Question

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).

• A. $\frac{(M-m)g}{\left(\frac{13}{6}\right)M + m}$
• B. $\frac{(M-m)g}{\left(\frac{8}{3}\right)M + m}$
• C. $\frac{(M-m)g}{2M + m}$
• D. $\frac{(M-m)g}{M + m}$

Hint:

For Atwood machines with massive pulleys, the effective mass added by the pulley is always $I/R^2$.

Answer 1

The Correct Option is B

To solve this problem, we need to determine the acceleration of the blocks using the principles of rotational dynamics and Newton's second law.

Firstly, consider the forces acting on the blocks. Let's denote the tension in the string on the side with mass \(M\) as \(T_1\) and on the side with mass \(m\) as \(T_2\).

For the block with mass \(M\), we have:

\(Mg - T_1 = Ma\) (Equation 1)

For the block with mass \(m\), we have:

\(T_2 - mg = ma\) (Equation 2)

The net torque on the pulley due to the tensions is given by:

\((T_1 - T_2)R = I\alpha\)

where \(I\) is the moment of inertia of the pulley, \(R\) is the radius of the pulley, and \(\alpha = \frac{a}{R}\) is the angular acceleration.

The moment of inertia of the pulley consists of the rim and the rods:

\(I_{\text{rim}} = MR^2\)

\(I_{\text{rod}} = 2 \times \frac{1}{3}M(2R)^2 = \frac{8}{3}MR^2\)

So, the total moment of inertia \(I\) is:\)

\(I = MR^2 + \frac{8}{3}MR^2 = \left(\frac{11}{3}\right)MR^2\)

Substituting this in the torque equation gives:

\((T_1 - T_2)R = \left(\frac{11}{3}\right)MR^2 \frac{a}{R}\)

Simplifying, we get:

\(T_1 - T_2 = \left(\frac{11}{3}\right)Ma\) (Equation 3)

Adding Equations 1 and 2, we get:

\((M-m)g = Ma + ma + T_2 - T_1\)

Rearranging and using Equation 3:

\((M-m)g = Ma + ma + \left(\frac{11}{3}\right)Ma\)

\((M-m)g = \left(\frac{11}{3}M + m\right)a\)

Therefore, the acceleration \(a\) is: 

\(a = \frac{(M-m)g}{\left(\frac{8}{3}\right)M + m}\)

This is the correct option, indicating the magnitudes of the acceleration experienced by the blocks.