Consider a 20 kg uniform circular disk of radius 0.2 m. It is pin supported at its center and is at rest initially. The disk is acted upon by a constant force F = 20 N through a massless string wrapped around is periphery as shown in the figure.
Suppose the disk makes n number of revolutions to attain an angular speed of 50 rad s⁻¹.
The value of n, to the nearest integer, is __________.
[Given : In one complete revolution, the disk rotates by 6.28 rad]

Solution

Physics sir · 4 steps

The string's pull gives a steady torque, so treat the spin like constant acceleration and find the angle.

1. Torque and inertia. The string leaves the rim tangentially, so the lever arm is the full radius: τ = FR = 20 × 0.2 = 4 N m. For a uniform disk about its centre, I = 1/2MR² = 1/2 × 20 × 0.04 = 0.4 kg m².
2. Angular acceleration. α = τ/I = 4/0.4 = 10 rad s⁻². It is constant because the force and lever arm do not change.

   ω² = ω₀² + 2αθ ⇒ θ = 50²/(2 × 10) = 125 rad

3. Angle to turns. The disk starts from rest, so ω₀ = 0. One revolution is 6.28 rad, so n = 125/6.28 ≈ 19.9.
4. So. The disk makes about 19.9 revolutions, which rounds to 20.

The common slip: 40. Taking the disk's moment of inertia as MR² (a ring) halves α to 5 rad s⁻², doubles the angle to 250 rad and gives about 40 revolutions.

The idea to keep. Constant torque means constant α, so the linear kinematics equations work with θ, ω, α. Always check which moment of inertia fits the body: a disk is 1/2MR².

Answer from the NTA final answer key. Solution by PrepWiser.