Table 7.2 lists quantities associated with linear motion and their analogues in rotational motion. We know that in rotational motion moment of inertia and torque play the same role as mass and force respectively in linear motion. For example, we know that in linear motion, work done is given by \(F d x\), in rotational motion about a fixed axis it should be \(\tau d \theta\), since we already know the correspondence \(\mathrm{d} x \rightarrow \mathrm{d} \theta\) and \(F \rightarrow \tau\).

Before we begin, we note a simplification that arises in the case of rotational motion about a fixed axis. Since the axis is fixed, only those components of torques, which are along the direction of the fixed axis need to be considered in our discussion. Only these components can cause the body to rotate about the axis. A component of the torque perpendicular to the axis of rotation will tend to turn the axis from its position. We specifically assume that there will arise necessary forces of constraint to cancel the effect of the perpendicular components of the (external) torques so that the fixed position of the axis will be maintained. The perpendicular components of the torques, therefore need not be taken into account. This means that for our calculation of torques on a rigid body:

(1) We need to consider only those forces that lie in planes perpendicular to the axis. Forces that are parallel to the axis will give torques perpendicular to the axis and need not be taken into account.

(2) We need to consider only those components of the position vectors which are perpendicular to the axis. Components of position vectors along the axis will result in torques perpendicular to the axis and need not be taken into account.

Work done by a torque

Figure 7.34 shows a cross-section of a rigid body rotating about a fixed axis, which is taken as the z-axis (perpendicular to the plane of the page; see F1g. 7.33). As said above we need to consider only those forces which lie in planes perpendicular to the axis. Let \(\mathbf{F}\) be one such typical force acting as shown on a particle of the body at point \(\mathrm{P}_1\) with its line of action in a plane perpendicular to the axis. For convenience, we call this to be the \(x^{\prime}-y^{\prime}\) plane (coincident with the plane of the page). The particle at \(P_1\) describes a circular path of radius \(r_1\) with centre \(\mathrm{C}\) on the axis; \(\mathrm{CP}_1=r_1\).

In time \(\Delta t\), the point moves to the position \(P_1^{\prime}\). The displacement of the particle \(d \mathbf{s}_1\), therefore, has magnitude \(\mathrm{d} s_1=r_1 \mathrm{~d} \theta\) and direction tangential at \(P_1\) to the circular path as shown. Here \(\mathrm{d} \theta\) is the angular displacement of the particle, \(\mathrm{d} \theta=\angle \mathrm{P}_1 \mathrm{CP}_1^{\prime}\). The work done by the force on the particle is

\(
\mathrm{d} W_1=\mathbf{F}_1, \mathrm{~d} \mathbf{s}_1=F_1 \mathrm{~d} s_1 \cos \phi_1=F_1\left(r_1 \mathrm{~d} \theta\right) \sin \alpha_1
\)
where \(\phi_1\) is the angle between \(\mathbf{F}_1\) and the tangent at \(\mathrm{P}_1\), and \(\alpha_1\) is the angle between \(\mathbf{F}_1\) and the radius vector \(\mathbf{O P}_1 ; \phi_1+\alpha_1=90^{\circ}\).

