PHYS 101 - ELEMENTS OF PHYSICS Moment of Inertia Question Bank Set 2

PHYS 101 - ELEMENTS OF PHYSICS

- Moment of Inertia

Question Bank - Set 2

Liberty University

Question 1

Question

A thin uniform rod of length Land mass Mis rotated about an axis perpen-

dicular to the rod that passes through one end of the rod. Find the moment of

inertia of the rod about this axis.

Solution

To find the moment of inertia of the rod, we need to consider the rotational

inertia contributed by each infinitesimally small mass element dm of the rod

and then integrate over the length of the rod.

Let m=M

Lbe the mass per unit length of the rod.

Step 1: Consider an infinitesimal mass element dm at a distance xfrom

the end of the rod.

The moment of inertia of this element about the axis of rotation is given by

dI =x2dm.

The mass element dm can be expressed in terms of dx as dm =m dx.

Substitute dm =m dx and x=xinto dI =x2dm to get dI =x2m dx.

Step 2: To find the total moment of inertia Iof the entire rod, integrate

dI from x= 0 to x=L.

I=ZL

x2m dx

Step 3:

I=mZL

x2dx =m1

3x3L

=mL3

3=ML2

Step 4: Therefore, the moment of inertia of the thin uniform rod about the

axis perpendicular to the rod that passes through one end is ML2

Question 2

Question

Find the moment of inertia of a thin rod of length Land mass Mabout an axis

perpendicular to the rod and passing through one of its ends.

Solution

To find the moment of inertia of the thin rod about an axis passing through

one of its ends and perpendicular to the rod, we can use the formula for the

moment of inertia of a continuous distribution of mass:

I=Zr2dm

To express dm in terms of dx (an element of length along the rod), we note

that dm

dx represents the mass per unit length of the rod. Since the mass of the

rod is Mand its length is L, we have dm

dx =M

L. Thus, dm =M

Ldx.

Now, consider a small element of the rod of length dx at a distance xfrom

the end of the rod. The distance of this element from the axis of rotation passing

through the end of the rod is r=x. Therefore, the moment of inertia of this

element is dI =r2×dm =x2×M

Ldx.

Since the rod extends from x= 0 to x=L, we integrate these contributions

over the length of the rod to find the total moment of inertia:

I=ZL

x2×M

Ldx

Step 1: Simplify the expression inside the integral.

The integral becomes:

I=M

LZL

x2dx

Step 2: Perform the integration.

I=M

Lx3

3L

I=M

LL3

3−03

3

I=M

L×L3

I=ML2

Therefore, the moment of inertia of the thin rod about an axis passing

through one of its ends and perpendicular to the rod is ML2

Question 3

Question

Find the moment of inertia about the x-axis of the region bounded by the curves

y=x2and y=x3.

Solution

Step 1: First, we need to find the points of intersection of the two curves. Setting

x2=x3, we find the points where the curves intersect:

x3−x2= 0

x2(x−1) = 0

This gives x= 0 and x= 1.

Step 2: To find the moment of inertia about the x-axis, we will use the

formula:

Ix=Zb

y2dx

Where aand bare the x-values of the points of intersection.

Step 3: We need to express yin terms of xfor the given curves. Since the

region is bounded by y=x2and y=x3,y=x2is the upper curve and y=x3

is the lower curve. Thus, y=x2and y=x3are the respective yexpressions.

Step 4: The moment of inertia about the x-axis is given by:

Ix=Z1

(x2−x3)2dx

Step 5: Simplify the integrand and then integrate:

Ix=Z1

(x4−2x5+x6)dx

Ix=1

5x5−1

3x6+1

7x71

Ix=1

5−1

3+1

7−(0)

Ix=24

105 =8

Therefore, the moment of inertia about the x-axis of the region bounded by

the curves y=x2and y=x3is 8

35 .

Question 4

Question

Calculate the moment of inertia of a uniform rod of mass Mand length Labout

an axis perpendicular to the rod and passing through one end of the rod.

Solution

Let’s consider a rod of mass Mand length L. The moment of inertia of the rod

about an axis perpendicular to the rod and passing through one end of 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 1: We can express dm in terms of Mand dx (the length element along

the rod). Since the rod is of uniform mass distribution, the mass per unit length

is M/L, so dm =M

Ldx.

Step 2: Next, we need to express rin terms of x(position along the rod).

The distance rof an element dx from the end of the rod is given by r=L−x.

Step 3: Substitute r=L−xand dm =M

Ldx into the formula for moment

of inertia:

I=Z(L−x)2M

Ldx

Step 4: Simplify the integrand:

I=ZM

L(L2−2Lx +x2)dx

I=Z(M−2mx +nx2)dx

where m=M

Land n=M

L2.

Step 5: Integrate the expression to find the moment of inertia:

I=Mx −mx2+n

3x3+C

Step 6: Evaluate the integral from 0 to L:

I=M(L)−m(L2) + n

3L3−(0 −m·02+n

3·03)

I=ML −ML +M

3L3

I=1

3ML2

Hence, the moment of inertia of the uniform rod about an axis perpendicular

to the rod and passing through one end of the rod is 1

3ML2.

Question 5

Question

Calculate the moment of inertia of a thin rod of length Land mass M, rotating

about an axis perpendicular to the rod and passing through one end.

Solution

Step 1: Consider an element of the rod of length dx at a distance xfrom the

end. The mass of this element is M

Ldx.

Step 2: The moment of inertia dI of this element about the end of the rod

is given by dI =M

Ldxx2.

Step 3: Integrating over the length of the rod, we get the total moment of

inertia I:

I=ZL

0M

Lx2dx

Step 4: Solving the integral, we have:

I=M

LZL

x2dx

L1

3x3L

L1

3L3−1

3×03

L1

3L3

3ML2

Step 5: Therefore, the moment of inertia of the rod rotating about an axis

perpendicular to the rod and passing through one end is 1

3ML2.

Question 6

Question

A thin uniform rod of length Land mass Mis rotated 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

Let’s denote the moment of inertia of the rod about the given axis as I.

Step 1: Divide the rod into infinitesimally small elements of length dx and

mass dm.

Mass of the element, dm =M

Ldx

Step 2: The moment of inertia of the element about the given axis is given

by dI =r2dm, where ris the perpendicular distance of the element from the

axis.

Step 3: For the element at a distance xfrom the end of the rod, r=L−x.

Step 4: Substitute dm =M

Ldx and r=L−xinto the equation for dI to

get dI = (L−x)2M

Ldx.

Step 5: Integrate dI from 0 to Lto find the total moment of inertia Iof

the rod about the given axis.

