inertia of a solid cylinder about its symmetry axis can be cal-
culated using the formula I=1
2MR2. To find the moment of inertia about an
axis passing through its center and perpendicular to the symmetry axis, we can
use the parallel axis theorem.
20
Step 1: Start with the moment of inertia formula for a solid cylinder about
its symmetry axis: Isymmetry =1
2MR2.
Step 2: Use the parallel axis theorem, which states that I=Isymmetry +
MD2, where Dis the distance between the two axes.
Step 3: In this case, the distance D=R
2since the axis passes through the
center of the cylinder.
Step 4: Substitute D=R
2into the parallel axis theorem: I=1
2MR2+
MR
22.
Step 5: Simplify the expression: I=1
2MR2+1
4MR2=3
4MR2.
Step 6: Therefore, the moment of inertia of the solid cylinder about an axis
passing through its center and perpendicular to its symmetry axis is I=3
4MR2.
Question 25
Question
A thin uniform rod of length Land mass Mis pivoted at one end and allowed to
swing freely in the vertical plane. Calculate the rod’s moment of inertia about
the pivot point.
Solution
Let’s consider a thin uniform rod of length Land mass Mpivoted at one end.
We are looking to find the moment of inertia of this rod about the pivot point.
Step 1: The moment of inertia of a point mass mat a distance rfrom the
pivot point is I=mr2.
Step 2: To find the moment of inertia of the entire rod, we need to sum up
the moments of inertia of all the infinitesimally small point masses that make
up the rod.
Step 3: Let’s consider an infinitesimally small mass dm at a distance xfrom
the pivot point. The mass dm can be written in terms of linear mass density λ
as dm =λdx, where dx is an infinitesimal length element along the rod.
Step 4: The moment of inertia of this small mass dm about the pivot point
is dI =dm ×x2. Substituting for dm and x, we get dI =λdx ×x2.
Step 5: To find the total moment of inertia of the rod, we need to integrate
dI from x= 0 to x=L. So, I=RL
0λx2dx.
Step 6: Substituting the expression for linear mass density λ=M
L, we get
I=RL
0
M
Lx2dx.
Step 7: Solving the integral, we find I=M
Lhx3
3iL
0=M
L×L3
3.
Step 8: Simplifying further, we get the moment of inertia of the rod about
the pivot point as I=1
3ML2.
21
Question 26
Question
A thin rod of mass Mand length Lis rotating about an axis perpendicular to
the rod passing through its midpoint. Find 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 passing through its midpoint can be calculated using the formula
I=1
12 ML2.
Step 2: Substituting the given values of mass Mand length Linto the
formula, we get:
I=1
12ML2=1
12M(L)2=1
12M(2L)2=1
12M(4L2) = 1
3ML
22
Step 3: Simplifying the expression gives us the moment of inertia as:
I=1
3M
2L
22
=1
3M
2L2
4=1
3·M
2·L2
4=1
3·ML2
8=1
24ML2
Therefore, the moment of inertia of the rod about the given axis is 1
24 ML2.
Question 27
Question
A thin uniform rod of mass Mand length Lrotates about an axis passing
through one end and perpendicular to the rod. Calculate the moment of inertia
of the rod about this axis.
Solution
Step 1: Define the necessary variables. Let mbe the mass per unit length of
the rod, which is given by m=M
L.
Step 2: Calculate the moment of inertia of a small element. Consider a small
element of the rod of length dx at a distance xfrom the end of the rod. The
mass of this element is dm =m·dx.
The moment of inertia of this small element about the axis passing through
one end is given by dI =dm ·x2.
Substitute dm and mto get dI = (M/L)·dx ·x2.
Step 3: Integrate to find the total moment of inertia. Integrate dI from
x= 0 to x=Lto find the total moment of inertia I:
I=ZL
0
M
L·x2dx
22
Step 4: Perform the integration.
I=M
LZL
0
x2dx
I=M
Lx3
3L
0
I=M
L·L3
3
Step 5: Simplify the expression.
I=M·L2
3
Therefore, the moment of inertia of the rod about the axis passing through
one end and perpendicular to the rod is M·L2
3.
Question 28
Question
A thin uniform rod of length Land mass Mis rotating about an axis perpen-
dicular to one end and passing through the other end. Calculate 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 end
of the rod. Step 2: The mass of this element is given by dm =M
Ldx. Step
3: The moment of inertia of this element about the axis is dI =dm ·x2.
Step 4: Substitute dm into the equation above to get dI =M
Lx2dx. Step 5:
Integrate dI from x= 0 to x=Lto find the total moment of inertia I. Step
6: I=RL
0
M
Lx2dx. Step 7: I=M
LRL
0x2dx. Step 8: I=M
Lhx3
3iL
0. Step 9:
I=M
L·L3
3. Step 10: I=M
3L2.
Therefore, the moment of inertia of the thin uniform rod about the given
axis is M
3L2.
Question 29
Question
A thin uniform rod of length Land mass Mis rotating about an axis passing
through one end and perpendicular to the rod. Calculate the moment of inertia
of the rod in terms of Mand L.
23
Solution
Step 1: Divide the rod into small elements of length dx such that the mass of
each element is dm =M
Ldx.
Step 2: The moment of inertia of each small element about the axis of
rotation is dI =dm ·x2, where xis the distance of the element from the axis of
rotation.
Step 3: Substituting the expression for dm into the equation for dI, we get
dI =M
Lx2dx.
Step 4: To find the total moment of inertia of the rod, integrate dI over the
length of the rod:
I=ZL
0
dI =ZL
0
M
Lx2dx
Step 5: Performing the integration, we get:
I=M
LZL
0
x2dx =M
Lx3
3L
0
Step 6: Evaluating the integral limits:
I=M
LL3
3−03
3=M
L·L3
3=1
3ML2
Therefore, the moment of inertia of the rod rotating about an axis passing
through one end and perpendicular to the rod is 1
3ML2.
Question 30
Question
A thin rod of length Land mass Mis rotated about an axis perpendicular to
the rod and passing through one end. Calculate the moment of inertia of the
rod about this axis.
