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JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • JEE Mains 2019
  • Medium
  • Numerical

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)

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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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