I=ZL

(L−x)2M

Ldx

Step 6: Simplify the integral and solve for I.

I=1

3ML2

Therefore, the moment of inertia of the rod about the given axis is 1

3ML2.

Question 7

Question

Calculate the moment of inertia of a thin uniform rod of mass mand length L

about an axis perpendicular to the rod passing through one end.

Solution

Let’s divide the rod into small elements ∆mof length ∆xat a distance xfrom

the end where the axis passes through. The mass of each element is given by

∆m=m

L∆x. The distance of this element from the axis passing through one

end is x.

The moment of inertia of each element is given by dI = ∆m·x2=m

L∆x·x2.

The total moment of inertia of the rod can be found by integrating this

expression from 0 to L:

I=ZL

Lx2dx

Step 1: Integrate the expression RL

Lx2dx.

I=m

LZL

x2dx

Step 2: Evaluate the integral:

I=m

L1

3x3L

Step 3: Substitute the limits of integration:

I=m

L1

3L3−0

Step 4: Simplify the expression:

I=m

L·1

3L3

Step 5:

I=1

3mL2

Therefore, the moment of inertia of the thin uniform rod about an axis

perpendicular to the rod passing through one end is 1

3mL2.

Question 8

Question

Find the moment of inertia of a thin rod of length Land mass Mabout an

axis passing through one end and perpendicular to the rod. Assume the rod has

constant density.

Solution

Given: Length of the rod, LMass of the rod, M

Let’s consider an elemental strip of width dx at a distance xfrom the end

of the rod.

Step 1: Determine the mass of the elemental strip. The mass of an elemental

strip dm is given by:

dm =M

Ldx

Step 2: Find the moment of inertia of the elemental strip. The moment

of inertia of an elemental strip about an axis perpendicular to the rod passing

through one end is:

dI =dm

L−x·x2

Substitute dm into the equation:

dI =M

L·dx

L·x2

L−x

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

L·x2

L−xdx

I=M

LZL

L−xdx

Integration becomes easier by using partial fractions.

Step 4: Perform partial fraction decomposition. Write x2

L−xas A

L−x+B

(L−x)2.

Solving for Aand B, we get A=−L2

Land B=L2

Step 5: Integrate the partial fractions. Now, integrate the partial fractions:

I=M

LZL

0−L2

L(L−x)+L2

L(L−x)2dx

I=−MZL

L−xdx +MZL

(L−x)2dx

I=−M[ln |L−x|]L

0+M−1

L−xL

I=−M[ln |0| − ln |L|] + M−1

0−L−−1

L

I=−Mln |L|+M1

L

I=ML

2

Question 9

Question

Find the moment of inertia of a thin rectangular plate of mass M, width a, and

length babout an axis passing through one of its corners and perpendicular to

the plate.

Solution

Step 1: Calculate the distance from the corner to the axis of rotation.

Let dbe the distance from the corner to the axis of rotation. This distance

can be calculated using the Pythagorean theorem:

d=pa2+b2

Step 2: Set up the integral for the moment of inertia.

The moment of inertia Ican be calculated by integrating over the mass

distribution of the rectangular plate. We can divide the plate into small rect-

angular strips parallel to the sides of length dx and thickness dy. The mass of

each strip dm is dm =M

ab dx ·dy. The moment of inertia of each strip about the

axis passing through the corner and perpendicular to the plate is dm ·d2.

Step 3: Set up the double integral for the moment of inertia.

The double integral for the moment of inertia can be set up as follows:

I=ZZ d2·dm

Step 4: Express dm in terms of dx and dy.

Since dx and dy are independent variables, dm can be expressed as follows:

dm =M

ab dx ·dy

Step 5: Substitute in the expression for dand integrate.

The moment of inertia can now be expressed as:

I=ZZ (pa2+b2)2·M

ab dx ·dy

I=ZZ M(a2+b2)

ab dx ·dy

Step 6: Evaluate the double integral to find the moment of inertia.

Integrating over the rectangle with sides of length aand b, we get:

I=M(a2+b2)

Therefore, the moment of inertia of the thin rectangular plate about the axis

passing through one of its corners and perpendicular to the plate is M(a2+b2).

Question 10

Question

Find the moment of inertia of a thin-walled hollow cylinder with mass M, inner

radius r1, and outer radius r2about 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

Step 2: Since the mass Mof the hollow cylinder is distributed uniformly, we

can express Min terms of its volume and density. The volume of the cylinder

is given by the difference in the volumes of the outer and inner cylinders:

V=πr2

2−r2

1h

where his the height of the cylinder.

Step 3: If the density of the cylinder is ρ, then the mass Mis:

M=ρV =ρπ r2

2−r2

1h

Step 4: Substitute the expression for Minto the formula for the moment of

inertia I:

I=1

2ρπ r2

2−r2

1hr2

1+r2

2

Step 5: Simplify the expression to get the final moment of inertia of the

thin-walled hollow cylinder:

I=1

2πρh r4

1+r4

2

Question 11

Question

Find the moment of inertia of a thin rod of length Land mass Mabout an axis

perpendicular to the rod and passing through one end of the rod.

Solution

Step 1: Determine the mass per unit length of the rod.

The mass per unit length of the rod is given by M

Step 2: Consider a small element of length dx and mass dm at a distance x

from the end where the axis passes through.

The mass of the element dm is given by dm =M

Ldx.

Step 3: Determine the moment of inertia of the element about the axis of

rotation.

The moment of inertia dI of the element about the axis of rotation is given by

dI =dm ·x2.

Step 4: Express the moment of inertia of the entire rod.

Integrating dI =M

Lx2dx over the length of the rod from 0 to L, we get

I=ZL

Lx2dx

Step 5: Evaluate the integral.

I=M

LZL

x2dx

L1

3x3L

L1

3L3−0

3ML2

Therefore, the moment of inertia of the thin rod about an axis perpendicular

to the rod and passing through one end of the rod is 1

3ML2.

Question 12

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 of its ends.

Solution

Step 1: Begin by defining the moment of inertia for a particle of mass dm at a

distance rfrom the axis of rotation. This can be expressed as dI =r2dm.

Step 2: Express the mass dm in terms of the linear mass density λand a

small length element dx. We have dm =λdx.

Step 3: Determine the distance rof the small mass element dx from the axis

of rotation in terms of xand L. Since we are measuring from one end, r=L−x.

Step 4: Substitute dm and rinto the expression for dI to get dI = (L−

x)2λdx.

