A cord of negligible mass is wound around the rim of a wheel supported by spokes with negligible mass. The mass of wheel is 10 kg and radius is 10 cm and it can freely rotate without any friction. Initially the wheel is at rest. If a steady pull of 20 N is applied on the cord, the angular velocity of the wheel, after the cord is unwound by 1 m , would be:

Solution

Physics sir · 4 steps

Spokes are massless, so all the mass sits on the rim: treat it as a ring and use work-energy.

1. Moment of inertia. The spokes have no mass, so all 10 kg is at distance R = 0.1 m from the axle. That makes the wheel a ring, so I = MR² = 10 × (0.1)² = 0.1 kg m².
2. Work done by pull. The cord unwinds by 1 m under a steady 20 N, so the pull does W = 20 × 1 = 20 J. There is no friction, so all of this becomes rotational kinetic energy.

   F s = 1/2 I ω² ⇒ 20 = 1/2 × 0.1 × ω² ⇒ ω² = 400

3. Cross-check with torque. Torque τ = FR = 2 N m, so α = τ / I = 20 rad s⁻². Unwinding 1 m turns the wheel by θ = s/R = 10 rad. Then ω² = 2αθ = 400, which gives the same result.
4. So. ω = 20 rad/s, which is option (a).

The idea to keep. A wheel with massless spokes is a ring, so use I = MR², not 1/2MR². For a steady pull through a known length of cord, the work done F s equals 1/2Iω².

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