Solution
Step 1: Let’s find an expression for the moment of inertia of an infinitesimal
element of the rod located at a distance xfrom the axis of rotation. The mass
of this infinitesimal element dm is given by dm =M
Ldx.
Step 2: The moment of inertia dI of this infinitesimal element about the axis
of rotation is given by dI =r2dm, where ris the distance of the element from the
axis of rotation. From the geometry of the situation, r=x, so dI =x2M
Ldx.
Step 3: The total moment of inertia Iof the rod can be obtained by inte-
grating dI over the entire length of the rod.
I=ZL
0
x2M
Ldx
24
Step 4: Simplifying the integral, we get
I=M
LZL
0
x2dx
Step 5: Integrating, we have
I=M
L1
3x3L
0
=M
L1
3L3=M
3L2
So, the moment of inertia of the rod about the given axis is M
3L2.
Question 31
Question
Find the moment of inertia of the region bounded by the curve y=√xand the
x-axis from x= 0 to x= 1 about the y-axis.
Solution
Step 1: Determine the area of the region bounded by the curve and the x-axis.
Area, A=Z1
0
√x dx =2
3x3/21
0
=2
3
Step 2: Calculate the distance from the centroid of the region to the y-axis.
The centroid of the region can be found using the formula:
¯y=1
AZ1
0
y·x dx
=1
AZ1
0
x3/2dx =1
A2
5x5/21
0
=2
5A
The distance from the centroid to the y-axis is then ¯y=2
5A.
Step 3: Use the Parallel Axis Theorem to find the moment of inertia.
Iy=Z1
0
y2dx =Z1
0
x dx =1
3x3
1
0=1
3
Step 4: Apply the Parallel Axis Theorem.
Iy=Icm +A·d2
1
3=2
5A+2
32
5A2
A=15
7,and Iy=17
21
25
Question 32
Question
A thin uniform rod of mass mand length Lis rotating about an axis passing
through one end perpendicular to the rod with angular velocity ω. Calculate
the moment of inertia of the rod about this axis.
Solution
Step 1: Determine the moment of inertia of a point mass about the axis passing
through one end perpendicular to the rod. The moment of inertia of a point
mass dm located at a distance rfrom the axis of rotation is given by dI =r2dm.
Step 2: Express dm in terms of dx, the length of a small element of the rod.
Since the rod is uniform, dm =m
Ldx.
Step 3: Integrate dI =r2dm over the entire length of the rod to find the
total moment of inertia.
I=ZL
0
r2·m
Ldx
Step 4: Define rin terms of x. The distance of a point on the rod from the
axis of rotation is r=x.
Step 5: Substitute r=xinto the integral.
I=ZL
0
x2·m
Ldx
Step 6: Solve the integral.
I=m
LZL
0
x2dx
Step 7: Integrate x2with respect to x.
I=m
Lx3
3L
0
Step 8: Evaluate the integral limits.
I=m
LL3
3−0
Step 9: Simplify the expression.
I=mL2
3
Therefore, the moment of inertia of the rod about the axis passing through
one end perpendicular to the rod is mL2
3.
26
Question 33
Question
A thin uniform rod of length Land mass Mis rotating about an axis that is
perpendicular to the rod and passing through one end. Find the moment of
inertia of the rod about this axis.
Solution
Step 1: Divide the rod into small elements of length dx. Step 2: The mass of
each small element dm can be expressed in terms of dx and M. Step 3: The
distance of each small element from the axis of rotation is x. Step 4: Express
the moment of inertia dI of each small element dm about the axis of rotation
in terms of xand dx. Step 5: Integrate dI over the entire length of the rod to
find the total moment of inertia I. Step 6: Substitute the values of M,L, and
the limits of integration into the expression for Ito obtain the final answer.
Question 34
Question
Find the moment of inertia of a solid cylinder of radius Rand mass Mabout an
axis that is perpendicular to the central axis of the cylinder and passes through
a point on the surface.
Solution
Step 1: The moment of inertia of a solid cylinder of radius Rand mass Mabout
an axis passing through its center and perpendicular to its central axis can be
calculated using the formula I=1
2MR2.
Step 2: To find the moment of inertia about an axis passing through a
point on the surface perpendicular to the central axis, we can use the parallel-
axis theorem. The theorem states that Iaxis =ICM +M d2, where ICM is the
moment of inertia about the center of mass axis and dis the perpendicular
distance between the two axes.
Step 3: The moment of inertia about the center of mass axis (ICM) can be
calculated using the formula ICM =1
2MR2.
Step 4: The distance dbetween the center of mass axis and the given axis
passing through a point on the surface can be calculated as d=R.
Step 5: Substituting ICM =1
2MR2and d=Rinto the parallel-axis theorem,
we get Iaxis =1
2MR2+MR2.
Step 6: Simplifying, we find Iaxis =3
2MR2.
Therefore, the moment of inertia of the solid cylinder about an axis passing
through a point on the surface perpendicular to the central axis is 3
2MR2.
27
Question 35
Question
Calculate the moment of inertia of a thin-walled hollow cylinder with inner
radius r1, outer radius r2, and height habout its central axis.
Solution
Step 1: The moment of inertia of a thin-walled hollow cylinder about its central
axis can be calculated using the formula:
I=1
2m(r2
1+r2
2)
where mis the mass of the cylinder. Since the cylinder is hollow, we need to
consider the difference in mass between the outer and inner cylinders.
Step 2: The mass of the thin-walled hollow cylinder can be calculated as
the mass of the outer cylinder minus the mass of the inner cylinder. The mass
mcan be calculated using the density ρof the material, the height hof the
cylinder, and the volume formula for a cylinder:
m=ρ×Volume
Step 3: The volume of the outer cylinder is (πr2
2−πr2
1)hand the volume of
the inner cylinder is πr2
1h, so the mass mof the hollow cylinder is:
m=ρπr2
2h−πr2
1h
Step 4: Substituting the mass expression back into the moment of inertia
formula gives:
I=1
2ρπr2
2h−πr2
1h(r2
1+r2
2)
Step 5: Simplifying the expression gives the final formula for the moment of
inertia of the thin-walled hollow cylinder about its central axis:
I=1
2ρhπ(r2
1+r2
2)(r2
2−r2
1)
28
Step 5: Simplifying the expression and solving the integral gives:
I=ZL
0
(L2−2Lx +x2)·M
Ldx
=ML2x
L−2Lx2
2L+x3
3L
L
0
=ML−L+L
3
=1
3ML2
Therefore, the moment of inertia of the thin uniform rod about the given
axis of rotation is 1
3ML2.
