Question

A uniform sphere of mass $m$ and radius $R$ rolls without slipping down an inclined plane set at an angle $\alpha$ to the horizontal. Find the kinetic energy of the sphere $t$ seconds after the beginning of motion.

• A. $K.E.=\frac{5}{14}m{g}^2{t}^2{\sin }^2\alpha$
• B. $K.E.=\frac{3}{14}m{g}^2{t}^2{\sin }^2\alpha$
• C. $K.E.=\frac{5}{14}m{g}^2{t}^2{\sin }^3\alpha$
• D. $K.E.=\frac{3}{14}m{g}^2{t}^2{\sin }^3\alpha$

Answer 1

The correct option is A $K.E.=\frac{5}{14}m{g}^2{t}^2{\sin }^2\alpha$

From figure: Moment of Inertia of sphere about point "P"

$I_P=\frac{2}{5}m{R}^2+m{r}^2=\frac{7}{5}m{r}^2$

${\Gamma }_P=r\ast mgsin\left(\alpha \right)$

Since there is no slipping:

${\Gamma }_P=I\alpha$

$I_P\frac{d\omega }{dt}=r\ast mgsin\left(\alpha \right)I_{P}d\omega =r\ast mgsin\left(\alpha \right){\int }_0^{\omega }{I}_{P}d\omega ={\int }_0^{t}r\ast mgsin\left(\alpha \right)dt{I}_P\omega =r\ast mgsin\left(\alpha \right)t$

$\omega =\frac{r\ast mgsin\left(\alpha \right)t}{\frac{7}{5}m{r}^2}\omega =\frac{5}{7}\frac{gsin\left(\alpha \right)t}{r}v=r\ast \omega v=\frac{5}{7}gsin\left(\alpha \right)t$

Now Kinetic Energy = Rotational K.E + Translational K.E

Kinetic Energy = $\frac{1}{2}m\left(\frac{5}{7}gsin\right(\alpha )t{)}^2+\frac{1}{2}(\frac{2}{5}m{r}^2\left)\right(\frac{5gsin\left(\alpha \right)t}{7r}{)}^2$

on solving

K.E. = $\frac{25}{98}{g}^{2}si{n}^2\left(\alpha \right)t^2+\frac{5}{49}{g}^{2}si{n}^2\left(\alpha \right)t^2$

Hence Kinetic Energy = $\frac{5}{14}m{g}^{2}si{n}^2\left(\alpha \right)t^2$