A cylinder is rolling down on an inclined plane of inclination 60 ° . Its acceleration during rolling down will be x 3 m s - 2 , where x = _______(use g = 10 m s - 2 ).

Question

Aicharya · Accepted Answer

For rolling motion, the acceleration of the cylinder along the inclined plane can be expressed as

$a = \frac{g s i n θ}{1 + \frac{I_{c m}}{M R^{2}}} . . . (1)$

The moment of inertia of the cylinder can be written as

$I_{c m} = \frac{1}{2} M R^{2} . . . (2)$

From equations (1) and (2), it follows that

$\begin{array}{rcl} a & = & \frac{g s i n θ}{1 + \frac{1}{2}} \\ = & \frac{2 \times 10 \times \frac{\sqrt{3}}{2}}{3} \\ = & \frac{10}{\sqrt{3}} m s^{- 2} \end{array}$

Therefore, $x = 10$ .

A cylinder is rolling down on an inclined plane of inclination 60 ° . Its acceleration during rolling down…

Solution