Question 2
Question
Find the moment of inertia of a thin uniform rod of length Land mass Mabout
an axis passing through one end perpendicular to the rod.
Solution
To find the moment of inertia of the thin uniform rod about an axis passing
through one end perpendicular to the rod, we can consider the rod as a collection
of infinitesimally small masses along its length. We can then sum up the moment
of inertia of each of these masses to find the total moment of inertia of the rod.
Step 1: Consider an infinitesimal mass dm at a distance xfrom the end
where the axis passes.
The mass dm can be expressed as:
dm =M
Ldx
Step 2: Find the moment of inertia dI of this infinitesimal mass dm.
The moment of inertia of an infinitesimal mass dm about the given axis is
given by:
dI =dm ·x2
Substitute the expression for dm:
dI =M
Ldx ·x2
Step 3: Integrate to find the total moment of inertia Iof the rod.
2
Integrating dI over the entire length of the rod gives the total moment of
inertia I:
I=ZdI =ZM
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
passing through one end perpendicular to the rod is M L2
3.
Question 3
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 of its ends.
Solution
Step 1: Choose an element of mass δm at a distance xfrom the axis passing
through one end of the rod.
Step 2: The mass δm of the element is given by δm =M
Ldx.
Step 3: The moment of inertia of this element about the axis is dI =δm·x2=
M
L·x2dx.
Step 4: To find the total moment of inertia, we need to integrate the expres-
sion for dI over the length of the rod.
I=ZL
0
M
Lx2dx
Step 5: Solving the integral gives:
I=M
Lx3
3L
0
=M
LL3
3−0
Step 6: Therefore, the moment of inertia of the rod about the axis passing
through one of its ends is:
I=1
3ML2
3
Question 4
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 end.
Solution
To find the moment of inertia of the rod, we will integrate the element of mass
over the entire length of the rod.
Step 1: Consider a small element of length dx at distance xfrom the end
of the rod. The mass of this element is dm =m
Ldx.
Step 2: The moment of inertia of this element about the axis is dI =
dm ·x2=m
Ldx·x2.
Step 3: Integrate dI over the entire length of the rod from 0 to L.
I=ZL
0
dI =ZL
0m
Ldx·x2
Step 4: Solve the integral.
I=m
LZL
0
x2dx =m
Lx3
3L
0
=m
L·L3
3=mL2
3
Step 5: Therefore, the moment of inertia of the thin uniform rod about the
given axis is mL2
3.
Question 5
Question
A thin rod of length Land mass Mis pivoted at one end. Find the moment of
inertia of the rod about an axis perpendicular to the rod and passing through
the free end.
Solution
To find the moment of inertia of the thin rod about an axis perpendicular to
the rod and passing through the free end, we can use the parallel axis theorem.
Step 1: Find the moment of inertia of the rod about its center
of mass The moment of inertia of the rod about its center of mass can be
calculated using the formula for a thin rod rotated about an axis perpendicular
to the rod and passing through its center:
ICM =1
12ML2
4
Step 2: Apply the parallel axis theorem The parallel axis theorem
states that the moment of inertia about any axis parallel to and at a distance d
from the axis passing through the center of mass is given by:
I=ICM +M d2
Step 3: Calculate the moment of inertia about the free end In this
case, the distance dis equal to L/2 (half the length of the rod). Substituting
into the parallel axis theorem formula:
I=1
12ML2+ML
22
I=1
12ML2+1
4ML2
I=1
3ML2
Therefore, the moment of inertia of the thin rod about an axis perpendicular
to the rod and passing through the free end is 1
3ML2.
Question 6
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 end.
Solution
Step 1: Divide the rod into infinitesimally small elements of length dx at a
distance xfrom the end where the axis passes through.
Step 2: The mass of each element dm is given by dm =M
Ldx.
Step 3: The moment of inertia of an elemental mass dm about the axis is
dI =dm ·x2.
Step 4: Substitute dm and xto obtain dI =M
Lx2dx.
Step 5: The total moment of inertia of the rod is obtained by integrating
over the entire length:
I=ZdI =ZL
0
M
Lx2dx
Step 6: Solve the integral:
I=M
Lx3
3L
0
Step 7: Substitute the limits of integration and simplify:
I=M
LL3
3−03
3
5
Step 8: Further simplify to obtain the moment of inertia:
I=M
L·L3
3=M
3L2
Therefore, the moment of inertia of the thin uniform rod of mass Mand
length Labout an axis perpendicular to the rod and passing through one end
is M
3L2.
Question 7
Question
Find the moment of inertia of the region bounded by the curve y=x2and the
y-axis, rotated about the line y=−2.
Solution
To find the moment of inertia, we will first need to determine the area of the
region of interest and its centroid.
Step 1: Find the area of the region
1. We need to find the points of intersection between the curve y=x2and
the y-axis.
2. Setting x= 0 in y=x2, we find that the parabola intersects the y-axis at
the point (0,0).
3. The area of the region can be expressed as an integral:
A=Z1
0
x2dx
4. Solving the integral, we have:
A=1
3x3
1
0
=1
3
Step 2: Find the x-coordinate of the centroid
1. The x-coordinate of the centroid is given by:
¯x=1
AZ1
0
x·x2dx
2. Simplifying the expression, we get:
¯x=R1
0x3dx
1
3
=1
4
6
Step 3: Find the moment of inertia
1. The moment of inertia about the line y=−2 is given by:
I=Z1
0
(x2)·(x−¯x)2dx
2. Substituting the value of ¯x, we get:
I=Z1
0
x2(x−1
4)2dx
3. Solving the integral will give us the moment of inertia.
Question 8
Question
A thin uniform rod of length Land mass Mis rotating about an axis passing
through one end and perpendicular to the rod. Find the moment of inertia of
the rod with respect to this axis.
