Question

A solid sphere of mass $M$and radius $R$is suspended from a thin rod, as shown in

$FIGURECP15.77.$The sphere can swing back and forth at the bottom of the rod. Find an expression for the frequency $f$of small angle oscillations.

Short answer

The expression for frequency$f=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{5g}{{\displaystyle 7R}}}$

Step-by-step solution

Given information

mass of the solid sphere$M$

radius of solid sphere$R$

The sphere can swing back and forth at the bottom of the rod.

Explanation

The frequency of oscillation of a solid sphere, the general equation is given by

$f=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{MgR}{I}}$

where,

$I$is the moment of inertia of a solid sphere.

The moment of inertia about center of mass of the sphere is given by

$I_{CM}=\frac{2}{5}M{R}^2$

The moment of Inertia about the pivot of the sphere is given by

$I_P=M{R}^2$

The moment of Inertia of the sphere is given by

$$\begin{gathered} I=I_{CM}+I_P \\ =\frac{2}{5}M{R}^2+M{R}^2 \\ =\frac{7}{5}M{R}^2 \end{gathered}$$

Subsitute the moment of inertia of sphere in the above frequency equation, we get

$$\begin{gathered} f=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{MgR}{{\displaystyle \frac{7}{5}}M{R}^2}} \\ =\frac{1}{2\mathrm{\pi }}\sqrt{\frac{5g}{{\displaystyle 7R}}} \end{gathered}$$

Therefore the expression for frequency of small angle oscillations of sphere is given by

$f=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{5g}{{\displaystyle 7R}}}$