JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A string is wound around a hollow cylinder of mass \(5\, kg\) and radius \(0.5\,m\). If the string is now pulled with a horizontal force of \(40\, N\), and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be ......... \(rad/s^2\) .(Neglect the mass and thickness of the string)
- A \(20\)
- B \(16\)
- C \(12\)
- D \(10\)
Answer & Solution
Correct Answer
(B) \(16\)
Step-by-step Solution
Detailed explanation
\(40 + f = m\left( {R\alpha } \right)\,\,\,\,\,\,\,\,...\left( i \right)\) \(40 \times R - f \times R = mR = {2^\alpha }\) \(40 - f = mR\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)\) From \((i)\) and \((ii)\) \(\alpha = \frac{{40}}{{mR}} = 16\)
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