Solution
Step 1: Consider an elemental mass δm at a distance xfrom the axis of rotation.
Step 2: The moment of inertia of this elemental mass about the axis of
rotation is δI = (δm)x2.
Step 3: We need to express δm in terms of xin order to integrate. Since the
rod is uniform, the linear mass density λ=M
L.
Step 4: The elemental mass δm =λdx.
Step 5: Substituting δm =λdx into δI = (δm)x2gives δI = (λdx)x2=
λx2dx.
Step 6: The total moment of inertia Iof the rod about the given axis is
obtained by integrating δI from 0 to L.
Step 7: Therefore, I=RL
0λx2dx.
Step 8: Evaluating the integral, we get I=hλx3
3iL
0.
Step 9: Plugging in the values, we get I=λL3
3.
Step 10: Finally, substituting λ=M
L, we find I=ML2
3.
Step 11: So, the moment of inertia of the rod with respect to the given axis
is ML2
3.
7
Question 9
Question
A thin uniform rod of mass Mand length Lis rotating about an axis passing
through one end and perpendicular to the rod. What is the moment of inertia
of the rod about this axis?
Solution
To find the moment of inertia of the rod about an axis passing through one
end and perpendicular to the rod, we can treat the rod as a collection of in-
finitesimally small particles, each with mass dm, at a distance rfrom the axis
of rotation.
Step 1: Let’s express dm in terms of the linear density λof the rod and
the length element dx. The mass of an infinitesimal length dx of the rod is
dm =λ dx.
Step 2: Next, we express rin terms of x, where xis the distance from
the end of the rod to the mass element dm. Since the rod is rotating along its
length, r=L−x.
Step 3: Now, we can express I(moment of inertia) as the sum of the
moments of inertia of all the mass elements dm along the length of the rod.
I=Zr2dm
Step 4: Substitute for dm and r:
I=ZL
0
(L−x)2λ dx
Step 5: Expand and solve the integral:
I=λZL
0
(L2−2Lx +x2)dx
=λL3
3−L2x
2+x3
3
L
0
=λL3
3−L3
2+L3
3
=1
3λL3
Step 6: Finally, substitute the linear density λ=M
Linto the expression for
I:
I=1
3M
LL3=1
3ML2
Therefore, the moment of inertia of the rod about an axis passing through
one end and perpendicular to the rod is 1
3ML2.
8
Question 10
Question
A solid cylinder of mass Mand radius Ris rotated about an axis passing through
its center and perpendicular to its length. Find the moment of inertia of the
cylinder with respect to this axis.
Solution
Let’s denote the moment of inertia of the cylinder with respect to the given axis
as I.
Step 1: Determine the moment of inertia of a small mass element.
Consider a small mass element of the cylinder at a distance rfrom the axis
of rotation. The moment of inertia of this mass element dm with respect to the
axis is r2dm.
Step 2: Express dm in terms of rand dr.
The mass of the small element dm can be expressed in terms of the linear
mass density λand the length dr as dm =λdr. Since the cylinder has a uniform
density, we have λ=M
πR2.
Step 3: Express the moment of inertia Iin terms of rand dr.
Integrating the moment of inertia of the small mass elements from r= 0 to
r=R, we get:
I=ZR
0
r2dm
Substitute dm =λdr into the expression:
I=ZR
0
r2λdr
Step 4: Integrate to find the moment of inertia I.
Now, substitute λ=M
πR2into the integral:
I=ZR
0
r2M
πR2dr
I=M
πR2ZR
0
r2dr
I=M
πR2r3
3R
0
I=M
πR2R3
3−0
I=MR2
3
Therefore, the moment of inertia of the cylinder with respect to the given
axis is M R2
3.
9
Question 11
Question
Find the moment of inertia of a uniform thin rod of mass Mand length Labout
an axis perpendicular to the rod and passing through its midpoint.
Solution
Step 1: Determine the mass per unit length λof the rod.
λ=M
L
Step 2: Calculate the moment of inertia of an infinitesimal element of the
rod at a distance xfrom the midpoint.
dI =λ dx ·x2
Step 3: Integrate the expression for dI from −L/2 to L/2 to find the total
moment of inertia I.
I=ZL/2
−L/2
λ·x2dx
I=ZL/2
−L/2
M
L·x2dx
I=M
LZL/2
−L/2
x2dx
I=M
Lx3
3L/2
−L/2
I=M
L(L/2)3
3−(−L/2)3
3
I=M
LL3
24 +L3
24
I=M
L·L3
12
I=1
12ML2
Therefore, the moment of inertia of the uniform thin rod of mass Mand
length Labout an axis perpendicular to the rod and passing through its mid-
point is 1
12 ML2.
10
Question 12
Question
A thin hoop of radius Rand mass Mis rotating about its central axis with
an angular velocity ω. Find the moment of inertia of the hoop about an axis
passing through its edge and perpendicular to its plane.
Solution
Step 1: The moment of inertia of the hoop rotating about its central axis is
Ic=MR2.
Step 2: Consider the parallel axis theorem, which states that the moment
of inertia about an axis parallel to and a distance daway from the center of
rotation is given by I=Ic+Md2. In this case, the distance from the central
axis to the edge is R, so the moment of inertia about an axis passing through
its edge and perpendicular to its plane is given by
I=Ic+MR2=MR2+MR2= 2M R2.
Question 13
Question
A thin rod of length Land mass Mis rotating about an axis perpendicular to
its length passing through one of its ends. Find the moment of inertia of the
rod about this axis.
Solution
Step 1: Consider a small element of the rod at a distance xfrom the axis of
rotation.
Step 2: The mass of this element can be considered as M
Ldx.
Step 3: The moment of inertia of this element about the axis of rotation is
dI =M
Lx2dx (using I=Rr2dm).
Step 4: To find the total moment of inertia of the rod, we need to integrate
dI from 0 to L.
Step 5: So, I=RL
0
M
Lx2dx.
Step 6: Solving the integral, we get I=M
Lhx3
3iL
0.
Step 7: Therefore, I=M
3LL3=1
3ML2.
