question_answer

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: JEE Main Online Paper (Held On 10 April 2015)

A) \[ma\]                                   

B) \[2 ma\]

C)  \[\frac{5}{3}\]                                  

D)  \[\frac{3}{2}ma\]

Correct Answer: D

Solution :

\[F=f=ma\]and \[fR=I\alpha \]for Rolling \[a=\alpha R\] \[F-f=ma\]                                          ??.(i) \[fR=\frac{m{{R}^{2}}}{2}.\frac{a}{R}\]   \[f=\frac{ma}{2}\]                           ?(ii) So, \[f-\frac{ma}{2}=ma\]            \[f=\frac{3}{2}ma\]