A uniform disc with mass M = 4 kg and radius R = 10 cm is mounted on a fixed horizontal axle as shown in figure. A block with mass m = 2 kg hangs from a massless cord that is wrapped around the rim of the disc. During the fall of the block, the cord does not slip and there is no friction at the axle. The tension in the cord is ____________ N.
(Take g = 10 ms⁻²)

Solution

Physics sir · 3 steps

The cord pulls the block up and spins the disc, so write one equation for each body and join them.

1. Block falls. Gravity pulls the block down and the tension pulls it up, so mg - T = ma.
2. Disc turns. The tension acts at the rim, so its torque is TR = Iα with I = 1/2MR². The cord does not slip, so a = Rα. This gives TR = 1/2MR² × a/R, that is, T = 1/2Ma. The radius cancels.

   mg = (m + M/2)a ⇒ a = (2 × 10)/(2 + 2) = 5 m s⁻²

3. So. T = 1/2Ma = 1/2 × 4 × 5 = 10 N, so the tension is 10.

The common slip: 20. Taking the tension equal to the block's weight, mg, gives 20 N. That ignores the fact that the block is accelerating downward.

The idea to keep. For a block hanging from a cord wrapped on a pulley, write Newton's law for the block and torque for the pulley, then link them with a = Rα. The tension is always less than mg.

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