Kinetic Energy of a Rigid Body Rotating About a Given Axis

Consider a rigid body rotating about a fixed axis with an angular speed \(\omega\). The \(i\) th particle is going in a circle of radius \(r_i\) with a linear speed \(v_i=\omega r_i\). 
For a particle at a distance from the axis, the linear velocity is \(v_i=r_i \omega\). The kinetic energy of motion of this particle is
\(
k_i=\frac{1}{2} m_i v_i^2=\frac{1}{2} m_i r_i^2 \omega^2
\)

where \(m_i\) is the mass of the particle. The total kinetic energy \(K\) of the body is then given by the sum of the kinetic energies of individual particles,
\(
K=\sum_{i=1}^n k_i=\frac{1}{2} \sum_{i=1}^n\left(m_i r_i^2 \omega^2\right)
\)
Here \(n\) is the number of particles in the body. Note \(\omega\) is the same for all particles. Hence, taking \(\omega\) out of the sum,
\(
K=\frac{1}{2} \omega^2\left(\sum_{i=1}^n m_i r_i^2\right)
\)
We define a new parameter characterising the rigid body, called the moment of inertia \(I\), given by
\(
I=\sum_{i=1}^n m_i r_i^2 \dots(7.34)
\)
With this definition,
\(
K=\frac{1}{2} I \omega^2 \dots(7.35)
\)
Note that the parameter \(I\) is independent of the magnitude of the angular velocity. It is a characteristic of the rigid body and the axis about which it rotates. Sometimes it is called rotational kinetic energy. It is not a new kind of kinetic energy as is clear from the derivation. It is the sum of \(\frac{1}{2} m v^2\) of all the particles.

Compare Eq. (7.35) for the kinetic energy of a rotating body with the expression for the kinetic energy of a body in linear (translational) motion,
\(K=\frac{1}{2} m v^2\)
Here, \(m\) is the mass of the body and \(v\) is its velocity. We have already noted the analogy between angular velocity \(\omega\) (in respect of rotational motion about a fixed axis) and linear velocity \(v\) (in respect of linear motion). 

Example 7.16: A wheel is rotating at an angular speed \(\omega\) about its axis which is kept vertical. An identical wheel initially at rest is gently dropped into the same axle and the two wheels start rotating with a common angular speed. Find this common angular speed.

Solution:

Let the moment of inertia of the wheel about the axis be \(I\). Initially, the first wheel is rotating at the angular speed \(\omega\) about the axle and the second wheel is at rest. Take both the wheels together as the system.
The total angular momentum of the system before the coupling is \(I \omega+0=I \omega\). When the second wheel is dropped into the axle, the two wheels slip on each other and exert forces of friction. The forces of friction have torques about the axis of rotation but these are torques of internal forces. No external torque is applied on the two-wheel system and hence the angular momentum of the system remains unchanged. If the common angular speed is \(\omega^{\prime}\), the total angular momentum of the two-wheel system is \(2 I \omega^{\prime}\) after the coupling. Thus,
\(
\begin{aligned}
& I \omega=2 I \omega^{\prime} \\
& \omega^{\prime}=\omega / 2
\end{aligned}
\)
\(
\text { or, } \quad \omega^{\prime}=\omega / 2 \text {. }
\)

Example 7.17: A wheel of moment of inertia \(I\) and radius \(r\) is free to rotate about its centre as shown in figure below. A string is wrapped over its rim and a block of mass \(m\) is attached to the free end of the string. The system is released from rest. Find the speed of the block as it descends through a height \(h\).