Step 5: Integrate dI over the length of the rod from x= 0 to x=Lto find

the total moment of inertia I:

I=ZL

dI =ZL

(L−x)2λdx

Step 6: Simplify the integral:

I=λZL

(L2−2Lx +x2)dx

Step 7: Integrate term by term:

I=λL2x

3−Lx2

2+x3

3

Step 8: Evaluate the integral at the limits of integration:

I=λL3

3−L3

2+L3

3=1

3λL3

Step 9: Finally, recall that λ=M

L, so substitute this back in to find the

moment of inertia Iin terms of Mand L:

I=1

3ML2

Question 13

Question

A thin uniform rod of mass Mand length Lis rotating about an axis passing

through one of its ends and perpendicular to its length. Find the moment of

inertia of the rod about this axis.

Solution

Let’s consider an elemental mass dm at a distance xfrom the end through which

the axis passes. The mass dm of the element is given by:

dm =M

Ldx

The moment of inertia of this elemental mass about the axis is given by:

dI =dm ·x2

To find the total moment of inertia, we integrate dI over the length of the

rod:

I=ZdI =ZL

Lx2dx

Step 1: Calculate dm in terms of dx.

dm =M

Ldx

Step 2: Find dI in terms of dx.

dI =M

Ldx ·x2

Step 3: Integrate dI to find the total moment of inertia.

I=ZL

Lx2dx =M

LZL

x2dx

Step 4: Evaluate the integral.

I=M

L1

3x3L

L1

3L3−0=1

3ML2

Therefore, the moment of inertia of the rod about the given axis is 1

3ML2.

Question 14

Question

Find the moment of inertia of a solid cone with height hand base radius R

about its central axis.

Solution

Step 1: Consider an elemental disk of radius xand thickness dx at a height x

from the vertex of the cone. The mass of this elemental disk can be expressed

in terms of its volume. Step 2: The volume of the elemental disk can be given

by dV =πx2dx. Step 3: The density of the cone is assumed to be constant,

so the mass of the elemental disk dm can be expressed as dm =ρdV , where

ρis the density of the cone. Step 4: Substituting dV into the expression for

dm, we get dm =ρπx2dx. Step 5: The moment of inertia dI of this elemental

disk about the central axis is given by dI =r2dm, where ris the distance of

the elemental disk from the axis. Step 6: Substituting r=xand dm =ρπx2dx

into the expression for dI, we get dI =ρπx4dx. Step 7: To find the total

moment of inertia Iof the cone, integrate dI from 0 to h. Step 8: Therefore,

I=Rh

0ρπx4dx. Step 9: Solving the integral, we get I=1

5ρπh5. Therefore,

the moment of inertia of a solid cone with height hand base radius Rabout its

central axis is 1

5ρπh5.

Question 15

Question

Find the moment of inertia of a solid cylinder of mass Mand radius Rabout

an axis passing through its center and perpendicular to its length.

Solution

Step 1: The moment of inertia for a solid cylinder of radius Rand mass M

about an axis passing through its center and perpendicular to its length can be

calculated using the formula I=1

2MR2.

Step 2: Given that the mass of the cylinder is Mand the radius is R, we

can substitute these values into the formula to find the moment of inertia:

I=1

2MR2

Therefore, the moment of inertia of a solid cylinder of mass Mand radius

Rabout an axis passing through its center and perpendicular to its length is

2MR2.

Question 16

Question

A thin rod of length Land mass Mis rotated around an axis perpendicular to

the rod and passing through one of its ends. 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 will consider

each infinitesimal element of the rod and sum up their individual moments of

inertia.

Step 1: Consider an infinitesimal element 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 of this element about the given axis is given

by dI =r2dm, where ris the distance of the element from the axis of rotation.

In this case, r=x.

Step 3: Substituting dm and rinto the expression for dI, we get:

dI =x2M

Ldx

Step 4: To find the total moment of inertia of the rod, we integrate dI from

x= 0 to x=L:

I=ZL

x2M

Ldx

Step 5: Solving the integral, we get:

I=M

LZL

x2dx

I=M

Lx3

3L

I=M

LL3

3−0

I=M

3L2

Therefore, the moment of inertia of the rod about the given axis is M

3L2.

Question 17

Question

Find the moment of inertia of a solid cylinder of radius Rand mass Mabout

an axis perpendicular to the cylinder passing through its center.

Solution

Step 1: The moment of inertia of a solid cylinder about its central axis can be

calculated using the formula I=1

2MR2.

Step 2: The moment of inertia for a solid cylinder through its central axis is calculated by integrating the infinitesimal masses.

Step 3: Let’s consider an infinitesimally thin disc of radius rand thickness dr within the cylinder.

Step 4: The mass dm of this disc is dm =M

πR2·πr2·dr =M

R2·r2·dr.

Step 5: To find the moment of inertia about the central axis, we need to express the mass element dm in terms of the distance rfrom the central axis.

Step 6: The moment of inertia of this elemental disc about the central axis is dI =dm·r2=M

R2·r2·dr·r2=M

R2·r4·dr.

Step 7: Now, we integrate this over the entire cylinder:

I=ZR

R2·r4·dr =M

R2r5

5R

R2·R5

5=1

5MR2.

Therefore, the moment of inertia of a solid cylinder of radius Rand mass Mabout an axis perpendicular to the cylinder passing through its center is 1

5MR2.

Question 18

Question

A thin uniform square plate of side length aand mass Mis rotated about an

axis through its center perpendicular to its plane. Calculate the moment of

inertia of the square plate with respect to this axis.

Solution

Let’s denote the moment of inertia of the square plate with respect to the axis

passing through its center and perpendicular to its plane as I. We can find I

by integrating the moment of inertia of each small element of the plate.

Step 1: Consider an infinitesimally small square element with side length dx

at a distance xfrom the center. The mass dm of this element can be expressed

as dm =ρ·dx ·dx, where ρ=M

a2is the mass per unit area of the plate.

Step 2: The moment of inertia dI of this small element about the axis is

given by dI =dm ·x2. Substituting the expression for dm into this equation,

we get dI =ρ·dx ·dx ·x2.

Step 3: To find the total moment of inertia I, we need to integrate dI over

the whole area of the plate.

I=Za/2

−a/2

dI =Za/2

−a/2

ρ·x2dx

Step 4: Substituting the expression for ρinto the integral, we have:

I=Za/2

−a/2

a2·x2dx

Step 5: Solving the integral, we get:

I=M

a2Za/2

−a/2

x2dx =M

a2x3

3a/2

−a/2

I=M

a2(a/2)3

3−(−a/2)3

3

I=M

a2·a3

24 =M

24 ·a2

a2=M

So, the moment of inertia of the square plate with respect to the axis passing

through its center and perpendicular to its plane is M

24 .

Question 19

Question

Find the moment of inertia of a thin uniform rod of length Land mass Mabout

an axis perpendicular to the rod passing through one of its ends.