Step 8: Hence, the moment of inertia of the rod about the given axis is
1
3ML2.
11
Question 14
Question
A thin uniform rod of mass mand length Lis rotating about an axis passing
through one end of the rod perpendicular to the length of the rod, with an
angular speed ω. Calculate the moment of inertia of the rod about this axis.
Solution
Step 1: The moment of inertia of the rod can be calculated using the formula
I=Rr2dm, where ris the distance of an element of mass dm from the axis of
rotation.
Step 2: Consider an element of mass dm located at a distance xfrom the
axis of rotation. The mass of this element can be expressed in terms of dx as
dm =m
Ldx.
Step 3: The distance rof this element dm from the axis of rotation is
r=L−x.
Step 4: Substitute dm and rinto the formula for moment of inertia: I=
Rr2dm =RL
0(L−x)2m
Ldx.
Step 5: Expand and simplify the integrand: I=mRL
0(L2−2Lx +x2)dx.
Step 6: Integrate term by term: I=mhL2x−Lx2+x3
3iL
0.
Step 7: Evaluate the definite integral: I=mhL3−L·L2+L3
3i.
Step 8: Simplify the expression: I=mL3−L3+L3
3.
Step 9: Final result: I=1
3mL2.
Therefore, the moment of inertia of the rod about the given axis is 1
3mL2.
Question 15
Question
A thin uniform rod of length Land mass Mis rotating about an axis passing
perpendicular to one end and parallel to the rod. Find the moment of inertia
of the rod about this axis.
Solution
To find the moment of inertia of the rod about the given axis, we need to use
the formula for the moment of inertia of a uniform rod rotating about an axis
perpendicular to one end:
I=1
3ML2
12
Step 1: Determine the moment of inertia of the rod about its center. First,
we consider the moment of inertia of the rod about the center (where the axis
passes). Let’s denote this as Icenter.
Icenter =1
12ML2
Step 2: Use the parallel axis theorem. We now need to use the parallel axis
theorem to find the moment of inertia of the rod about the given axis passing
perpendicular to one end. The parallel axis theorem states that the moment
of inertia about any axis parallel to an axis through the center of mass can be
found by:
I=Icenter +Md2
where dis the perpendicular distance between the two axes, which in this case
is L/2 since the axis of rotation is perpendicular to one end and parallel to the
rod.
Step 3: Apply the formula. Substitute Icenter and dinto the formula to find
the moment of inertia of the rod about the given axis:
I=1
12ML2+ML
22
I=1
12ML2+1
4ML2
I=1
12ML2+3
12ML2
I=1
4ML2
Therefore, the moment of inertia of the thin uniform rod about the given
axis is 1
4ML2.
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 of its ends.
Solution
Let’s consider the rod as being made up of infinitesimally small mass elements
along its length. Let δm be the mass of an infinitesimal mass element at a
distance xfrom the axis passing through one end of the rod. The mass of this
element is δm =M
Ldx.
The moment of inertia of this infinitesimal mass element about the given
axis is dI = (δm)(x2) = M
Ldx(x2).
13
To find the total moment of inertia Iof the entire rod, we integrate dI from
0 to L:
I=ZL
0M
Ldx(x2)
Step 1: Simplify the integrand
I=ZL
0
Mx2
Ldx
Step 2: Integrate with respect to x
I=Mx3
3LL
0
I=ML2
3
Step 3: Final result Therefore, the moment of inertia of the thin uniform
rod of length Land mass Mabout an axis perpendicular to the rod and passing
through one of its ends is M L2
3.
Question 17
Question
A thin rod of length Land mass Mis rotated about an axis perpendicular to
the rod and passing through its center. Find 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 center
of the rod. The mass of this element can be approximated as dm =M
Ldx.
Step 2: The moment of inertia of this element about the axis of rotation is
given by dI =r2dm, where ris the distance of the element from the axis of
rotation. In this case, r=x.
Step 3: Thus, the moment of inertia of the entire rod can be found by
summing up the moments of inertia of all such small elements:
I=ZdI =ZL/2
−L/2
x2M
Ldx
Step 4: Solving the integral gives:
I=M
LZL/2
−L/2
x2dx
14
Step 5: Evaluating the integral, we get:
I=M
Lx3
3L/2
−L/2
Step 6: Substituting the limits of integration and simplifying further yields:
I=1
3ML3
8−(−L)3
8
Step 7: Simplifying the expression gives:
I=1
12ML2
Thus, the moment of inertia of the thin rod about the axis passing through
its center is 1
12 ML2.
Question 18
Question
A thin spherical shell with radius Rand mass Mis rotating about an axis
passing through its center. Determine the moment of inertia of the shell with
respect to the axis of rotation.
Solution
We can calculate the moment of inertia of a thin spherical shell by considering
the shell as a collection of thin rings. The moment of inertia of each ring is
given by Iring = dm ·r2, where dm is the mass of the ring and ris the distance
of the ring from the axis of rotation.
Step 1: Divide the shell into thin rings. Since the spherical shell is thin, we
can consider it as a collection of infinitesimally thin rings. Let’s choose a ring
of radius rand thickness ∆r.
Step 2: Determine the mass of the ring. The mass of the ring can be
calculated using the surface density σof the shell:
dm = σ·2πr ·∆r
Given that the mass Mis distributed over the entire surface area 4πR2of the
shell, we have σ=M
4πR2.
Step 3: Calculate the moment of inertia of the ring. The moment of inertia
of the thin ring is:
Iring = dm ·r2=M
4πR2·2πr ·∆r·r2
15
Step 4: Integrate to find the total moment of inertia. To find the total
moment of inertia of the shell, we need to sum up the moments of inertia of all
the thin rings. Integrating from r= 0 to r=R:
I=ZR
0
Iring dr =ZR
0M
2Rr3dr
Step 5: Perform the integration.
I=M
2RZR
0
r3dr =M
2Rr4
4R
0
=1
8MR2
Therefore, the moment of inertia of the thin spherical shell with respect to
the axis of rotation is 1
8MR2.
