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:

A) \[ma\]               

B)        \[\frac{5}{3}\,ma\]

C) \[\frac{3}{2}\,ma\]        

D)        \[2\,ma\]

Correct Answer: C

Solution :

[c] From figure, \[ma=F-f\]                                 ?(i) Mass = m   And, torque \[\tau =I\alpha \]             \[\frac{m{{R}^{2}}}{2}\alpha =fR\]             \[\frac{m{{R}^{2}}}{2}\frac{a}{R}=fR\]          \[\left[ \therefore \alpha \frac{a}{R} \right]\]             \[\frac{ma}{2}=f\]                                 ?(ii) Put this value in equation (i), \[ma=F-\frac{ma}{2}\] or \[F-\frac{3ma}{2}\]