A uniform solid cylindrical roller of mass m | vedclass

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(C) From the free body diagram of the cylinder:$1$. The equation of translational motion is:$ma = F - f$  ---$(i)$where $f$ is the frictional force ac

Vedclass · Accepted Answer

$\frac{3}{2}ma$

Vedclass · Answer

(C) From the free body diagram of the cylinder:$1$. The equation of translational motion is:$ma = F - f$  ---$(i)$where $f$ is the frictional force acting at the point of contact.$2$. The equation of rotational motion about the centre of mass is:$	au = I\alpha$$fR = \left(\frac{1}{2}mR^2ight)\alpha$Since the cylinder is rolling without slipping,the condition $a = R\alpha$ holds,so $\alpha = \frac{a}{R}$.Substituting this into the torque equation:$fR = \left(\frac{1}{2}mR^2ight)\left(\frac{a}{R}ight)$$f = \frac{1}{2}ma$  ---(ii)$3$. Substituting the value of $f$ from equation (ii) into equation $(i)$:$ma = F - \frac{1}{2}ma$$F = ma + \frac{1}{2}ma = \frac{3}{2}ma$

$A$ 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:

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