Question 19
Question
A thin uniform rod of mass mand length Lis 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: To find the moment of inertia of the rod, we need to consider the
rotational inertia of each infinitesimal mass element along the rod and integrate
over the entire length. Let’s consider an infinitesimal mass element dm at a
distance xfrom the end of the rod.
Step 2: The rotational inertia of an infinitesimal mass element dm at a
distance xfrom the axis of rotation is given by dI =r2dm, where ris the
distance of dm from the axis of rotation.
Step 3: Using the geometry of the rod, the mass element dm is located at a
distance xfrom the end of the rod. Therefore, the distance ris equal to L−x.
Step 4: Substituting r=L−xand dm =m
Ldx into the expression dI =
r2dm, we have dI = (L−x)2m
Ldx.
Step 5: Now, we can integrate dI over the entire length of the rod. The total
moment of inertia Iof the rod is given by:
I=ZL
0
(L−x)2m
Ldx
Step 6: Simplifying the integral, we have:
I=mL3
3−L2
2x+L
3x2
L
0
16
Step 7: Evaluating the integral limits, we get:
I=mL3
3−L3
2+L3
3
Step 8: Simplifying further, we find:
I=mL2
3
Step 9: Therefore, the moment of inertia of the thin uniform rod rotating
about an axis perpendicular to the rod and passing through one end is mL2
3.
Question 20
Question
A thin uniform rod of length Land mass Mis pivoted at one end. Find the
moment of inertia of the rod about an axis perpendicular to the rod and passing
through the end of the rod where it is pivoted.
Solution
To find the moment of inertia Iof the rod about the given axis, we will consider
the rod to be made up of infinitesimally small mass elements.
Step 1: Identify the mass element Consider a small mass element dm
at a distance xfrom the pivot point. The mass of this element can be expressed
as:
dm =m
Ldx
where mis the total mass of the rod.
Step 2: Calculate the moment of inertia of the mass element The
moment of inertia dI of the mass element dm about the pivot point is given by:
dI =r2dm =x2dm
where ris the distance of the mass element from the pivot point.
Substitute the expression for dm:
dI =x2m
Ldx=m
Lx2dx
Step 3: Integrate to find the total moment of inertia Integrate dI
from 0 to Lto find the total moment of inertia I:
I=ZL
0
m
Lx2dx
I=m
LZL
0
x2dx
17
I=m
Lx3
3L
0
I=m
LL3
3−0
I=mL2
3
Therefore, the moment of inertia of the rod about the given axis is mL2
3.
Question 21
Question
A uniform rod of mass Mand length Lis rotating about an axis passing through
one end perpendicular to the rod with angular speed ω. Calculate the moment
of inertia of the rod about this axis.
Solution
Step 1: Determine the moment of inertia of an infinitesimal element of the rod.
Let’s consider an infinitesimal element of length dx at distance xfrom the
axis of rotation. The mass of this element can be written as dm =M
Ldx. The
moment of inertia of this element about the given axis is dI =x2dm.
Step 2: Integrate to find the total moment of inertia.
Integrating from 0 to L:
I=ZL
0
x2M
Ldx
Step 3: Simplify and solve the integral.
I=M
LZL
0
x2dx =M
Lx3
3L
0
I=M
LL3
3−0=M
L·L3
3=M
3L2
Therefore, the moment of inertia of the rod about the given axis is 1
3ML2.
Question 22
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 end of the rod.
18
Solution
Step 1: Determine the mass per unit length of the rod. Step 2: Express the
moment of inertia of an infinitesimal element of the rod. Step 3: Integrate the
moment of inertia over the entire length of the rod to find the total moment of
inertia. Step 4: Simplify the expression for the moment of inertia and express
it in terms of the rod’s mass Mand length L. Step 5: Hence, the moment of
inertia of the thin uniform rod about the given axis is calculated.
Let’s solve it step by step.
Step 1: Determine the mass per unit length of the rod.
The mass per unit length, λ, of the rod is given by:
λ=M
L
Step 2: Express the moment of inertia of an infinitesimal element of the
rod.
Consider an infinitesimal element of length dx at a distance xfrom the end
of the rod. The mass of this element is dm =λdx.
The moment of inertia of this element about the axis perpendicular to the
rod and passing through one end is given by:
dI =dm ·x2=λdx ·x2
Step 3: Integrate the moment of inertia over the entire length of the rod to
find the total moment of inertia.
The total moment of inertia Iis obtained by integrating dI over the length
of the rod:
I=ZL
0
λx2dx
Step 4: Simplify the expression for the moment of inertia and express it in
terms of the rod’s mass Mand length L.
Substitute λ=M
Linto the integral:
I=ZL
0M
Lx2dx =M
L·x3
3L
0
I=M
L·L3
3=ML2
3
Step 5: Hence, the moment of inertia of the thin uniform rod about the
given axis is M L2
3.
Question 23
Question
A thin rod of length Land mass Mis rotating about an axis that passes through
one end of the rod perpendicular to its length. Find the moment of inertia of
the rod about this axis.
19
Solution
To find the moment of inertia of the rod about an axis passing through one end
perpendicular to its length, we can treat the rod as a collection of infinitesimally
small point masses. We will integrate the contributions to the moment of inertia
from each of these small masses.
Step 1: Choose a small mass element δm at a distance xfrom the axis of
rotation. The mass of the rod can be written as δm =M
Ldx.
Step 2: The moment of inertia dI of this mass element about the axis is
given by dI = (δm)x2.
Step 3: Substitute the expression for δm into the equation for dI:
dI =M
Ldxx2
Step 4: Integrate dI from 0 to Lto find the total moment of inertia Iof
the rod:
I=ZL
0M
Lx2dx
Step 5: Solve the integral:
I=M
LZL
0
x2dx
=M
Lx3
3L
0
=M
LL3
3−0
=1
3ML2
Therefore, the moment of inertia of the rod about the axis passing through
one end perpendicular to its length is 1
3ML2.
Question 24
Question
Find the moment of inertia of a solid cylinder of radius Rand mass M, about
an axis that passes through its center and is perpendicular to its symmetry axis.
