The pulley shown in figure is made using a thin rim and two rods of length equal to diameter of the rim. The rim and each rod have a mass of M . Two blocks of mass of M and m are attached to two ends of a light string passing over the pulley, which is hinged to rotate freely in vertical plane about its center. The magnitudes of the acceleration experienced by the blocks is ________ (assume no slipping of string on pulley).

Solution

Physics sir · 4 steps

Find the pulley's I first; it joins the masses in the denominator as I/R².

1. Moment of inertia. The thin rim gives MR². Each rod has length 2R and turns about its middle, so it gives M(2R)²/12 = MR²/3. Two rods give 2MR²/3, so in total I = MR² + 2MR²/3 = 5/3MR².
2. Equations of motion. Take M moving down with acceleration a. For M: Mg - T₁ = Ma. For m: T₂ - mg = ma. For the pulley: (T₁ - T₂)R = Iα, and no slipping means α = a/R, so T₁ - T₂ = I/R²a = 5/3Ma.
3. Add them up. Adding all three equations cancels both tensions. This leaves (M - m)g = (M + m + 5/3M)a.

   a = (M-m)g/(8/3M + m)

4. So. The blocks accelerate at (M-m)g/((8/3)M + m), which is option (a).

The idea to keep. With a heavy pulley, a = (net hanging weight)/(m₁ + m₂ + I/R²). Build I piece by piece: a thin rim is MR², and a rod about its centre is ML²/12.

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