Question 606: Centre of Mass and Rotational Motion

Question

A solid sphere rolls down from top of inclined plane, 7 m high, without slipping. Its linear speed at the foot of plane is $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$

Accepted Answer

Given, height of inclined plane, $h=7 \mathrm{~m}$
$$g=10 \mathrm{~m} / \mathrm{s}^2$$
By conservation of energy,
Potential energy lost by the solid sphere in rolling down the inclined plane $=$ Kinetic energy gained by the sphere.
$$\begin{aligned}
m g h & =\frac{1}{2} m v^2+\frac{1}{2} l \omega^2 \
& =\frac{1}{2} m v^2+\frac{1}{2} \cdot \frac{2}{5} m R^2 \cdot\left(\frac{v}{R}ight)^2 \
& =\frac{1}{2} m v^2+\frac{1}{5} m v^2 \
m g h & =\frac{7}{10} m v^2 \
	herefore \quad v^2 & =\frac{10}{7} g h \
\Rightarrow \quad v & =\sqrt{\frac{10}{7} g h}=\sqrt{\frac{10}{7} 	imes 10 	imes 7}=\sqrt{100} \mathrm{~m} / \mathrm{s}
\end{aligned}$$

Answer

$\sqrt{70} \mathrm{~m} / \mathrm{s}$

Answer

$\sqrt{\frac{140}{3}} \mathrm{~m} / \mathrm{s}$

Answer

$\sqrt{\frac{280}{3}} \mathrm{~m} / \mathrm{s}$

Answer

$\sqrt{100} \mathrm{~m} / \mathrm{s}$

Centre of Mass and Rotational Motion

A solid sphere rolls down from top of inclined plane, 7 m high, without slipping. Its linear speed at the foot of plane is \left(g=10 \mathrm{~m} / \mathrm{s}^2\right)

Choose the correct answer: