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).
Show Hint
For Atwood machines with massive pulleys, the effective mass added by the pulley is always $I/R^2$.
Step 1: Understanding the Concept:
In a real pulley system where the pulley has mass, we must consider its moment of inertia ($I$). The net driving force $(M-m)g$ must accelerate the two blocks as well as provide angular acceleration to the pulley.
Step 2: Key Formula or Approach:
Acceleration $a = \frac{\text{Net Force}}{\text{Total Effective Mass}} = \frac{(M-m)g}{M + m + \frac{I}{R^2}}$.
Step 3: Detailed Explanation:
Calculate the moment of inertia $I$ of the pulley:
1. $I_{rim} = M R^2$.
2. Each rod has mass $M$ and length $L = 2R$. Since they act as diameters, they rotate about their midpoint.
$I_{rod} = \frac{1}{12} M L^2 = \frac{1}{12} M (2R)^2 = \frac{1}{3} M R^2$.
There are two such rods.
Total $I = I_{rim} + 2 I_{rod} = M R^2 + 2 (\frac{1}{3} M R^2) = M R^2 + \frac{2}{3} M R^2 = \frac{5}{3} M R^2$.
Substitute into the acceleration formula:
\[ a = \frac{(M - m) g}{M + m + \frac{\frac{5}{3} M R^2}{R^2}} \]
\[ a = \frac{(M - m) g}{M + m + \frac{5}{3} M} \]
Combine terms with $M$: $M + \frac{5}{3} M = \frac{8}{3} M$.
\[ a = \frac{(M - m) g}{\frac{8}{3} M + m} \]
Step 4: Final Answer:
The acceleration is $\frac{(M-m)g}{\left(\frac{8}{3}\right)M + m}$.
