A uniform solid cylinder of mass $M$ and radius$R$ rolls on a rough inclined plane with its axis perpendicular to the line of the greatest slope. As the cylinder rolls, it winds up a light string that passes over a light pulley and a small mass $m$ is hanging from the free end of the string . The part of the string between the pulley and the cylinder is parallel to the line of greatest slope. If $\theta$ is the inclination of the plane with the horizontal, calculate the tension in the string.

Energy Method:

Let $v$ be the velocity of the c.m. of the cylinder when it travelled a distance y along the plane. The velocity and the distance travelled by the mass m in this interval will be $2v$ and $2y$ respectively. Decrease in P. E. of cylinder$=$ Increase in $($rolling K. E. of the cylinder$+$translational K.E. of the mass m $+$ PE. of the mass$)$

$Mg\left(y\sin \theta \right)=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2+\frac{1}{2}m{\left(2v\right)}^2+mg\left(2y\right)$

where $\omega =\frac{v}{R}$ and $I=\frac{M{R}^2}{2}$

Substituting these value in above $e{q}^n$ and solving for $v$, we get

$v^2=\frac{4g\times (M\sin \theta -2m)}{8m+3M}$

Using third equation of motion $v^2=u^2+2ay$, where $u=0$

$\therefore a=\frac{v^2}{2y}=\frac{\left(2Mg\sin \theta -4mg\right)}{\left(8m+3M\right)}$

Substituting these value in above $e{q}^n$ $\left(3\right)$, we can get $T=\frac{\left(3+4\sin \theta \right)mMg}{8m+3M}$