JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A uniform solid cylindrical roller of mass \('m'\) is being pulled on a horizontal surface with force \(F\) parallel to the surface and applied at its centre. If the acceleration of the cylinder is \('a'\) and it is rolling without slipping then the value of \('F'\) is
- A \(ma\)
- B \(\frac {5}{3}\,ma\)
- C \(\frac {3}{2}\,ma\)
- D \(2\,ma\)
Answer & Solution
Correct Answer
(C) \(\frac {3}{2}\,ma\)
Step-by-step Solution
Detailed explanation
From figure, \(ma = F - f \) ...(i) And, torque \(\tau = I\alpha \)…
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