Question

If a rope wound around a solid cylinder is allowed to unwind by holding the free end of the rope, then the acceleration with which the cylinder falls down is

• A. \(6.53\,\text{m s}^{-2}\)
• B. \(9.8\,\text{m s}^{-2}\)
• C. \(3.27\,\text{m s}^{-2}\)
• D. \(19.6\,\text{m s}^{-2}\)

Hint:

For a solid cylinder unwinding from a string, \[ \boxed{ a=\frac{2g}{3} } \] since \[ I=\frac12 mR^2. \]

Answer 1

The Correct Option is A

Step 1: Apply translational motion. For the falling cylinder, \[ mg-T=ma. \]

Step 2: Apply rotational motion. Torque due to tension, \[ TR=I\alpha. \] For a solid cylinder, \[ I=\frac12 mR^2, \qquad a=\alpha R. \] Hence, \[ TR=\frac12 mR^2\left(\frac{a}{R}\right), \] or \[ T=\frac12 ma. \]

Step 3: Calculate acceleration. Substituting into the translational equation, \[ mg-\frac12 ma=ma, \] \[ mg=\frac32 ma, \] \[ a=\frac{2g}{3}. \] Using \[ g=9.8\,\text{m s}^{-2}, \] \[ a=\frac{2\times9.8}{3} =6.53\,\text{m s}^{-2}. \] Thus, \[ \boxed{a=6.53\,\text{m s}^{-2}} \] Hence, \[ \boxed{(A)} \] is the correct answer.