Question 521: Centre of Mass and Rotational Motion

Question

An inclined plane makes an angle $30^{\circ}$ with horizontal. A solid sphere rolls down from the top of the inclined plane from rest without slipping has a linear acceleration along the plane equal to (where $g$ is acceleration due to gravity) (given $\sin 30^{\circ}=0.5$)

Accepted Answer

$$\begin{aligned}
& \quad a=\frac{g \sin 	heta}{\left(1+\frac{K^2}{\mathrm{R}^2}ight)}=\frac{g \sin 30^{\circ}}{\left(1+\frac{2}{5}ight)} \
& 	herefore \quad a=\frac{5 g}{7} 	imes\left(\frac{1}{2}ight)=\frac{5 g}{14}
\end{aligned}$$

Answer

$\frac{5 \mathrm{g}}{14}$

Answer

$\frac{5 g}{4}$

Answer

$\frac{2 \mathrm{g}}{3}$

Answer

$\frac{\mathrm{g}}{3}$

Centre of Mass and Rotational Motion

An inclined plane makes an angle 30^{\circ} with horizontal. A solid sphere rolls down from the top of the inclined plane from rest without slipping has a linear acceleration along the plane equal to (where g is acceleration due to gravity) (given \sin 30^{\circ}=0.5)

Choose the correct answer: