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Question

Physics Question on System of Particles & Rotational Motion

A uniform solid cylindrical roller of mass 'mm' is being pulled on a horizontal surface with force FF parallel to the surface and applied at its centre. If the acceleration of the cylinder is 'aa' and it is rolling without slipping then the value of 'FF' is :

A

mama

B

2ma2\, ma

C

32\frac{3}{2} ma

D

52\frac{5}{2} ma

Answer

32\frac{3}{2} ma

Explanation

Solution


Ff=maF-f=m a...(1)
fr=(mr22)(ar)f^{r}=\left(\frac{m r^{2}}{2}\right)\left(\frac{a}{r}\right)...(2)
f=ma2.f=\frac{m a}{2}. Therefore, F=3ma2F=\frac{3 m a}{2}