Solution:

Let the speed of the block be \(v\) when it descends through a height \(h\). So is the speed of the string and hence of a particle at the rim of the wheel.
The angular velocity of the wheel is \(v / r\) and its kinetic energy at this instant is \(\frac{1}{2} I(v / r)^2\).
Using the principle of conservation of energy, the gravitational potential energy lost by the block must be equal to the kinetic energy gained by the block and the wheel. Thus,

\(
\begin{aligned}
& m g h=\frac{1}{2} m v^2+\frac{1}{2} I \frac{v^2}{r^2} \\
& \text { or, } \quad v=\left[\frac{2 m g h}{m+I / r^2}\right]^{1 / 2} \text { } \\
&
\end{aligned}
\)

Angular Impulse

The angular impulse of a torque in a given time interval is defined as
\(
J=\int_{t_1}^{t_2} \tau d t
\)
If \(\tau\) be the resultant torque acting on a body
\(
\tau=\frac{d L}{d t}, \text { or, } \tau d t=d L
\)
Integrating this
\(
J=L_2-L_1
\)
Thus, the change in angular momentum is equal to the angular impulse of the resultant torque.

Power Delivered and Work done by a Torque

Consider a rigid body rotating about a fixed axis on which a torque acts. The torque produces angular acceleration and the kinetic energy increases. The rate of increase of the kinetic energy equals the rate of doing work on it, i.e., the power delivered by the torque.
\(
\begin{aligned}
P & =\frac{d W}{d t}=\frac{d K}{d t} \\
& =\frac{d}{d t}\left(\frac{1}{2} I \omega^2\right)=I \omega \frac{d \omega}{d t}=I \alpha \omega=\tau \omega .
\end{aligned}
\)
The work done in an infinitesimal angular displacement \(d \theta\) is
\(
d W=P d t=\tau \omega d t=\tau d \theta .
\)
The work done in a finite angular displacement \(\theta_1\) to \(\theta_2\) is
\(
W=\int_{\theta_1}^{\theta_2} \tau d \theta
\)

Calculation Of Moment Of Inertia

We have defined the moment of inertia of a system about a given line as
\(
I=\sum_{i=1}^n m_i r_i^2
\)
where \(m_i\) is the mass of the \(i\) th particle and \(r_i\) is its perpendicular distance from the given line. If the system is considered to be a collection of discrete particles, this definition may directly be used to calculate the moment of inertia.

Example 7.18: Consider a light rod with two heavy mass particles at its ends. Let \(A B\) be a line perpendicular to the rod as shown in the figure below). What is the moment of inertia of the system about \(A B\)?

Solution:

Moment of inertia of the particle on the left is \(m_1 r_1^2\)
Moment of inertia of the particle on the right is \(m_2 r_2^2\)
The moment of inertia of the rod is negligible as the rod is light.
Thus, the moment of inertia of the system about \(A B\) is \(m_1 r_1^2+m_2 r_2^2\)

Example 7.19: Three particles, each of mass \(m\), are situated at the vertices of an equilateral triangle \(A B C\) of side \(L\) (figure below). Find the moment of inertia of the system about the line \(A X\) perpendicular to \(A B\) in the plane of \(A B C\).

Solution:

Perpendicular distance of \(A\) from \(A X=0\)
Perpendicular distance of \(B\) from \(A X=L\)
Perpendicular distance of \(C\) from \(A X=L/2\)
Thus, the moment of inertia of the particle at \(A=0\), of the particle at \(B=m L^2\), and of the particle at \(C=m(L / 2)^2\). The moment of inertia of the three-particle system about \(A X\) is
\(
0+m L^2+m(L / 2)^2=\frac{5 m L^2}{4} .
\)
Note that the particles on the axis do not contribute to the moment of inertia.

Moment of Inertia of Continuous Mass Distributions

If the body is assumed to be continuous, one can use the technique of integration to obtain its moment of inertia about a given line. Consider a small element of the body. The element should be so chosen that the perpendiculars from different points of the element to the given line differ only by infinitesimal amounts. Let its mass be \(d m\) and its perpendicular distance from the given line be \(r\). Evaluate the product \(r^2 d m\) and integrate it over the appropriate limits to cover the whole body. Thus,
\(
I=\int r^2 d m
\)
under proper limits.
We can call \(r^2 d m\) the moment of inertia of the small element. Moment of inertia of the body about the given line is the sum of the moments of inertia of its constituent elements about the same line.