Solution

Let’s consider the rod to be along the x-axis with one end at the origin (0,0)

and the other end at (L,0). To find the moment of inertia, we need to integrate

over the length of the rod.

Step 1: Identify the mass element. Let’s consider a small mass element δm

at a distance xfrom the origin. The mass of this element is proportional to its

length, so δm =M

Ldx.

Step 2: Find the moment of inertia of the mass element. The moment of

inertia of a mass element is given by dI = (δm)x2. Substituting δm into the

equation gives dI =M

Lx2dx.

Step 3: Integrate to find the total moment of inertia. Integrating over the

length of the rod gives the total moment of inertia:

I=ZL

Lx2dx

Step 4: Evaluate the integral.

I=M

LZL

x2dx =M

Lx3

3L

LL3

3−0=M

3L2

Therefore, the moment of inertia of the thin uniform rod about an axis

perpendicular to the rod passing through one of its ends is M

3L2.

Question 20

Question

A thin rod of length Land mass Mis rotating about an axis passing through

one end of the rod perpendicular to the rod. Find the moment of inertia of the

rod about this axis.

Solution

Let’s consider a thin rod of length Land mass Mrotating about an axis passing

through one end of the rod perpendicular to the rod. We will determine the

moment of inertia of the rod about this axis.

Step 1: Identify the formula for moment of inertia of a rod rotating about

an axis passing through one end perpendicular to the rod. The formula for

the moment of inertia of a rod rotating about an axis passing through one end

perpendicular to the rod is given by:

I=1

3ML2

Step 2: Substitute the values into the formula. Given: Length of the rod,

L=LMass of the rod, M=M

Substitute these values into the formula:

I=1

3ML2

Step 3: Calculate the moment of inertia.

I=1

3ML2

Therefore, the moment of inertia of the rod about the axis passing through

one end perpendicular to the rod is 1

3ML2.

Question 21

Question

A thin rod of length Land mass Mis bent at its midpoint to form an L-shape.

The moment of inertia of the bent rod about an axis perpendicular to the plane

of the rod and passing through one end is given by:

I=1

6ML2

Determine the moment of inertia of the bent rod about an axis perpendicular

to the plane of the rod and passing through the corner where the rod is bent.

Solution

Step 1: The moment of inertia of the bent rod can be split into two parts: one

about the axis passing through the corner where the rod is bent (denoted as

Ibent) and the other about the axis passing through one end of the rod (denoted

as Iend). We are given the moment of inertia about the end axis to be 1

6ML2.

Step 2: According to the parallel axis theorem, the moment of inertia about

an axis parallel to and at a distance dfrom an axis passing through the center

of mass is given by:

Ibent =Icm +Md2

Step 3: The center of mass of the bent rod lies at the midpoint of the rod

where the rod is bent. The distance from this center of mass to the corner where

the rod is bent is half the length of the rod, L

Step 4: Substitute the values into the parallel axis theorem equation:

Ibent =1

6ML2+ML

22

Step 5: Simplify the expression:

Ibent =1

6ML2+1

4ML2

Step 6: Combine the terms:

Ibent =5

12ML2

Step 7: Therefore, the moment of inertia of the bent rod about an axis

perpendicular to the plane of the rod and passing through the corner where the

rod is bent is 5

12 ML2.

Question 22

Question

A thin rod of length Land mass Mis rotating about an axis perpendicular to

the rod and passing through one end with angular velocity ω. 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 axis of

rotation. The mass of this element dm is given by:

dm =M

Ldx

Step 2: The moment of inertia dI of this element about the axis of rotation

is given by:

dI =dm ·x2

Step 3: Substituting the expression for dm into the moment of inertia equa-

tion gives:

dI =M

Ldx ·x2

Step 4: To find the total moment of inertia I, we integrate dI from 0 to L:

I=ZL

dI =ZL

Lx2dx

Step 5: Solving the integral gives:

I=M

LZL

x2dx

Step 6: Integrating x2with respect to x:

I=M

Lx3

3L

Step 7: Evaluating the integral at the limits:

I=M

LL3

3−03

3=M

LL3

3

Step 8: Simplifying the expression gives the moment of inertia of the rod

about the given axis as:

I=ML2

Question 23

Question

A uniform rod of length Land mass Mis rotated about an axis perpendicular

to the rod and passing through one end of the rod. 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 axis of

rotation. The mass of this element dm can be expressed as:

dm =M

Ldx

Step 2: The moment of inertia of this element about the axis of rotation is

given by:

dI =dm ·x2=M

Lx2dx

Step 3: To find the total moment of inertia of the rod, we integrate dI from

x= 0 to x=L:

I=ZL

Lx2dx

Step 4: Solving the integral:

I=M

LZL

x2dx

I=M

Lx3

3L

I=M

LL3

3

I=ML2

Step 5: Therefore, the moment of inertia of the rod about the given axis is

ML2

Question 24

Question

A thin rectangular plate of dimensions 2 m by 3 m is rotated about an axis

passing through one of its corners and perpendicular to the plane of the plate.

Find the moment of inertia of the plate about this axis.

Solution

Step 1: Let’s label the axes as follows: the x-axis goes along the 3m side of

the plate, the y-axis goes along the 2m side of the plate, and the z-axis is

perpendicular to the plate and passes through the corner of the plate. The

plate will rotate about the z-axis, which passes through the corner where the

x-axis and y-axis intersect.

Step 2: The moment of inertia of a thin rectangular plate about an axis

passing through one of its corners and perpendicular to the plane of the plate

is given by the formula:

I=1

3m(a2+b2)

where: - mis the mass of the plate, - ais the length of the side of the plate

perpendicular to the axis of rotation (in this case, 3 m), and - bis the length of

the side of the plate parallel to the axis of rotation (in this case, 2 m).

Step 3: To find the mass mof the plate, we need to know the area density

σ(mass per unit area) of the plate.

Step 4: The total area of the plate is given by:

A= 2 ×3 = 6 m2

Step 5: Assuming the plate has uniform density, the mass mof the plate is

given by:

m=σ×A

Step 6: Let’s substitute the values of A,a, and binto the formula for the

moment of inertia:

I=1

3(σ×A)(a2+b2) = 1

3(σ×6)(32+ 22)

Step 7: Simplifying further, we get:

I=1

3(σ×6)(9 + 4) = 1

3(σ×6)(13) = 26σ

Therefore, the moment of inertia of the plate about the given axis is 26σ.

Question 25

Question

Find the moment of inertia of a solid cylinder with radius Rand height habout

an axis passing through its center and perpendicular to its circular base.

