Question 240: Oscillations

Question

A small spherical ball of radius ' $r$ ' is rolling on a curved surface which is frictionless and has a radius of curvature ' R '. Its motion is simple harmonic. Then its tine period of oscillation is proportional to ( $\mathrm{g}=$ acceleration due to gravity)

Accepted Answer

$$ 	ext { From the figure, } $$
$$ \begin{aligned} &	ext { Restoring force } \mathrm{F}=-\mathrm{mg} \sin 	heta\ &\begin{array}{ll}  & \sin 	heta=	heta=\frac{x}{R-x} \ 	herefore & F=-m g\left(\frac{x}{R-r}ight) \quad \Rightarrow F=-\left(\frac{m g}{R-r}ight) x \ 	herefore & K=\frac{m g}{R-r} \ 	herefore & m \omega^2=\frac{m g}{R-r} \ 	herefore & \omega=\sqrt{\frac{g}{R-r}} \ 	herefore & T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{R-r}{g}} \end{array} \end{aligned} $$

Answer

$\sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$

Answer

$\sqrt{\frac{\mathrm{r}}{\mathrm{g}}}$

Answer

$\sqrt{\frac{R-r}{g}}$

Answer

$\sqrt{\frac{R+r}{g}}$

Oscillations

A small spherical ball of radius ' r ' is rolling on a curved surface which is frictionless and has a radius of curvature ' R '. Its motion is simple harmonic. Then its tine period of oscillation is proportional to ( \mathrm{g}= acceleration due to gravity)

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