Rotational Motion: Real-World Scenarios (2)

medium • 3 min read
Tags Rotational Motion

Question

A potter’s wheel of radius R=0.5R = 0.5 m and moment of inertia I=2I = 2 kg·m2^2 is initially at rest. The potter applies a constant tangential force F=10F = 10 N at the rim for 55 seconds. Find the angular velocity at the end of 55 s and the linear speed of a clay piece sitting at the rim.

Solution — Step by Step

Torque τ=F⋅R=10×0.5=5\tau = F \cdot R = 10 \times 0.5 = 5 N·m. The force is tangential, so sin⁡θ=1\sin\theta = 1.

From τ=Iα\tau = I \alpha:

α=τI=52=2.5 rad/s2\alpha = \frac{\tau}{I} = \frac{5}{2} = 2.5 \text{ rad/s}^2

Starting from rest, ω=αt=2.5×5=12.5\omega = \alpha t = 2.5 \times 5 = 12.5 rad/s.

v=ωR=12.5×0.5=6.25v = \omega R = 12.5 \times 0.5 = 6.25 m/s.

Final answers: ω=12.5\omega = 12.5 rad/s, v=6.25v = 6.25 m/s.

Why This Works

Rotational motion mirrors linear motion equation by equation: τ\tau replaces FF, II replaces mm, α\alpha replaces aa. Once you internalise this analogy, every rotational kinematics question becomes a linear one in disguise.

The clay at the rim moves in a circle; its linear speed is the tangential speed of the rim. Anything stuck on the wheel at radius rr moves at v=ωrv = \omega r.

Alternative Method

Use the work-energy theorem in rotational form. Work done by torque = τθ\tau \theta. Angular displacement in 55 s is θ=12αt2=31.25\theta = \frac{1}{2}\alpha t^2 = 31.25 rad. So W=5×31.25=156.25W = 5 \times 31.25 = 156.25 J. This equals 12Iω2=12(2)(12.5)2=156.25\frac{1}{2} I \omega^2 = \frac{1}{2}(2)(12.5)^2 = 156.25 J. Confirms our ω\omega.

For any “started from rest, constant torque” problem, ω=τtI\omega = \frac{\tau t}{I} in one shot. Memorise this — it appears in JEE Main almost every year in some disguise.

Common Mistake

Students plug rr for RR when the question gives multiple radii (e.g., a stepped pulley). The lever arm depends on where the force is applied, not on the radius of the body. Always re-read which radius the force acts at.

Another trap — using τ=Iα\tau = I \alpha with II about the wrong axis. The wheel rotates about its central axis, so II given is already correct. But if a problem says “uniform disc, mass MM, radius RR” without giving II, you need I=12MR2I = \frac{1}{2} M R^2 for the central axis, not 14MR2\frac{1}{4} M R^2 (that is for a diameter axis).

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next

A solid sphere rolls down incline without slipping — find acceleration

hard

Angular Momentum Conservation — Ice Skater Spinning

easy

Conservation of angular momentum — figure skater spinning faster

medium

A cylinder rolls down an incline without slipping — compare with sliding

hard

Calculate moment of inertia of uniform disc about diameter using perpendicular axis theorem

medium