Solution

Let’s consider the solid cylinder with radius Rand height h. We can find the

moment of inertia about the central axis by integrating over the volume of the

cylinder.

Step 1: Divide the cylinder into thin disk-shaped slices along its height.

Let’s consider a thin disk-shaped slice with radius r, thickness dr, and dis-

tance yfrom the central axis. The mass of this slice can be represented as

dm =ρdV =ρA dr, where ρis the density of the cylinder and Ais the area of

the disk slice.

Step 2: Find the moment of inertia dI of this disk slice about the central

axis.

The moment of inertia of a thin disk-shaped slice about an axis passing

through its center and perpendicular to its circular base is dI =1

2r2dm.

Substitute dm into the equation and simplify:

dI =1

2r2ρA dr =1

2r2ρπr2dr

Step 3: Integrate dI to find the total moment of inertia Iof the cylinder.

Integrate dI over the entire volume of the cylinder:

I=ZdI =ZR

2r2ρπr2dr

I=1

2ρπ ZR

r4dr

I=1

10ρπR5

Hence, the moment of inertia of the solid cylinder about the central axis is

10 ρπR5.

Question 26

Question

A thin rod of length Land mass Mis rotating about an axis passing through

one of its ends perpendicular to its length, with an angular velocity ω. Find

the moment of inertia of the rod about an axis passing through its center and

perpendicular to its length.

Solution

Step 1: Let’s denote the moment of inertia of the rod about an axis passing

through its center and perpendicular to its length as I.

Step 2: We know that the moment of inertia of a point mass mat a distance

rfrom an axis of rotation is given by Ipoint =mr2.

Step 3: To find the moment of inertia of the rod, we need to consider each

infinitesimally small mass element dm of the rod and sum up its contributions

to the total moment of inertia.

Step 4: Let’s consider a small mass element dm of the rod at a distance x

from the center of the rod. The mass of this element dm can be expressed as

dm =M

Ldx.

Step 5: The moment of inertia dI of this mass element about the center of

the rod is dI =dm ·x2=M

Ldx·x2.

Step 6: Now, we can integrate dI from −L/2 to L/2 to find the total moment

of inertia I:

I=ZL/2

−L/2

dI =ZL/2

−L/2M

Ldx·x2.

Step 7: Evaluating the integral, we get:

I=M

LZL/2

−L/2

x2dx.

Step 8: Solving the integral, we find:

I=M

Lx3

3L/2

−L/2

3LL3

8−(−L)3

8.

Step 9: Simplifying further, we get:

I=M

3LL3

8+L3

= M L2

12.

Step 10: Therefore, the moment of inertia of the rod about an axis passing

through its center and perpendicular to its length is 1

12 ML2.

Question 27

Question

A thin uniform rod of length Land mass Mis bent into a quarter circle, with

the radius of the quarter circle being R. Find the moment of inertia of this bent

rod about an axis passing through one end of the rod and perpendicular to the

plane of the quarter circle.

Solution

Step 1: The bent rod can be thought of as a combination of two shapes: a

line segment and a quarter circle. The moment of inertia of the line segment

is 1

3ML2, and the moment of inertia of the quarter circle about its diameter is

2MR2. We will now find the distance between the axis passing through one

end of the rod and each shape’s center of mass.

Step 2: The center of mass of the line segment is located at L

2from the end

of the rod. The center of mass of the quarter circle lies at a distance of 2R

πalong

the diameter from the point where the axis passes through.

Step 3: To find the effective moment of inertia about the given axis, we

need to use the parallel axis theorem. The moment of inertia of the two shapes

summed together is I=1

3ML2+1

2MR2.

Step 4: The distance between the axis and the center of mass of the line

segment is L

2, so the contribution of the line segment to the total moment of

inertia is 1

3ML2×L

22.

Step 5: The distance between the axis and the center of mass of the quarter

circle is 2R

π, so the contribution of the quarter circle to the total moment of

inertia is 1

2MR2×2R

π2.

Step 6: Adding the contributions of the line segment and quarter circle, we

have: I=1

3ML2×L

22+1

2MR2×2R

π2.

Step 7: Simplifying the expression, we get: I=1

3ML2×L2

4+1

2MR2×4R2

π2.

Step 8: Further simplifying gives us the final result:

I=1

12ML2+2

π2MR2

Question 28

Question

Find the moment of inertia of a thin rod of mass mand length Labout an axis

passing through one end of the rod and perpendicular to the rod.

Solution

Step 1: The moment of inertia of a continuous system can be calculated by

integrating the product of the square of the distance from the axis of rotation

to the mass element and the mass element. Let’s consider a small mass element

of length dx at a distance xfrom the axis passing through one end of the rod.

Step 2: The mass dm of the element is given by dm =m

Ldx, where m

Lis the

linear mass density of the rod.

Step 3: The moment of inertia element dI for this mass element is dI =

x2dm.

Step 4: Substituting dm =m

Ldx into the expression for dI, we get:

dI =x2m

Ldx

Step 5: Integrate the expression for dI from x= 0 to x=Lto find the total

moment of inertia I:

I=ZL

=ZL

x2m

Ldx

=m

3L

3L(L3−0)

=mL2

Step 6: Therefore, the moment of inertia of the thin rod about an axis

passing through one end of the rod and perpendicular to the rod is mL2

Question 29

Question

A thin uniform rod of length Land mass Mis rotating 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 an elemental mass dm at a distance xfrom the axis of rotation.

The mass of this element can be expressed as dm =M

Ldx.

Step 1: Write the expression for the moment of inertia of the elemental

mass. The moment of inertia dI of the elemental mass about the given axis is

dI =dm ·x2=M

Lx2dx.

Step 2: Integrate to find the total moment of inertia. To find the total

moment of inertia I, we integrate the expression for dI over the length of the

rod:

I=ZL

Lx2dx

Step 3: Simplify the integral.

I=M

LZL

x2dx =M

Lx3

3L

3LL3−03=M

3L2

Step 4: Evaluate the final expression. Therefore, the moment of inertia of

the rod about the given axis is M

3L2.

Question 30

Question

A thin rod of length Lis bent at its midpoint to form a right angle. Find the

moment of inertia of this bent rod about an axis passing through one end of the

rod and perpendicular to the plane of the bent rod.

Solution

Step 1: Consider the two parts of the bent rod separately. We will calculate the

moment of inertia for each part of the rod and then sum the results to find the

total moment of inertia.

Step 2: Moment of inertia for the straight part of the rod. Let’s denote the

mass per unit length of the rod as λ.

The moment of inertia for a uniform rod of length L/2 and mass per unit

length λabout an axis passing through one end and perpendicular to the rod

is given by:

I1=1

3ml2=1

3(λL

2)(L

2)2=1

24λL3

Step 3: Moment of inertia for the bent part of the rod. Since the bent part

is a right triangle with dimensions L

2and L

2, we can consider it as two rectangles

with dimensions L

2and L

The moment of inertia for each rectangle about an axis through its center

and perpendicular to its plane is given by:

I2=1

12ml2=1

12(λL

2)(L

4)2=1

96λL3

Step 4: Total moment of inertia. The total moment of inertia is the sum of

the moment of inertia for each part:

Itotal =I1+ 2I2=1

24λL3+ 2( 1

96λL3) = 1

12λL3

Therefore, the moment of inertia of the bent rod about the specified axis is

12 λL3.

Question 31

Question

Find the moment of inertia of a thin rod of length Land mass M, rotating

about an axis perpendicular to the rod and passing through one end of the rod.

Solution

Step 1: Divide the rod into small elements ∆miof length ∆xi.

Step 2: The moment of inertia of each element ∆Iiis given by dIi=

∆mi(∆xi)2.

Step 3: To find ∆mi, express it in terms of the linear mass density λ=M

and ∆xi:

∆mi=λ∆xi.

Step 4: Substitute ∆mi=λ∆xiinto the expression for ∆Iito get:

∆Ii=λ(∆xi)3.

Step 5: The total moment of inertia Iis the sum of the moments of inertia

of all small elements:

I=X∆Ii=λX(∆xi)3.

Step 6: Replace the sum with an integral to find the total moment of inertia:

I=ZL

λx2dx.

Step 7: Evaluate the integral:

I=λx3

3L

Step 8: Simplify the expression:

I=1

3λL3.

Step 9: Substitute λ=M

Lto find the moment of inertia I:

I=1

3M

LL3=1

3ML2.

Therefore, the moment of inertia of the thin rod rotating about an axis

perpendicular to the rod and passing through one end of the rod is 1

3ML2.

Question 32

Question

Determine the moment of inertia of a solid cylinder of radius Rand mass M

about an axis passing through its center perpendicular to its length.

Solution

Let’s consider a solid cylinder of radius Rand mass M. The moment of iner-

tia Iabout an axis passing through its center perpendicular to its length can

be calculated by integrating the area density function over the volume of the

cylinder.

Step 1: First, we need to determine the area density function of the cylinder.

The volume density function ρcan be defined as M

V, where V=πR2his the

volume of the cylinder and his the height of the cylinder.

Step 2: The elemental mass dm for an elemental volume dV =πR2dy is

given by dm =ρdV . Substituting the values, we get dm =M

πR2hπR2dy =M

hdy.

Step 3: The elemental moment of inertia dI for this elemental mass dm

about an axis passing through its center and perpendicular to its length is given

by dI =r2dm, where ris the perpendicular distance between the elemental

mass dm and the axis of rotation.

Step 4: Since the elemental mass dm is located at a distance yfrom the

axis of rotation, the perpendicular distance is r=pR2+y2. Substituting

r=pR2+y2and dm =M

hdy into the formula for dI, we get

dI = (pR2+y2)2M

hdy.

Step 5: Integrating dI from −h/2 to h/2 to consider the whole volume of

the cylinder, we have

I=Zh/2

−h/2

(pR2+y2)2M

hdy.

Step 6: Solving the integral, we get

I=M

hZh/2

−h/2

(R2+y2)dy.

I=M

h R2Zh/2

−h/2

dy +Zh/2

−h/2

y2dy!.

I=M

h R2(h) + 1

3y3

h/2

−h/2!.

I=M

h R2(h) + 1

3h

23

−1

3−h

23!.

I=M

hR2h+h3

24 +h3

24 .

I=M

hR2h+h3

12 .

I=MR2

h+h

12.

Therefore, the moment of inertia of the solid cylinder about the given axis

is MR2

h+h

12 .

Question 33

Question

A thin, uniform rod of length Land mass Mis rotating about an axis per-

pendicular to the rod and passing through one end. Determine the moment of

inertia of the rod about this axis.

Solution

Let’s consider the rod as a collection of small mass elements dm. The moment

of inertia of the entire rod can then be found by summing up the moments of

inertia of all these small mass elements.

Step 1: Express the differential mass element dm in terms of dx. The mass

of the rod is uniformly distributed, so the linear mass density λis given by

λ=M

The differential mass element dm is given by

dm =λ dx =M

Ldx.

Step 2: Express the moment of inertia of the differential mass element dI.

The moment of inertia of a differential mass element dI about the axis passing

through one end is given by

dI =r2dm,

where ris the distance of dm from the axis of rotation. For this rod, r=x(the

distance from the end of the rod), so

dI =x2M

Ldx.

Step 3: Integrate the expression for dI to find the total moment of inertia

Iof the rod. Integrating both sides, we get

ZdI =Zx2·M

Ldx.

I=M

LZL

x2dx.

I=M

Lx3

3L

I=M

LL3

3−0.

I=M

L·L3

I=1

3ML2.

Therefore, the moment of inertia of the rod about the axis passing through

one end is 1

3ML2.

Question 34

Question

A thin rod of mass Mand length Lis rotating 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 an elemental mass dm of the rod located at a distance xfrom the

end of the rod where the axis of rotation passes. The length of this elemental

mass is dx. The moment of inertia dI of this elemental mass about the axis can

be calculated as dm ·x2.

Step 1: Setting up the integral

I=ZdI

=Zx2dm

Step 2: Expressing dm in terms of dx We know that mass per unit

length λ=M

L. So, dm =λ·dx.

Step 3: Substituting into the integral

I=Zx2(λ dx)

=λZx2dx

=λ1

3x3L

=λ1

3L3

3λL3

3M

LL3

3ML2

Therefore, the moment of inertia of the thin rod about the given axis is

3ML2.

Question 35

Question

A thin rod of length Land mass Mis rotated about one end with an angular

speed ω. Find the moment of inertia of the rod about an axis perpendicular to

the rod passing through its center of mass.

Solution

Step 1: The moment of inertia of a straight rod rotating about an axis perpen-

dicular to the rod through one end is given by the formula I=1

3ML2.

Step 2: To find the moment of inertia about an axis perpendicular to the rod

passing through its center of mass, we will use the parallel axis theorem which

states Inew =Icm +MD2, where Inew is the moment of inertia about a new

axis, Icm is the moment of inertia about the center of mass, Dis the distance

between the two axes, and Mis the mass of the object.

Step 3: Since the axis passing through the center of mass is located at the

midpoint of the rod, the distance Dbetween the two axes is L/2.

