Question

A hollow cylinder with inner radius R, outer radius $2$R and mass M is rolling with speed of its axial. Its kinetic energy is?

Answer 1

$K.E.=\frac{1}{2}M{v}^2+\frac{1}{2}I{w}^2$

here, $I=\frac{M}{2}(4R^2+R^2)=\frac{5M{R}^2}{2}$

$V=2Rw$

$\Rightarrow K.E.=\frac{1}{2}M{v}^2+\frac{1}{2}\times \frac{5M{R}^2}{2}\times \frac{v^2}{4R^2}$

$\Rightarrow K.E.=\left(\frac{8+5}{16}\right)M{R}^2=\frac{13}{16}M{R}^2$

Hence, the answer is $\frac{13}{16}M{R}^2.$