Case-A: Uniform rod about a perpendicular bisector

Consider a uniform rod of mass \(M\) and length \(l\) (figure below) and suppose the moment of inertia is to be calculated about the bisector \(A B\). Take the origin at the middle point \(O\) of the rod. Consider the element of the rod between a distance \(x\) and \(x+d x\) from the origin. As the rod is uniform,
Mass per unit length of the \(\operatorname{rod}=M / l\)
so that the mass of the element \(=(M / l) d x\)

The perpendicular distance of the element from the line \(A B\) is \(x\). The moment of inertia of this element about \(A B\) is
\(
d I=\frac{M}{l} d x x^2
\)
When \(x=-l / 2\), the element is at the left end of the rod. As \(x\) is changed from \(-l / 2\) to \(l / 2\), the elements cover the whole rod.

Thus, the moment of inertia of the entire rod about \(A B\) is
\(
I=\int_{-l / 2}^{l / 2} \frac{M}{l} x^2 d x=\left[\frac{M}{l} \frac{x^3}{3}\right]_{-l / 2}^{l / 2}=\frac{M l^2}{12}
\)

Case-B: Moment of inertia of a rectangular plate about a line parallel to an edge and passing through the centre

The situation is shown in the figure below. Draw a line parallel to \(A B\) at a distance \(x\) from it and another at a distance \(x+d x\). We can take the strip enclosed between the two lines as the small element.

It is “small” because the perpendiculars from different points of the strip to \(A B\) differ by not more than \(d x\). As the plate is uniform, its mass per unit area \(=\frac{M}{b l}\). Mass of the strip \(=\frac{M}{b l} b d x=\frac{M}{l} d x\).
The perpendicular distance of the strip from \(A B=x\). The moment of inertia of the strip about \(A B=d I=\frac{M}{l} d x x^2\). The moment of inertia of the given plate is, therefore,
\(
I=\int_{-l / 2}^{l / 2} \frac{M}{l} x^2 d x=\frac{M l^2}{12} .
\)
The moment of inertia of the plate about the line parallel to the other edge and passing through the centre may be obtained from the above formula by replacing \(l\) by \(b\) and thus,
\(
I=\frac{M b^2}{12}
\)

Case-C: Moment of inertia of a circular ring about its axis (the line perpendicular to the plane of the ring through its centre)

Suppose the radius of the ring is \(R\) and its mass is \(M\). As all the elements of the ring are at the same perpendicular distance \(R\) from the axis, the moment of inertia of the ring is
\(
I=\int r^2 d m=\int R^2 d m=R^2 \int d m=M R^2
\)

Case-D: Moment of inertia of a uniform circular plate about its axis

Let the mass of the plate be \(M\) and its radius \(R\) (figure below). The centre is at \(O\) and the axis \(O X\) is perpendicular to the plane of the plate.

Draw two concentric circles of radii \(x\) and \(x+d x\), both centred at \(O\) and consider the area of the plate in between the two circles.

This part of the plate may be considered to be a circular ring of radius \(x\). As the periphery of the ring is \(2 \pi x\) and its width is \(d x\), the area of this elementary ring is \(2 \pi x d x\). The area of the plate is \(\pi R^2\). As the plate is uniform,
\(
\begin{aligned}
& \text { its mass per unit area }=\frac{M}{\pi R^2} . \\
& \text { Mass of the ring } \quad=\frac{M}{\pi R^2} 2 \pi x d x=\frac{2 M x d x}{R^2}
\end{aligned}
\)
Using the result obtained above for a circular ring, the moment of inertia of the elementary ring about \(O X\) is
\(
d I=\left[\frac{2 M x d x}{R^2}\right] x^2
\)
The moment of inertia of the plate about \(O X\) is
\(
I=\int_0^R \frac{2 M}{R^2} x^3 d x=\frac{M R^2}{2}
\)

Case-E: Moment of inertia of a hollow cylinder about its axis

Suppose the radius of the cylinder is \(R\) and its mass is \(M\). As every element of this cylinder is at the same perpendicular distance \(R\) from the axis, the moment of inertia of the hollow cylinder about its axis is
\(
I=\int R^2 d m=R^2 \int d m=M R^2
\)

Case-F: Moment of inertia of a uniform solid cylinder about its axis

Let the mass of the cylinder be \(M\) and its radius \(R\). Draw two cylindrical surfaces of radii \(x\) and \(x+d x\) coaxial with the given cylinder. Consider the part of the cylinder in between the two surfaces (figure below ). This part of the cylinder may be considered to be a hollow cylinder of radius \(x\). The area of cross-section of the wall of this hollow cylinder is \(2 \pi x d x\). If the length of the cylinder is \(l\), the volume of the material of this elementary hollow cylinder is \(2 \pi x d x l\).

The volume of the solid cylinder is \(\pi R^2 l\) and it is uniform, hence its mass per unit volume is
\(
\rho=\frac{M}{\pi R^2 l} \text { }
\)
The mass of the hollow cylinder considered is
\(
\frac{M}{\pi R^2 l} 2 \pi x d x l=\frac{2 M}{R^2} x d x 
\)
As its radius is \(x\), its moment of inertia about the given axis is
\(
d I=\left[\frac{2 M}{R^2} x d x\right] x^2 
\)
The moment of inertia of the solid cylinder is, therefore,
\(
I=\int_0^R \frac{2 M}{R^2} x^3 d x=\frac{M R^2}{2} 
\)
Note that the formula does not depend on the length of the cylinder.

Case-G: Moment of inertia of a uniform hollow sphere about a diameter

Let \(M\) and \(R\) be the mass and the radius of the sphere, \(O\) it’s centre and \(O X\) the given axis (figure below). The mass is spread over the surface of the sphere and the inside is hollow.

Let us consider a radius \(O A\) of the sphere at an angle \(\theta\) with the axis \(O X\) and rotate this radius about \(O X\). The point \(A\) traces a circle on the sphere. Now change \(\theta\) to \(\theta+d \theta\) and get another circle of somewhat larger radius on the sphere. The part of the sphere between these two circles, shown in the figure, forms a ring of radius \(R \sin \theta\). The width of this ring is \(R d \theta\) and its periphery is \(2 \pi R \sin \theta\). Hence,
the area of the ring \(=(2 \pi R \sin \theta)(R d \theta)\)
Mass per unit area of the sphere \(=\frac{M}{4 \pi R^2}\)
The mass of the ring
\(
=\frac{M}{4 \pi R^2}(2 \pi R \sin \theta)(R d \theta)=\frac{M}{2} \sin \theta d \theta
\)
The moment of inertia of this elemental ring about \(O X\) is
\(
\begin{aligned}
d I & =\left(\frac{M}{2} \sin \theta d \theta\right)(R \sin \theta)^2 \\
& =\frac{M}{2} R^2 \sin ^3 \theta d \theta
\end{aligned}
\)
As \(\theta\) increases from 0 to \(\pi\), the elemental rings cover the whole spherical surface. The moment of inertia of the hollow sphere is, therefore,
\(
\begin{aligned}
I & =\int_0^\pi \frac{M}{2} R^2 \sin ^3 \theta d \theta=\frac{M R^2}{2}\left[\int_0^\pi\left(1-\cos ^2 \theta\right) \sin \theta d \theta\right] \\
& =\frac{M R^2}{2}\left[\int_{\theta=0}^\pi-\left(1-\cos ^2 \theta\right) d(\cos \theta)\right] \\
& =\frac{-M R^2}{2}\left[\cos \theta-\frac{\cos ^3 \theta}{3}\right]_0^\pi=\frac{2}{3} M R^2
\end{aligned}
\)

Alternate method

Consider any particle \(P\) of the surface, having coordinates \(\left(x_i, y_i, z_i\right)\) with respect to the centre \(O\) as the origin (figure below) and \(O X\) as the \(X\)-axis. Let \(P Q\) be perpendicular to \(O X\). Then \(O Q=x_i\). That is the definition of \(x\)-coordinate.

