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If a spherical ball rolls with out slipping, the fraction of its total energy associated with rotation is

\[\begin {array} {1 1} (a)\;\frac{3}{5} & \quad (b)\;\frac{2}{7} \\ (c)\;\frac{2}{5} & \quad  (d)\;\frac{3}{7} \end {array}\]
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The moment of inertia of a sphere about its central axis and a solid spherical shell of mass M and radius R as shown in the figure above is $I_{\text{sphere}} = \large\frac{2}{5}$$MR^2$
Kinetic energy of the rotation of the sphere KE$_{\text{rotation}} = \large\frac{1}{2} $$I\omega^2 $ where $\omega $ is the angular velocity.
$\Rightarrow$ KE$_{\text{rotation}} = \large\frac{1}{2} $$I\omega^2 = \large\frac{1}{2} \frac{2}{5}$$ MR^2 \omega^2$
Now, angular velocity $\omega = \large\frac{v}{R}$, where $v$ is the linear velocity of the sphere.
$\Rightarrow$ KE$_{\text{rotation}} = \large\frac{1}{2} $$I\omega^2 = \large\frac{1}{2} \frac{2}{5}$$ MR^2 $$\large\frac{v^2}{R^2}$$ = \large\frac{1}{5} $$Mv^2$
Kinetic energy of the linear motion KE$_{\text{linear motion}} = \large\frac{1}{2}$$Mv^2$
Total Kinetic energy $=$ KE$_{\text{rotation}} + $KE$_{\text{linear motion}} = \large\frac{1}{5}$$Mv^2 + \large\frac{1}{2}$$ Mv^2$$ = \large\frac{7}{10} $$Mv^2$
Now, the fraction of its total energy associated with rotation $= \large\frac {\text{KE of Rotation}}{\text{Total KE}}$$ = \Large\frac{ \large\frac{1}{5} \normalsize Mv^2}{\large\frac{7}{10} \normalsize Mv^2}$
$\qquad = \large\frac{2}{7}$
answered Nov 29, 2013 by meena.p
edited Mar 25, 2014 by balaji.thirumalai
 

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