Question

A uniform solid cylinder of mass m and radius R is released from the top of a fixed rough inclined plane of inclination $\theta$. The coefficient of friction between the cylinder and the inclined plane is $\mu =\left(1/4\right)\tan \theta$. Find total kinetic energy of the cylinder at the bottom.

Answer 1

$v^2{u}^2+2as$

$V=\sqrt{2\times \frac{3gsin\theta }{4}\times l}=\sqrt{\frac{3}{2}}glsin\theta$

net $\frac{1/2\times m\times 3/2glsin\theta }{3mglsin\theta /4}$

$\mu mgcos\theta \times R=\frac{{mR}^2}{2}\times \gamma$

$\mu gcos\theta =\frac{R^{}}{2}\gamma$

$2\mu gcos\theta /R=\gamma$

$\omega ={\omega }^0+\gamma t$

$\frac{2\mu cos\theta }{R}\times \sqrt{\frac{8l}{3gsin\theta }}$

$\frac{2\times sin\theta g}{cos\theta \times 4R}\times \sqrt{\frac{8l}{3gsin\theta }}$