Thus, \(P Q^2=O P^2-O Q^2\)
\(
=\left(x_i^2+y_i^2+z_i^2\right)-x_i^2=y_i^2+z_i^2
\)
The moment of inertia of the particle \(P\) about the \(X\)-axis
\(
=m_i\left(y_i^2+z_i^2\right) \text {. }
\)
The moment of inertia of the hollow sphere about the \(X\)-axis is, therefore,
\(
I_x=\sum_i m_i\left(y_i^2+z_i^2\right)
\)
Similarly, the moment of inertia of the hollow sphere about the \(Y\)-axis is
\(
I_y=\sum_i m_i\left(z_i^2+x_i^2\right)
\)
and about the \(Z\)-axis it is
\(
I_z=\sum_i m_i\left(x_i^2+y_i^2\right)
\)
Adding these three equations we get
\(
\begin{aligned}
I_x+I_y+I_z & =\sum_i 2 m_i\left(x_i^2+y_i^2+z_i^2\right) \\
& =\sum_i 2 m_i R^2=2 M R^2 .
\end{aligned}
\)
As the mass is uniformly distributed over the entire surface of the sphere, all diameters are equivalent. Hence \(I_x, I_y\) and \(I_z\) must be equal.
\(
\text { Thus, } \quad I=\frac{I_x+I_y+I_z}{3}=\frac{2}{3} M R^2 \text { }
\)

Case-H: Moment of inertia of a uniform solid sphere about a diameter

Let \(M\) and \(R\) be the mass and radius of the given solid sphere. Let \(O\) be the centre and \(O X\) the given axis. Draw two spheres of radii \(x\) and \(x+d x\) concentric with the given solid sphere. The thin spherical shell trapped between these spheres may be treated as a hollow sphere of radius \(x\).

The mass per unit volume of the solid sphere
\(
=\frac{M}{\frac{4}{3} \pi R^3}=\frac{3 M}{4 \pi R^3} \text {. }
\)
The thin hollow sphere considered above has a surface area \(4 \pi x^2\) and thickness \(d x\). Its volume is \(4 \pi x^2 d x\) and hence its mass is
\(
\begin{aligned}
& =\left(\frac{3 M}{4 \pi R^3}\right)\left(4 \pi x^2 d x\right) \\
& =\frac{3 M}{R^3} x^2 d x .
\end{aligned}
\)
Its moment of inertia about the diameter \(O X\) is, therefore,
\(
d I=\frac{2}{3}\left[\frac{3 M}{R^3} x^2 d x\right] x^2=\frac{2 M}{R^3} x^4 d x .
\)
If \(x=0\), the shell is formed at the centre of the solid sphere. As \(x\) increases from 0 to \(R\), the shells cover the whole solid sphere.

The moment of inertia of the solid sphere about \(O X\) is, therefore,
\(
I=\int_0^R \frac{2 M}{R^3} x^4 d x=\frac{2}{5} M R^2
\)

Case-I: Moment of Inertia of a thin ring

Consider a thin ring of radius \(R\) and mass \(M\), rotating in its own plane around its centre with angular velocity \(\omega\). Each mass element of the ring is at a distance \(\mathrm{R}\) from the axis, and moves with a speed \(R \omega\). The kinetic energy is therefore,
\(
K=\frac{1}{2} M v^2=\frac{1}{2} M R^2 \omega^2
\)
Comparing with Eq. (7.35) we get \(I=M R^2\) for the ring.

Case-J: A light rod rotating about an axis

Next, take a rigid rod of negligible mass of length of length \(l\) with a pair of small masses, rotating about an axis through the centre of mass perpendicular to the rod (F1g. 7.28). Each mass \(M / 2\) is at a distance \(l / 2\) from the axis. The moment of inertia of the masses is therefore given by
\(
(M / 2)(l / 2)^2+(M / 2)(l / 2)^2
\)
Thus, for the pair of masses, rotating about the axis through the centre of mass perpendicular to the rod
\(
I=M l^2 / 4
\)

Table 7.1 simply gives the moment of inertia of various familiar regular-shaped bodies about specific axes.