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In the figure shown mass of both, the spherical body and block is $\mathrm{m}$. Moment of inertia of the spherical body about centre of mass is $2 m R^2$. The spherical body rolls on the horizontal surface. There is no slipping at any surfaces in contact. The ratio of kinetic energy of the spherical body to that of block is
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$2 / 3$
Let $v$ be the linear velocity of centre of mass of the spherical body and $w$ its angular velocity about centre of mass.
Then $\left.\omega=\frac{v}{2 R} \right\rvert\, \mathrm{KE}$ of spherical body
$K_1=\frac{1}{2} m v^2+\frac{1}{2} I \omega^2\left|K_1=\frac{1}{2} m v^2+\frac{1}{2}(2 m R)^2\left(\frac{v^2}{4 R^2}\right)=\frac{3}{4} m v^2\right|$ ?.
(i) Speed of the block will be $\left.v^{\prime}=(\omega)(3 R)=3 R \omega=(3 R)\left(\frac{v}{2 R}\right)=\frac{3}{2} v \right\rvert\,$
$\therefore K E$ of block
$K_2=\frac{1}{2} m v^2\left|=\frac{1}{2} m\left(\frac{3}{2} v\right)^2=\frac{9}{8} m v^2\right|$
From equations (i) and (ii),
$| \frac{K_1}{K_2}=\frac{2}{3} |$
Then $\left.\omega=\frac{v}{2 R} \right\rvert\, \mathrm{KE}$ of spherical body
$K_1=\frac{1}{2} m v^2+\frac{1}{2} I \omega^2\left|K_1=\frac{1}{2} m v^2+\frac{1}{2}(2 m R)^2\left(\frac{v^2}{4 R^2}\right)=\frac{3}{4} m v^2\right|$ ?.
(i) Speed of the block will be $\left.v^{\prime}=(\omega)(3 R)=3 R \omega=(3 R)\left(\frac{v}{2 R}\right)=\frac{3}{2} v \right\rvert\,$
$\therefore K E$ of block
$K_2=\frac{1}{2} m v^2\left|=\frac{1}{2} m\left(\frac{3}{2} v\right)^2=\frac{9}{8} m v^2\right|$
From equations (i) and (ii),
$| \frac{K_1}{K_2}=\frac{2}{3} |$
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