Question

A mass m supported by massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?
\(\{g\) = acceleration due to gravity\(\}\)

• A. \(\frac{5g}{6}\)
• B. \(\frac{2g}{3}\)
• C. \(\frac{g}{2}\)
• D. \(g\)

Hint:

Write Newton's second law for the block and the torque equation for the cylinder, then use the no-slip condition.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The block falls while the string unwinds from the hollow cylinder of moment of inertia \(I=mR^{2}\).
No slipping gives \(a=R\alpha\).

Step 2: Block
\(mg-T=ma\)

Step 3: Cylinder
\(TR=I\alpha=mR^{2}\cdot\dfrac{a}{R}\), so \(T=ma\).

Step 4: Solve
Substitute: \(mg-ma=ma\), so \(a=\dfrac{g}{2}\), option (C).

Final Answer:
The mass falls with acceleration \(\dfrac{g}{2}\). \[ \boxed{\text{(C) }\dfrac{g}{2}} \]