A hollow sphere is released from the top of an inclined plane of...

For a hollow sphere, the moment of inertia is $I = \dfrac{2}{3}mR^2$.

Let $a$ be the linear acceleration and $\alpha$ be the angular acceleration. For pure rolling, $a = \alpha R$.

The equations of motion are:
$mg\sin\theta - f = ma$
$fR = I\alpha = \left(\dfrac{2}{3}mR^2\right)\left(\dfrac{a}{R}\right) \Rightarrow f = \dfrac{2}{3}ma$

Substituting $f$ in the first equation:
$mg\sin\theta - \dfrac{2}{3}ma = ma \Rightarrow a = \dfrac{3}{5}g\sin\theta$

The required friction for pure rolling is:
$f = \dfrac{2}{3}m\left(\dfrac{3}{5}g\sin\theta\right) = \dfrac{2}{5}mg\sin\theta$

To prevent sliding, $f \le \mu N$, where $N = mg\cos\theta$:
$\dfrac{2}{5}mg\sin\theta \le \mu mg\cos\theta \Rightarrow \mu \ge \dfrac{2}{5}\tan\theta$

Thus, the minimum coefficient of friction is \mu_\min = \dfrac{2}{5}\tan\theta.

If the friction coefficient is half of this minimum value, $\mu = \dfrac{1}{5}\tan\theta$. The sphere will slide while rolling, and kinetic friction acts on it:
$f_k = \mu N = \left(\dfrac{1}{5}\tan\theta\right)(mg\cos\theta) = \dfrac{1}{5}mg\sin\theta$

The new linear acceleration $a'$ is:
$mg\sin\theta - f_k = ma' \Rightarrow a' = g\sin\theta - \dfrac{1}{5}g\sin\theta = \dfrac{4}{5}g\sin\theta$

The translational kinetic energy after traveling a distance $l$ is:
$K_t = \dfrac{1}{2}mv^2 = ma'l = m\left(\dfrac{4}{5}g\sin\theta\right)l = \dfrac{4}{5}mgl\sin\theta$

The angular acceleration $\alpha'$ is:
$\alpha' = \dfrac{f_k R}{I} = \dfrac{\dfrac{1}{5}mgR\sin\theta}{\dfrac{2}{3}mR^2} = \dfrac{3g\sin\theta}{10R}$

The time $t$ taken to travel distance $l$ is given by $l = \dfrac{1}{2}a't^2$:
$t^2 = \dfrac{2l}{a'} = \dfrac{2l}{\dfrac{4}{5}g\sin\theta} = \dfrac{5l}{2g\sin\theta}$

The rotational kinetic energy after distance $l$ is:
$K_r = \dfrac{1}{2}I\omega^2 = \dfrac{1}{2}I(\alpha' t)^2 = \dfrac{1}{2}$\left(\dfrac{2}{3}mR^2\right)$\left(\dfrac{3g\sin\theta}{10R}\right)^2 \left(\dfrac{5l}{2g\sin\theta}\right)<br><br>$K_r = \left(\dfrac{1}{3}mR^2\right)\left(\dfrac{9g^2\sin^2\theta}{100R^2}\right)\left(\dfrac{5l}{2g\sin\theta}\right) = \dfrac{3}{40}mgl\sin\theta$<br><br>The total kinetic energy is:<br>$K = K_t + K_r = $\dfrac{4}{5}$mgl\sin\theta + $\dfrac{3}{40}$mgl\$sin\theta =$\dfrac{35}{40}mgl\$sin\theta = \dfrac{7}{8}$mgl\$sin\theta