Solution
The moment of inertia of a solid cylinder about its symmetry axis can be cal-
culated using the formula I=1
2MR2. To find the moment of inertia about an
axis passing through its center and perpendicular to the symmetry axis, we can
use the parallel axis theorem.
20
Step 1: Start with the moment of inertia formula for a solid cylinder about
its symmetry axis: Isymmetry =1
2MR2.
Step 2: Use the parallel axis theorem, which states that I=Isymmetry +
MD2, where Dis the distance between the two axes.
Step 3: In this case, the distance D=R
2since the axis passes through the
center of the cylinder.
Step 4: Substitute D=R
2into the parallel axis theorem: I=1
2MR2+
MR
22.
Step 5: Simplify the expression: I=1
2MR2+1
4MR2=3
4MR2.
Step 6: Therefore, the moment of inertia of the solid cylinder about an axis
passing through its center and perpendicular to its symmetry axis is I=3
4MR2.
Question 25
Question
A thin uniform rod of length Land mass Mis pivoted at one end and allowed to
swing freely in the vertical plane. Calculate the rod’s moment of inertia about
the pivot point.
Solution
Let’s consider a thin uniform rod of length Land mass Mpivoted at one end.
We are looking to find the moment of inertia of this rod about the pivot point.
Step 1: The moment of inertia of a point mass mat a distance rfrom the
pivot point is I=mr2.
Step 2: To find the moment of inertia of the entire rod, we need to sum up
the moments of inertia of all the infinitesimally small point masses that make
up the rod.
Step 3: Let’s consider an infinitesimally small mass dm at a distance xfrom
the pivot point. The mass dm can be written in terms of linear mass density λ
as dm =λdx, where dx is an infinitesimal length element along the rod.
Step 4: The moment of inertia of this small mass dm about the pivot point
is dI =dm ×x2. Substituting for dm and x, we get dI =λdx ×x2.
Step 5: To find the total moment of inertia of the rod, we need to integrate
dI from x= 0 to x=L. So, I=RL
0λx2dx.
Step 6: Substituting the expression for linear mass density λ=M
L, we get
I=RL
0
M
Lx2dx.
Step 7: Solving the integral, we find I=M
Lhx3
3iL
0=M
L×L3
3.
Step 8: Simplifying further, we get the moment of inertia of the rod about
the pivot point as I=1
3ML2.
21
Question 26
Question
A thin rod of mass Mand length Lis rotating about an axis perpendicular to
the rod passing through its midpoint. Find 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 passing through its midpoint can be calculated using the formula
I=1
12 ML2.
Step 2: Substituting the given values of mass Mand length Linto the
formula, we get:
I=1
12ML2=1
12M(L)2=1
12M(2L)2=1
12M(4L2) = 1
3ML
22
Step 3: Simplifying the expression gives us the moment of inertia as:
I=1
3M
2L
22
=1
3M
2L2
4=1
3·M
2·L2
4=1
3·ML2
8=1
24ML2
Therefore, the moment of inertia of the rod about the given axis is 1
24 ML2.
Question 27
Question
A thin uniform rod of mass Mand length Lrotates about an axis passing
through one end and perpendicular to the rod. Calculate the moment of inertia
of the rod about this axis.
Solution
Step 1: Define the necessary variables. Let mbe the mass per unit length of
the rod, which is given by m=M
L.
Step 2: Calculate the moment of inertia of a small element. Consider a small
element of the rod of length dx at a distance xfrom the end of the rod. The
mass of this element is dm =m·dx.
The moment of inertia of this small element about the axis passing through
one end is given by dI =dm ·x2.
Substitute dm and mto get dI = (M/L)·dx ·x2.
Step 3: Integrate to find the total moment of inertia. Integrate dI from
x= 0 to x=Lto find the total moment of inertia I:
I=ZL
0
M
L·x2dx
22
Step 4: Perform the integration.
I=M
LZL
0
x2dx
I=M
Lx3
3L
0
I=M
L·L3
3
Step 5: Simplify the expression.
I=M·L2
3
Therefore, the moment of inertia of the rod about the axis passing through
one end and perpendicular to the rod is M·L2
3.
Question 28
Question
A thin uniform rod of length Land mass Mis rotating about an axis perpen-
dicular to one end and passing through the other end. Calculate 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 end
of the rod. Step 2: The mass of this element is given by dm =M
Ldx. Step
3: The moment of inertia of this element about the axis is dI =dm ·x2.
Step 4: Substitute dm into the equation above to get dI =M
Lx2dx. Step 5:
Integrate dI from x= 0 to x=Lto find the total moment of inertia I. Step
6: I=RL
0
M
Lx2dx. Step 7: I=M
LRL
0x2dx. Step 8: I=M
Lhx3
3iL
0. Step 9:
I=M
L·L3
3. Step 10: I=M
3L2.
Therefore, the moment of inertia of the thin uniform rod about the given
axis is M
3L2.
Question 29
Question
A thin uniform rod of length Land mass Mis rotating about an axis passing
through one end and perpendicular to the rod. Calculate the moment of inertia
of the rod in terms of Mand L.
23
Solution
Step 1: Divide the rod into small elements of length dx such that the mass of
each element is dm =M
Ldx.
Step 2: The moment of inertia of each small element about the axis of
rotation is dI =dm ·x2, where xis the distance of the element from the axis of
rotation.
Step 3: Substituting the expression for dm into the equation for dI, we get
dI =M
Lx2dx.
Step 4: To find the total moment of inertia of the rod, integrate dI over the
length of the rod:
I=ZL
0
dI =ZL
0
M
Lx2dx
Step 5: Performing the integration, we get:
I=M
LZL
0
x2dx =M
Lx3
3L
0
Step 6: Evaluating the integral limits:
I=M
LL3
3−03
3=M
L·L3
3=1
3ML2
Therefore, the moment of inertia of the rod rotating about an axis passing
through one end and perpendicular to the rod is 1
3ML2.
Question 30
Question
A thin rod of length Land mass Mis rotated about an axis perpendicular to
the rod and passing through one end. Calculate the moment of inertia of the
rod about this axis.
Solution
Step 1: Let’s find an expression for the moment of inertia of an infinitesimal
element of the rod located at a distance xfrom the axis of rotation. The mass
of this infinitesimal element dm is given by dm =M
Ldx.
