Q.A cord of negligible mass is wound round the rim of a fly wheel of mass 20 kg and radius 20 cm. A steady pull of 25 N is applied on the cord as shown in Fig. 6.31. The flywheel is mounted on a horizontal axle with frictionless bearings.

$I=\tfrac12MR^2=\tfrac12\times20\times(0.2)^2=0.4$ kg m$^2$. Torque $\tau=FR=25\times0.2=5.0$ N m, so $\alpha=\tau/I=12.5$ rad/s$^2$. …

The figure shows a flywheel — a heavy wheel of mass $M = 20\ \text{kg}$ and radius $R = 20\ \text{cm}$ — mounted on a fixed axle through its centre. A cord is wound around the rim of the wheel. One end of the cord is attached to the rim; the other end hangs vertically downward, and a steady force $F = 25\ \text{N}$ pulls on it. The cord leaves the rim tangentially, so the force is always perpendicular to the radius at the point of contact.

The physical idea is straightforward: the pull of the cord exerts a torque about the axle, causing the flywheel to rotate. Because the force is constant and always tangential, the torque is constant, and the wheel undergoes uniform angular acceleration. The figure is used to work out the angular acceleration, the tension in the cord (if the cord were massless and the pull were applied directly, the tension equals the applied force), and the resulting motion.

The key relation is the rotational analogue of Newton’s second law:

Here $\tau$ is the net torque about the axis, $I$ is the moment of inertia of the flywheel about that axis, and $\alpha$ is the angular acceleration.

For a solid disc or cylinder of mass $M$ and radius $R$, the moment of inertia about its central axis is

The torque from the tangential force $F$ is

because the lever arm is exactly $R$ (the force is perpendicular to the radius). There is no other torque acting about the axle (we ignore friction in this idealised problem). So

Cancelling one factor of $R$ gives

and solving for $\alpha$:

Plugging in the numbers — $F = 25\ \text{N}$, $M = 20\ \text{kg}$, $R = 0.20\ \text{m}$ — yields

The figure also sets up the idea that the cord unwinds without slipping. The linear acceleration of a point on the rim — and therefore the acceleration of the cord as it is pulled — is $a = \alpha R$. So

This connects the rotational dynamics to the linear motion of the cord, a link that appears repeatedly in problems involving pulleys, yo-yos, and rolling objects.

In short, the figure is a clean, minimal illustration of how a tangential force causes rotational acceleration, and it provides the numbers for a concrete calculation that reinforces the formula $\tau = I\alpha$. …

Concept: Rotational Dynamics and the Work-Energy Theorem for a Rigid Body

Step 1: Find the moment of inertia of the flywheel (treated as a uniform disc)

Step 2: (a) Find the torque and angular acceleration

The pull $F=25$ N acts tangentially at the rim: $\tau = FR = 25\times0.20 = 5.0$ N m.

Step 3: (b) Find the work done by the pull

The force is constant and tangential, so work $=$ force $\times$ distance unwound:

Step 4: (c) Find the kinetic energy gained

Angular displacement for 2 m unwound: $\theta = s/R = 2/0.20 = 10$ rad.

Starting from rest: $\omega^2=2\alpha\theta = 2(12.5)(10)=250$. …