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 \[F-\frac{3ma}{2}\]