A uniform disc of mass 0.5 kg and radius r is projected with velocity 18 m / s at t=0 s on a rough horizontal surface. It starts off with a purely sliding motion at t=0 s. After 2 s it acquires a purely rolling motion (see figure). The total kinetic energy of the disc after 2 s will be __________ J (given, coefficient of friction is 0.3 and g=10 m / s² ).

Solution

Physics sir · 3 steps

Friction slows the centre and spins the disc up; find v at 2 s, then add both kinetic energies.

1. Slowing the centre. Kinetic friction f = μ mg acts backwards on the sliding disc, so the centre decelerates at a = μ g = 0.3 × 10 = 3 m s⁻². After 2 s, v = 18 - 3 × 2 = 12 m s⁻¹.
2. Check the rolling. The same friction gives a torque μ mg r, so α = (μ mg r)/(1/2mr²) = (2μ g)/r = 6/r. After 2 s, ω = 12/r, so ω r = 12 m s⁻¹ = v. Pure rolling has started, as the figure says.

   K = 1/2mv² + 1/2(1/2mr²)ω² = 3/4mv²

3. So. K = 3/4 × 0.5 × 12² = 0.375 × 144 = 54 J. The answer is 54.

The common slip: 36. Counting only the translational part, 1/2mv², gives 36 J and leaves out the rotational energy of the spinning disc.

The idea to keep. For a body that slides and then starts rolling, the same friction force slows v and builds up ω. Once v = ω r, a disc's total kinetic energy is 3/4mv², not 1/2mv².

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