Step 4: The moment of inertia about the center of mass is Icm =1

12 ML2for

a rod rotating about its center.

Step 5: Substituting Icm =1

12 ML2,M, and D=L/2 into the parallel axis

theorem equation, we get

Inew =1

12ML2+ML2/4 = 1

3ML2.

Question 2

Question

Find the moment of inertia of a thin rod of length Land mass Mabout an axis

perpendicular to the rod and passing through one of its ends.

Solution

To find the moment of inertia of the thin rod about an axis passing through

one of its ends and perpendicular to the rod, we can use the formula for the

moment of inertia of a continuous distribution of mass:

I=Zr2dm

To express dm in terms of dx (an element of length along the rod), we note

that dm

dx represents the mass per unit length of the rod. Since the mass of the

rod is Mand its length is L, we have dm

dx =M

L. Thus, dm =M

Ldx.

Now, consider a small element of the rod of length dx at a distance xfrom

the end of the rod. The distance of this element from the axis of rotation passing

through the end of the rod is r=x. Therefore, the moment of inertia of this

element is dI =r2×dm =x2×M

Ldx.

Since the rod extends from x= 0 to x=L, we integrate these contributions

over the length of the rod to find the total moment of inertia:

I=ZL

x2×M

Ldx

Step 1: Simplify the expression inside the integral.

The integral becomes:

I=M

LZL

x2dx

Step 2: Perform the integration.

I=M

Lx3

3L

I=M

LL3

3−03

3

I=M

L×L3

I=ML2

Therefore, the moment of inertia of the thin rod about an axis passing

through one of its ends and perpendicular to the rod is ML2

Question 3

Question

Find the moment of inertia about the x-axis of the region bounded by the curves

y=x2and y=x3.

Solution

Step 1: First, we need to find the points of intersection of the two curves. Setting

x2=x3, we find the points where the curves intersect:

x3−x2= 0

x2(x−1) = 0

This gives x= 0 and x= 1.

Step 2: To find the moment of inertia about the x-axis, we will use the

formula:

Ix=Zb

y2dx

Where aand bare the x-values of the points of intersection.

Step 3: We need to express yin terms of xfor the given curves. Since the

region is bounded by y=x2and y=x3,y=x2is the upper curve and y=x3

is the lower curve. Thus, y=x2and y=x3are the respective yexpressions.

Step 4: The moment of inertia about the x-axis is given by:

Ix=Z1

(x2−x3)2dx

Step 5: Simplify the integrand and then integrate:

Ix=Z1

(x4−2x5+x6)dx

Ix=1

5x5−1

3x6+1

7x71

Ix=1

5−1

3+1

7−(0)

Ix=24

105 =8

Therefore, the moment of inertia about the x-axis of the region bounded by

the curves y=x2and y=x3is 8

35 .

Question 4

Question

Calculate the moment of inertia of a uniform rod of mass Mand length Labout

an axis perpendicular to the rod and passing through one end of the rod.

Solution

Let’s consider a rod of mass Mand length L. The moment of inertia of the rod

about an axis perpendicular to the rod and passing through one end of 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 1: We can express dm in terms of Mand dx (the length element along

the rod). Since the rod is of uniform mass distribution, the mass per unit length

is M/L, so dm =M

Ldx.

Step 2: Next, we need to express rin terms of x(position along the rod).

The distance rof an element dx from the end of the rod is given by r=L−x.

Step 3: Substitute r=L−xand dm =M

Ldx into the formula for moment

of inertia:

I=Z(L−x)2M

Ldx

Step 4: Simplify the integrand:

I=ZM

L(L2−2Lx +x2)dx

I=Z(M−2mx +nx2)dx

where m=M

Land n=M

L2.

Step 5: Integrate the expression to find the moment of inertia:

I=Mx −mx2+n

3x3+C

Step 6: Evaluate the integral from 0 to L:

I=M(L)−m(L2) + n

3L3−(0 −m·02+n

3·03)

I=ML −ML +M

3L3

I=1

3ML2

Hence, the moment of inertia of the uniform rod about an axis perpendicular

to the rod and passing through one end of the rod is 1

3ML2.

Question 5

Question

Calculate the moment of inertia of a thin rod of length Land mass M, rotating

about an axis perpendicular to the rod and passing through one end.

Solution

Step 1: Consider an element of the rod of length dx at a distance xfrom the

end. The mass of this element is M

Ldx.

Step 2: The moment of inertia dI of this element about the end of the rod

is given by dI =M

Ldxx2.

Step 3: Integrating over the length of the rod, we get the total moment of

inertia I:

I=ZL

0M

Lx2dx

Step 4: Solving the integral, we have:

I=M

LZL

x2dx

L1

3x3L

L1

3L3−1

3×03

L1

3L3

3ML2

Step 5: Therefore, the moment of inertia of the rod rotating about an axis

perpendicular to the rod and passing through one end is 1

3ML2.

Question 6

Question

A thin uniform rod of length Land mass Mis rotated 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

Let’s denote the moment of inertia of the rod about the given axis as I.

Step 1: Divide the rod into infinitesimally small elements of length dx and

mass dm.

Mass of the element, dm =M

Ldx

Step 2: The moment of inertia of the element about the given axis is given

by dI =r2dm, where ris the perpendicular distance of the element from the

axis.

Step 3: For the element at a distance xfrom the end of the rod, r=L−x.

Step 4: Substitute dm =M

Ldx and r=L−xinto the equation for dI to

get dI = (L−x)2M

Ldx.

Step 5: Integrate dI from 0 to Lto find the total moment of inertia Iof

the rod about the given axis.

I=ZL

(L−x)2M

Ldx

Step 6: Simplify the integral and solve for I.

I=1

3ML2

Therefore, the moment of inertia of the rod about the given axis is 1

3ML2.

Question 7

Question

Calculate the moment of inertia of a thin uniform rod of mass mand length L

about an axis perpendicular to the rod passing through one end.

Solution

Let’s divide the rod into small elements ∆mof length ∆xat a distance xfrom

the end where the axis passes through. The mass of each element is given by

∆m=m

L∆x. The distance of this element from the axis passing through one

end is x.

The moment of inertia of each element is given by dI = ∆m·x2=m

L∆x·x2.

The total moment of inertia of the rod can be found by integrating this

expression from 0 to L:

I=ZL

Lx2dx

Step 1: Integrate the expression RL

Lx2dx.

I=m

LZL

x2dx

Step 2: Evaluate the integral:

I=m

L1

3x3L

Step 3: Substitute the limits of integration:

I=m

L1

3L3−0

Step 4: Simplify the expression:

I=m

L·1

3L3

Step 5:

I=1

3mL2

Therefore, the moment of inertia of the thin uniform rod about an axis

perpendicular to the rod passing through one end is 1

3mL2.

