Question 543: Centre of Mass and Rotational Motion

Question

An inclined plane makes an angle of $30^{\circ}$ with the horizontal. A solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration ( $\mathrm{g}=$ acceleration due to gravity, $\sin 30^{\circ}=0.5$ )

Accepted Answer

$$
\begin{aligned}
& a=\frac{g \sin 	heta}{\left(1+\frac{K^2}{R^2}ight)}=\frac{g \sin 30^{\circ}}{\left(1+\frac{2}{5}ight)} \
	herefore \quad & a=\frac{5 g}{7} 	imes 0.5=\frac{5 g}{14}
\end{aligned}$$

Answer

$\frac{2 g}{3}$

Answer

$\frac{5 g}{14}$

Answer

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

Answer

$\frac{5 g}{7}$

Centre of Mass and Rotational Motion

An inclined plane makes an angle of 30^{\circ} with the horizontal. A solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration ( \mathrm{g}= acceleration due to gravity, \sin 30^{\circ}=0.5 )

Choose the correct answer: