When a sphere with moment of inertia I rolls | vedclass

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(A) Let the mass of the sphere be $M$ and its radius be $R$.The moment of inertia of a solid sphere is $I = \frac{2}{5}MR^2$.For pure rolling,the rela

Vedclass · Accepted Answer

$28\%$

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(A) Let the mass of the sphere be $M$ and its radius be $R$.The moment of inertia of a solid sphere is $I = \frac{2}{5}MR^2$.For pure rolling,the relationship between linear velocity and angular velocity is $v = R\omega$,which implies $\omega = \frac{v}{R}$.The total kinetic energy $(K_{total})$ is the sum of rotational kinetic energy $(K_{rot})$ and translational kinetic energy $(K_{trans})$:$K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} 	imes (\frac{2}{5}MR^2) 	imes (\frac{v}{R})^2 = \frac{1}{5}Mv^2$.$K_{trans} = \frac{1}{2}Mv^2$.$K_{total} = K_{rot} + K_{trans} = \frac{1}{5}Mv^2 + \frac{1}{2}Mv^2 = \frac{2+5}{10}Mv^2 = \frac{7}{10}Mv^2$.The percentage of rotational kinetic energy is given by:$	ext{Percentage} = \frac{K_{rot}}{K_{total}} 	imes 100 = \frac{\frac{1}{5}Mv^2}{\frac{7}{10}Mv^2} 	imes 100 = \frac{1}{5} 	imes \frac{10}{7} 	imes 100 = \frac{2}{7} 	imes 100 \approx 28.57\%$.Rounding to the nearest integer,we get $28\%$.

When a sphere with moment of inertia $I$ rolls down an inclined plane,what percentage of its total energy is rotational kinetic energy?

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