Ratio of Translational to Total Kinetic Energy of Solid Sphere (2018) – Rankers Physics

Ratio of Translational to Total Kinetic Energy of Solid Sphere (2018)

A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy ($K_t$) as well as rotational kinetic energy ($K_r$) simultaneously. The ratio $K_t : (K_t + K_r)$ for the sphere is: (2018)

$10 : 7$
$5 : 7$
$7 : 10$
$2 : 5$

Solution:

For a solid sphere, $K_t = \frac{1}{2}mv^2$ and $K_r = \frac{1}{5}mv^2$. The total kinetic energy is $K_t + K_r = \frac{7}{10}mv^2$. The ratio $K_t : (K_t + K_r)$ evaluates to $5 : 7$.

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