The torque due to \(\mathbf{F}_1\) about the origin is \(\mathbf{O P}_1 \times \mathbf{F}_1\). Now \(\mathbf{O P_1}=\mathbf{O C}+\mathbf{O P}_1\). [Refer to Fig. 7.17(b).] Since \(\mathbf{O C}\) is along the axis, the torque resulting from it is excluded from our consideration. The effective torque due to \(\mathbf{F}_1\) is \(\boldsymbol\tau_1=\mathbf{C P} \times \mathbf{F}_1\); it is directed along the axis of rotation and has a magnitude \(\tau_1=r_1 F_1 \sin \alpha\), Therefore,
\(
\mathrm{d} W_1=\tau_1 \mathrm{~d} \theta
\)
If there are more than one forces acting on the body, the work done by all of them can be added to give the total work done on the body. Denoting the magnitudes of the torques due to the different forces as \(\tau_1, \tau_2, \ldots\), etc,
\(
\mathrm{d} W=\left(\tau_1+\tau_2+\ldots\right) \mathrm{d} \theta
\)
Remember, the forces giving rise to the torques act on different particles, but the angular displacement \(\mathrm{d} \theta\) is the same for all particles. Since all the torques considered are parallel to the fixed axis, the magnitude \(\tau\) of the total torque is just the algebraic sum of the magnitudes of the torques, i.e., \(\tau=\tau_1+\tau_2+\ldots\). We, therefore, have
\(
\mathrm{d} W=\tau \mathrm{d} \theta \dots(7.41)
\)
This expression gives the work done by the total (external) torque \(\tau\) which acts on the body rotating about a fixed axis. Its similarity with the corresponding expression
\(
\mathrm{d} W=F \mathrm{~d} s
\)
for linear (translational) motion is obvious.
Dividing both sides of Eq. (7.41) by dt gives
\(
\begin{aligned}
& P=\frac{\mathrm{d} W}{\mathrm{~d} t}=\tau \frac{\mathrm{d} \theta}{\mathrm{d} t}=\tau \omega \\
& \text { or } P=\tau \omega \dots(7.42)
\end{aligned}
\)
This is the instantaneous power. Compare this expression for power in the case of rotational motion about a fixed axis with that of power in the case off linear motion,
\(
P=F v
\)
In a perfectly rigid body there is no internal motion. The work done by external torques is, therefore, not dissipated and goes on to increase the kinetic energy of the body. The rate at which work is done on the body is given by Eq. (7.42). This is to be equated to the rate at which kinetic energy increases. The rate of increase of kinetic energy is
\(
\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{I \omega^2}{2}\right)=I \frac{(2 \omega)}{2} \frac{\mathrm{d} \omega}{\mathrm{d} t}
\)
We assume that the moment of inertia does not change with time. This means that the mass of the body does not change, the body remains rigid, and also the axis does not change its position with respect to the body.
Since \(\alpha=\mathrm{d} \omega / \mathrm{d} t\), we get
\(
\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{I \omega^2}{2}\right)=I \omega \alpha
\)
Equating rates of work done and of increase in kinetic energy,
\(
\tau \omega=I \omega \alpha
\)
\(
\tau=I \alpha \dots(7.43)
\)
Eq. (7.43) is similar to Newton’s second law for linear motion expressed symbolically as \(F=m a\)
Just as force produces acceleration, torque produces angular acceleration in a body. The angular acceleration is directly proportional to the applied torque and is inversely proportional to the moment of inertia of the body. In this respect, Eq.(7.43) can be called Newton’s second law for rotational motion about a fixed axis.

Example 7.30: A cord of negligible mass is wound round the \(\mathrm{rim}\) of a flywheel of mass \(20 \mathrm{~kg}\) and radius \(20 \mathrm{~cm}\). A steady pull of \(25 \mathrm{~N}\) is applied on the cord as shown in Fig. 7.35. The flywheel is mounted on a horizontal axle with frictionless bearings.
(a) Compute the angular acceleration of the wheel.
(b) Find the work done by the pull, when \(2 \mathrm{~m}\) of the cord is unwound.
(c) Find also the kinetic energy of the wheel at this point. Assume that the wheel starts from rest.
(d) Compare answers to parts (b) and (c).

Solution:

\(
\begin{array}{ll}
\text { (a) We use } & I \alpha=\tau \\
\text { the torque } & \tau=F R \\
& =25 \times 0.20 \mathrm{Nm}(\text { as } R=0.20 \mathrm{~m}) \\
& =5.0 \mathrm{Nm}
\end{array}
\)
\(I=\) Moment of inertia of flywheel about 1ts
\(
\begin{aligned}
& \text { axis }=\frac{M R^2}{2} \\
& =\frac{20.0 \times(0.2)^2}{2}=0.4 \mathrm{~kg} \mathrm{~m}^2 \\
& \alpha=\text { angular acceleration } \\
& =5.0 \mathrm{~N} \mathrm{~m} / 0.4 \mathrm{~kg} \mathrm{~m}^2=12.5 \mathrm{~s}^{-2} \\
&
\end{aligned}
\)
(b) Work done by the pull unwinding \(2 \mathrm{~m}\) of the cord
\(
=25 \mathrm{~N} \times 2 \mathrm{~m}=50 \mathrm{~J}
\)
(c) Let \(\omega\) be the final angular velocity. The kinetic energy gained \(=\frac{1}{2} I \omega^2\), since the wheel starts from rest. Now, \(\omega^2=\omega_0^2+2 \alpha \theta, \quad \omega_0=0\)
The angular displacement \(\theta\) = length of unwound string / radius of wheel \(=2 \mathrm{~m} / 0.2 \mathrm{~m}=10 \mathrm{rad}\) \(\omega^2=2 \times 12.5 \times 10.0=250(\mathrm{rad} / \mathrm{s})^2\) \(\therefore\) K.E.gained \(=\frac{1}{2} \times 0.4 \times 250=50 \mathrm{~J}\)
(d) The answers are the same, i.e. the kinetic energy gained by the wheel = work done by the force. There is no loss of energy due to friction.