A solid circular disc of mass 50 kg rolls along a horizontal floor so that its center of mass has a speed of 0.4 m / s. The absolute value of work done on the disc to stop it is ________ J.

Solution

Physics sir · 4 steps

A rolling disc stores energy in its spin too, so the work to stop it equals all of its kinetic energy.

1. Work-energy theorem. The work done on the disc equals the change in its kinetic energy. It ends at rest, so the size of the work is its whole starting kinetic energy.
2. Add both parts. A rolling body has translational energy 1/2mv² and rotational energy 1/2Iω². For a solid disc I = 1/2mR², and rolling without slipping gives ω = v/R, so the rotational part is 1/4mv².

   K = 1/2mv² + 1/4mv² = 3/4mv²

3. Put in numbers. K = 3/4 × 50 × (0.4)² = 3/4 × 50 × 0.16 = 6 J.
4. So. The work done is -6 J, and its absolute value is 6 J.

The common slip: 4. Taking only the translational energy 1/2mv² gives 4 J and leaves out the rotational energy.

The idea to keep. For rolling, write K = 1/2mv²(1 + k²/R²). A solid disc has k²/R² = 1/2, a ring 1 and a solid sphere 2/5. Never drop the spin part.

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