Question 600: Centre of Mass and Rotational Motion

Question

A body slides down a smooth inclined plane having angle $$	heta$$ and reaches the bottom with velocity $$v$$. If a body is a sphere, then its linear velocity at the bottom of the plane is

Accepted Answer

The linear velocity of the body, $$v=\sqrt{2 g h}$$
The velocity of the sphere about its centre,
$$v_{\mathrm{CM}}=\sqrt{\frac{2 g h}{1+\frac{K^2}{R^2}}}=\frac{v} {\sqrt{1+\frac{K^2}{R^2}}} \quad 	ext{... (i)}$$
For uniform solid sphere,
$$\frac{K^2}{R^2}=\frac{2}{5}$$
$$\begin{aligned}
&	ext { Substituting value in Eq. (i), we get }\
&v_{\mathrm{CM}}=\frac{v}{\sqrt{1+\left(\frac{2}{5}ight)}}=\sqrt{\frac{5}{7}} v
\end{aligned}$$

Answer

$$\sqrt{\frac{9}{7}} v$$

Answer

$$\sqrt{\frac{5}{7}} v$$

Answer

$$\sqrt{\frac{2}{7}} v$$

Answer

$$\sqrt{\frac{3}{7}} v$$

Centre of Mass and Rotational Motion

A body slides down a smooth inclined plane having angle $\theta and reaches the bottom with velocity v$. If a body is a sphere, then its linear velocity at the bottom of the plane is

Choose the correct answer: