Rotational Motion Question 221
The spherical body and block have mass m. Moment of inertia of the spherical body about centre of mass is $ \frac{2}{5}mR^{2} $ . The spherical body rolls on the horizontal surface. There is no slipping at any surfaces in contact. The ratio of kinetic than kinetic energy of the spherical body to that of block is
Options:
3/4
1/3
2/3
1/2
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Answer:
Correct Answer: C
Solution:
[c] Let v be the linear velocity of centre of mass of the spherical body and w its angular velocity about centre of mass.
Then $\omega =\frac{v}{R}$
KE of spherical body
$K _{1}=\frac{1}{2}mv^{2}+\frac{1}{2}I{{\omega }^{2}}$
$K _{1}=\frac{1}{2}mv^{2}+\frac{1}{2}{{(2mR)}^{2}}\left( \frac{v^{2}}{4R^{2}} \right)=\frac{3}{4}mv^{2}$?.
(i) Speed of the block will be determined by the forces acting upon it.
$v’=(\omega )(3R)=3R\omega =(3R)\left( \frac{v}{2R} \right)=\frac{3}{2}v$
$\therefore $KE of block $K _{2}=\frac{1}{2}mv^{2}$
$=\frac{1}{2}m{{\left( \frac{3}{2}v \right)}^{2}}=\frac{9}{8}mv^{2}$
From equations (i) and (ii), $
$\frac{K _{1}}{K _{2}}=\frac{2}{3}$
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