Question

A cylinder is released from rest from the top of an inclined plane of inclination $\theta$ and length 'l'. If the cylinder rolls without slipping, what will be its speed when it reaches the bottom?

• A. $\sqrt{\frac{3}{4}glcos\theta }$
• B. $\sqrt{\frac{3}{4}glsin\theta }$
• C. $\sqrt{\frac{3}{4}gltan\theta }$
• D. $\sqrt{\frac{3}{4}glcot\theta }$

Answer 1

The correct option is B

$\sqrt{\frac{3}{4}glsin\theta }$

Let the mass of the cylinder be m and its radius r. Suppose the linear speed of the cylinder when it reaches the bottom is v. As the cylinder rolls without slipping, its angular speed about its axis is $\omega =\frac{v}{r}$. The kinetic energy at the bottom will be= $\frac{3}{4}m{v}^2$
Note: Try reducing this result on your own for practice.

Thus, $\frac{3}{4}m{v}^2=mglsin\theta$

or, v=$\sqrt{\frac{4}{3}glsin\theta }$