Question

A uniform cylinder of mass $M$ and radius $R$ is released from rest on a rough inclined surface of inclination $\theta$ with the horizontal as shown in figure. As the cylinder rolls down the inclined surface, the maximum elongation in the spring of stiffness $k$ is

Initially spring is in its natural length.

• A. $\frac{3}{4}\frac{Mg\sin \theta }{k}$
• B. $\frac{2Mg\sin \theta }{k}$
• C. $\frac{Mg\sin \theta }{k}$
• D. $\text{None of these}$

Answer 1

The correct option is B $\frac{2Mg\sin \theta }{k}$

In pure rolling,
Work done by static friction is zero.
Applying energy conservation principle at maximum elongation,
Work done by gravity =Change in potential energy of spring
$\therefore$$\frac{1}{2}k{x}^2=Mgx\sin \theta$
$\Rightarrow x=\frac{2Mg\sin \theta }{k}$