A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should be unbinding the strings to achieve a speed of 4 ms⁻¹, is ____________ cm. (take g = 10 ms⁻²)

Solution

Physics sir · 4 steps

The cylinder falls and spins at the same time, so the lost potential energy feeds both kinds of KE.

1. No slipping. The strings are fixed at the top and unwind from the surface, so the cylinder rolls down the strings without slipping. Its speed and spin are linked by v = ω R. The tension does no work, because the point of contact with the string is momentarily at rest.
2. Energy balance. For a solid cylinder, I = 1/2MR². The rotational KE is 1/2 × 1/2MR² × v²/R² = 1/4Mv². This adds to the translational 1/2Mv².

   Mgh = 1/2Mv² + 1/4Mv² = 3/4Mv² ⇒ h = 3v²/4g

3. Put in numbers. h = (3 × 4²)/(4 × 10) = 48/40 = 1.2 m.
4. So. The cylinder must unwind through 1.2 m, which is 120 cm.

The common slip: 80. Leaving out the rotational KE and using v² = 2gh gives 0.8 m, which is 80 cm, not 120.

The idea to keep. For any body unwinding from a string, write Mgh = 1/2Mv²(1 + k²/R²). For a solid cylinder, k²/R² = 1/2, so the body has fallen h = 3v²/4g.

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