Question 418: Centre of Mass and Rotational Motion

Question

A solid sphere is rolling on a horizontal plane without slipping. If the ratio of angular momentum about axis of rotation of the sphere to the total energy of moving sphere is $$\pi: 22$$ then, the value of its angular speed will be ____________ $$\mathrm{rad} / \mathrm{s}$$.

Accepted Answer

Given that the solid sphere is rolling without slipping, we have:
Angular momentum $$L = \left(I_{	ext{com}}ight)(\omega)$$
Kinetic energy $$K = \frac{1}{2}(I_{	ext{com}})(\omega^2) + \frac{1}{2}MV_{	ext{com}}^2$$
For a solid sphere, the moment of inertia is $$I_{	ext{com}} = \frac{2}{5}MR^2$$, and the relationship between linear and angular velocity is $$V_{	ext{com}} = R\omega$$.
Substituting these values into the expressions for $$L$$ and $$K$$:
$$L = \frac{2}{5}MR^2 \frac{V_{	ext{com}}}{R} = \frac{2MRV_{	ext{com}}}{5}$$
$$K = \frac{1}{2}\left(\frac{2}{5}MR^2ight) \frac{V_{	ext{com}}^2}{R^2} + \frac{1}{2}MV_{	ext{com}}^2 = \frac{7}{10}MV_{	ext{com}}^2$$
Now, the given ratio of $$\frac{L}{K}$$ is $$\frac{\pi}{22}$$:
$$\frac{L}{K} = \frac{4}{7} \frac{R}{V_{	ext{com}}} = \frac{\pi}{22}$$
Since $$V_{	ext{com}} = R\omega$$, we can substitute this relationship into the equation and solve for $$\omega$$:
$$\frac{4}{7} \frac{R}{R\omega} = \frac{\pi}{22}$$
$$\frac{4}{7\omega} = \frac{\pi}{22}$$
$$\omega = \frac{4}{7} 	imes \frac{22}{\pi} 	imes 7 = 4$$
Thus, the value of the angular speed is $$4 \ 	ext{rad/s}$$.

Centre of Mass and Rotational Motion

A solid sphere is rolling on a horizontal plane without slipping. If the ratio of angular momentum about axis of rotation of the sphere to the total energy of moving sphere is $\pi: 22 then, the value of its angular speed will be ____________ \mathrm{rad} / \mathrm{s}$.

Integer / Numeric Answer Type