Question

A solid sphere is rolling witout slipping.the ratio of rotational kinetic energy to total kinetic energy is given by

Answer 1

Consider a rolling body of,

• mass M

• radius R

• radius of gyration K

• moment of inertia I

• linear velocity v

• angular velocity ω

The translational kinetic energy of the rolling body is given by,

$\longrightarrow\sf{K_t=\dfrac{1}{2}Mv^2}$

The rotational kinetic energy of the rolling body is given by,

$\longrightarrow\sf{K_r=\dfrac{1}{2}I\omega^2\quad\quad\dots(1)}$

But we know that,

• $\sf{I=MK^2}$

• $\sf{\omega=\dfrac{v}{R}}$

Then (1) becomes,

$\longrightarrow\sf{K_r=\dfrac{1}{2}\cdot MK^2\left(\dfrac{v}{R}\right)^2}$

$\longrightarrow\sf{K_r=\dfrac{1}{2}Mv^2\cdot \dfrac{K^2}{R^2}}$

So the total kinetic energy of the rolling body is,

$\longrightarrow\sf{K=K_t+K_r}$

$\longrightarrow\sf{K=\dfrac{1}{2}Mv^2+\dfrac{1}{2}Mv^2\cdot\dfrac{K^2}{R^2}}$

$\longrightarrow\sf{K=\dfrac{1}{2}Mv^2\left[1+\dfrac{K^2}{R^2}\right]}$

Now the ratio of rotational kinetic energy to total kinetic energy is,

$\longrightarrow\sf{\dfrac{K_r}{K}=\dfrac{\dfrac{1}{2}Mv^2\cdot\dfrac{K^2}{R^2}}{\dfrac{1}{2}Mv^2\left[1+\dfrac{K^2}{R^2}\right]}}$

$\longrightarrow\sf{\underline{\underline{\dfrac{K_r}{K}=\dfrac{\dfrac{K^2}{R^2}}{1+\dfrac{K^2}{R^2}}}}}$

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For a solid sphere,

• $\sf{\dfrac{K^2}{R^2}=\dfrac{2}{5}}$

Then the ratio is,

$\longrightarrow\sf{\dfrac{K_r}{K}=\dfrac{\dfrac{2}{5}}{1+\dfrac{2}{5}}}$

$\longrightarrow\sf{\dfrac{K_r}{K}=\dfrac{\quad\dfrac{2}{5}\quad}{\quad\dfrac{7}{5}\quad}}$

$\longrightarrow\sf{\underline{\underline{\dfrac{K_r}{K}=\dfrac{2}{7}}}}$