Question

A block of mass \(m\) is hanging by a rope tied to a rotating solid disc of mass \(M\) and radius \(R\) as shown in the figure. If \(\alpha\) is the angular acceleration of the disc, then the linear acceleration of the block is

• A. \(\frac{m}{m+M}g\)
• B. \(\frac{m}{2m+M}g\)
• C. \(\frac{m}{m+\frac{M}{2}}g\)
• D. \(\frac{2ma}{m+M}g\)

Hint:

For a block attached to a rotating pulley or disc, combine translational motion \(mg-T=ma\) with rotational motion \(TR=I\alpha\), and use the no-slip condition \(a=R\alpha\).

Answer 1

The Correct Option is C

Step 1: Set up the free body diagram for the hanging block.
Let $a$ be the linear acceleration of the block (downward) and $T$ be the tension in the rope. For the block of mass $m$, applying Newton's second law in the vertical direction (taking downward as positive): \[ mg - T = ma \] The weight $mg$ drives the block downward while tension $T$ opposes it.
Step 2: Relate the linear acceleration of the block to the angular acceleration of the disc.
The rope unwinds from the rim of the disc of radius $R$. Therefore, the linear acceleration $a$ of the block equals the tangential acceleration at the rim of the disc: \[ a = R\alpha \] where $\alpha$ is the angular acceleration of the disc. This kinematic constraint links the translational and rotational parts of the system.
Step 3: Find the moment of inertia of the solid disc.
For a uniform solid disc of mass $M$ and radius $R$, the moment of inertia about the central axis is: \[ I = \frac{1}{2}MR^2 \] This standard result comes from integrating $r^2 dm$ over the disc; the factor of 1/2 reflects how the mass is distributed closer to the center on average.
Step 4: Apply Newton's rotational law to the disc.
The tension $T$ in the rope acts tangentially at the rim, creating a torque $TR$ on the disc. Using $\tau = I\alpha$: \[ TR = \frac{1}{2}MR^2 \alpha \] Dividing both sides by $R$: \[ T = \frac{1}{2}MR\alpha \] Using $a = R\alpha$: \[ T = \frac{1}{2}Ma \]
Step 5: Substitute tension into the block's equation.
From Step 1, $mg - T = ma$. Substituting $T = Ma/2$: \[ mg - \frac{1}{2}Ma = ma \] \[ mg = a\!\left(m + \frac{M}{2}\right) \] Solving for $a$: \[ a = \frac{mg}{m + \frac{M}{2}} = \frac{m}{m + \frac{M}{2}} g \]
Step 6: Interpret the result and verify limiting cases.
If $M \to 0$ (massless disc), then $a \to g$ (block falls freely). If $M \to \infty$ (very heavy disc), then $a \to 0$ (the block barely accelerates). The disc's rotational inertia effectively adds $M/2$ to the system's effective mass, which is the rotational contribution to resist acceleration. \[ \boxed{a = \frac{m}{m + \frac{M}{2}} g} \]