Step 2: The moment of inertia dI of this infinitesimal element about the axis
of rotation is given by dI =r2dm, where ris the distance of the element from the
axis of rotation. From the geometry of the situation, r=x, so dI =x2M
Ldx.
Step 3: The total moment of inertia Iof the rod can be obtained by inte-
grating dI over the entire length of the rod.
I=ZL
0
x2M
Ldx
24
Step 4: Simplifying the integral, we get
I=M
LZL
0
x2dx
Step 5: Integrating, we have
I=M
L1
3x3L
0
=M
L1
3L3=M
3L2
So, the moment of inertia of the rod about the given axis is M
3L2.
Question 31
Question
Find the moment of inertia of the region bounded by the curve y=√xand the
x-axis from x= 0 to x= 1 about the y-axis.
Solution
Step 1: Determine the area of the region bounded by the curve and the x-axis.
Area, A=Z1
0
√x dx =2
3x3/21
0
=2
3
Step 2: Calculate the distance from the centroid of the region to the y-axis.
The centroid of the region can be found using the formula:
¯y=1
AZ1
0
y·x dx
=1
AZ1
0
x3/2dx =1
A2
5x5/21
0
=2
5A
The distance from the centroid to the y-axis is then ¯y=2
5A.
Step 3: Use the Parallel Axis Theorem to find the moment of inertia.
Iy=Z1
0
y2dx =Z1
0
x dx =1
3x3
1
0=1
3
Step 4: Apply the Parallel Axis Theorem.
Iy=Icm +A·d2
1
3=2
5A+2
32
5A2
A=15
7,and Iy=17
21
25
Question 32
Question
A thin uniform rod of mass mand length Lis rotating about an axis passing
through one end perpendicular to the rod with angular velocity ω. Calculate
the moment of inertia of the rod about this axis.
Solution
Step 1: Determine the moment of inertia of a point mass about the axis passing
through one end perpendicular to the rod. The moment of inertia of a point
mass dm located at a distance rfrom the axis of rotation is given by dI =r2dm.
Step 2: Express dm in terms of dx, the length of a small element of the rod.
Since the rod is uniform, dm =m
Ldx.
Step 3: Integrate dI =r2dm over the entire length of the rod to find the
total moment of inertia.
I=ZL
0
r2·m
Ldx
Step 4: Define rin terms of x. The distance of a point on the rod from the
axis of rotation is r=x.
Step 5: Substitute r=xinto the integral.
I=ZL
0
x2·m
Ldx
Step 6: Solve the integral.
I=m
LZL
0
x2dx
Step 7: Integrate x2with respect to x.
I=m
Lx3
3L
0
Step 8: Evaluate the integral limits.
I=m
LL3
3−0
Step 9: Simplify the expression.
I=mL2
3
Therefore, the moment of inertia of the rod about the axis passing through
one end perpendicular to the rod is mL2
3.
26
Question 33
Question
A thin uniform rod of length Land mass Mis rotating about an axis that is
perpendicular to the rod and passing through one end. Find the moment of
inertia of the rod about this axis.
Solution
Step 1: Divide the rod into small elements of length dx. Step 2: The mass of
each small element dm can be expressed in terms of dx and M. Step 3: The
distance of each small element from the axis of rotation is x. Step 4: Express
the moment of inertia dI of each small element dm about the axis of rotation
in terms of xand dx. Step 5: Integrate dI over the entire length of the rod to
find the total moment of inertia I. Step 6: Substitute the values of M,L, and
the limits of integration into the expression for Ito obtain the final answer.
Question 34
Question
Find the moment of inertia of a solid cylinder of radius Rand mass Mabout an
axis that is perpendicular to the central axis of the cylinder and passes through
a point on the surface.
Solution
Step 1: The moment of inertia of a solid cylinder of radius Rand mass Mabout
an axis passing through its center and perpendicular to its central axis can be
calculated using the formula I=1
2MR2.
Step 2: To find the moment of inertia about an axis passing through a
point on the surface perpendicular to the central axis, we can use the parallel-
axis theorem. The theorem states that Iaxis =ICM +M d2, where ICM is the
moment of inertia about the center of mass axis and dis the perpendicular
distance between the two axes.
Step 3: The moment of inertia about the center of mass axis (ICM) can be
calculated using the formula ICM =1
2MR2.
Step 4: The distance dbetween the center of mass axis and the given axis
passing through a point on the surface can be calculated as d=R.
Step 5: Substituting ICM =1
2MR2and d=Rinto the parallel-axis theorem,
we get Iaxis =1
2MR2+MR2.
Step 6: Simplifying, we find Iaxis =3
2MR2.
Therefore, the moment of inertia of the solid cylinder about an axis passing
through a point on the surface perpendicular to the central axis is 3
2MR2.
27
Question 35
Question
Calculate the moment of inertia of a thin-walled hollow cylinder with inner
radius r1, outer radius r2, and height habout its central axis.
Solution
Step 1: The moment of inertia of a thin-walled hollow cylinder about its central
axis can be calculated using the formula:
I=1
2m(r2
1+r2
2)
where mis the mass of the cylinder. Since the cylinder is hollow, we need to
consider the difference in mass between the outer and inner cylinders.
Step 2: The mass of the thin-walled hollow cylinder can be calculated as
the mass of the outer cylinder minus the mass of the inner cylinder. The mass
mcan be calculated using the density ρof the material, the height hof the
cylinder, and the volume formula for a cylinder:
m=ρ×Volume
Step 3: The volume of the outer cylinder is (πr2
2−πr2
1)hand the volume of
the inner cylinder is πr2
1h, so the mass mof the hollow cylinder is:
m=ρπr2
2h−πr2
1h
Step 4: Substituting the mass expression back into the moment of inertia
formula gives:
I=1
2ρπr2
2h−πr2
1h(r2
1+r2
2)
Step 5: Simplifying the expression gives the final formula for the moment of
inertia of the thin-walled hollow cylinder about its central axis:
I=1
2ρhπ(r2
1+r2
2)(r2
2−r2
1)
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