Question 8

Question

Find the moment of inertia of a thin rod of length Land mass Mabout an

axis passing through one end and perpendicular to the rod. Assume the rod has

constant density.

Solution

Given: Length of the rod, LMass of the rod, M

Let’s consider an elemental strip of width dx at a distance xfrom the end

of the rod.

Step 1: Determine the mass of the elemental strip. The mass of an elemental

strip dm is given by:

dm =M

Ldx

Step 2: Find the moment of inertia of the elemental strip. The moment

of inertia of an elemental strip about an axis perpendicular to the rod passing

through one end is:

dI =dm

L−x·x2

Substitute dm into the equation:

dI =M

L·dx

L·x2

L−x

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

L·x2

L−xdx

I=M

LZL

L−xdx

Integration becomes easier by using partial fractions.

Step 4: Perform partial fraction decomposition. Write x2

L−xas A

L−x+B

(L−x)2.

Solving for Aand B, we get A=−L2

Land B=L2

Step 5: Integrate the partial fractions. Now, integrate the partial fractions:

I=M

LZL

0−L2

L(L−x)+L2

L(L−x)2dx

I=−MZL

L−xdx +MZL

(L−x)2dx

I=−M[ln |L−x|]L

0+M−1

L−xL

I=−M[ln |0| − ln |L|] + M−1

0−L−−1

L

I=−Mln |L|+M1

L

I=ML

2

Question 9

Question

Find the moment of inertia of a thin rectangular plate of mass M, width a, and

length babout an axis passing through one of its corners and perpendicular to

the plate.

Solution

Step 1: Calculate the distance from the corner to the axis of rotation.

Let dbe the distance from the corner to the axis of rotation. This distance

can be calculated using the Pythagorean theorem:

d=pa2+b2

Step 2: Set up the integral for the moment of inertia.

The moment of inertia Ican be calculated by integrating over the mass

distribution of the rectangular plate. We can divide the plate into small rect-

angular strips parallel to the sides of length dx and thickness dy. The mass of

each strip dm is dm =M

ab dx ·dy. The moment of inertia of each strip about the

axis passing through the corner and perpendicular to the plate is dm ·d2.

Step 3: Set up the double integral for the moment of inertia.

The double integral for the moment of inertia can be set up as follows:

I=ZZ d2·dm

Step 4: Express dm in terms of dx and dy.

Since dx and dy are independent variables, dm can be expressed as follows:

dm =M

ab dx ·dy

Step 5: Substitute in the expression for dand integrate.

The moment of inertia can now be expressed as:

I=ZZ (pa2+b2)2·M

ab dx ·dy

I=ZZ M(a2+b2)

ab dx ·dy

Step 6: Evaluate the double integral to find the moment of inertia.

Integrating over the rectangle with sides of length aand b, we get:

I=M(a2+b2)

Therefore, the moment of inertia of the thin rectangular plate about the axis

passing through one of its corners and perpendicular to the plate is M(a2+b2).

Question 10

Question

Find the moment of inertia of a thin-walled hollow cylinder with mass M, inner

radius r1, and outer radius r2about 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

Step 2: Since the mass Mof the hollow cylinder is distributed uniformly, we

can express Min terms of its volume and density. The volume of the cylinder

is given by the difference in the volumes of the outer and inner cylinders:

V=πr2

2−r2

1h

where his the height of the cylinder.

Step 3: If the density of the cylinder is ρ, then the mass Mis:

M=ρV =ρπ r2

2−r2

1h

Step 4: Substitute the expression for Minto the formula for the moment of

inertia I:

I=1

2ρπ r2

2−r2

1hr2

1+r2

2

Step 5: Simplify the expression to get the final moment of inertia of the

thin-walled hollow cylinder:

I=1

2πρh r4

1+r4

2

Question 11

Question

Find the moment of inertia of a thin rod of length Land mass Mabout an axis

perpendicular to the rod and passing through one end of the rod.

Solution

Step 1: Determine the mass per unit length of the rod.

The mass per unit length of the rod is given by M

Step 2: Consider a small element of length dx and mass dm at a distance x

from the end where the axis passes through.

The mass of the element dm is given by dm =M

Ldx.

Step 3: Determine the moment of inertia of the element about the axis of

rotation.

The moment of inertia dI of the element about the axis of rotation is given by

dI =dm ·x2.

Step 4: Express the moment of inertia of the entire rod.

Integrating dI =M

Lx2dx over the length of the rod from 0 to L, we get

I=ZL

Lx2dx

Step 5: Evaluate the integral.

I=M

LZL

x2dx

L1

3x3L

L1

3L3−0

3ML2

Therefore, the moment of inertia of the thin rod about an axis perpendicular

to the rod and passing through one end of the rod is 1

3ML2.

Question 12

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 of its ends.

Solution

Step 1: Begin by defining the moment of inertia for a particle of mass dm at a

distance rfrom the axis of rotation. This can be expressed as dI =r2dm.

Step 2: Express the mass dm in terms of the linear mass density λand a

small length element dx. We have dm =λdx.

Step 3: Determine the distance rof the small mass element dx from the axis

of rotation in terms of xand L. Since we are measuring from one end, r=L−x.

Step 4: Substitute dm and rinto the expression for dI to get dI = (L−

x)2λdx.

Step 5: Integrate dI over the length of the rod from x= 0 to x=Lto find

the total moment of inertia I:

I=ZL

dI =ZL

(L−x)2λdx

Step 6: Simplify the integral:

I=λZL

(L2−2Lx +x2)dx

Step 7: Integrate term by term:

I=λL2x

3−Lx2

2+x3

3

Step 8: Evaluate the integral at the limits of integration:

I=λL3

3−L3

2+L3

3=1

3λL3

Step 9: Finally, recall that λ=M

L, so substitute this back in to find the

moment of inertia Iin terms of Mand L:

I=1

3ML2

Question 13

Question

A thin uniform rod of mass Mand length Lis rotating about an axis passing

through one of its ends and perpendicular to its length. Find the moment of

inertia of the rod about this axis.

Solution

Let’s consider an elemental mass dm at a distance xfrom the end through which

the axis passes. The mass dm of the element is given by:

dm =M

Ldx

The moment of inertia of this elemental mass about the axis is given by:

dI =dm ·x2

To find the total moment of inertia, we integrate dI over the length of the

rod:

I=ZdI =ZL

Lx2dx

Step 1: Calculate dm in terms of dx.

dm =M

Ldx

Step 2: Find dI in terms of dx.

dI =M

Ldx ·x2