Question

A 95 kgsolid sphere with a 15 cmradius is suspended by a vertical wire. A torque of 0.20 N.mis required to rotate the sphere through an angle of 0.85 radand then maintain that orientation. What is the period of the oscillations that result when the sphere is then released?

Short answer

Period of oscillation that results when the sphere is released is $12\mathrm{s}$.

Step-by-step solution

The given data

1. Mass of solid sphere,$M=95\mathrm{kg}$.
2. Radius of solid sphere,$R=15\mathrm{cm}\mathrm{or}0.15\mathrm{m}$.
3. Torque required for rotation,$t=0.20\mathrm{N}$

4. Angle of rotation,$\theta =0.85\mathrm{rad}$

Understanding the concept of moment of inertia

Using the concept of the torsion pendulum, we can find the torsion constant. With the help of this constant and the equation for rotational inertia of the solid sphere, we can find the required time.

Formula:

Torsion constant of oscillation by torque, $\mathrm{k}=\frac{\mathrm{t}}{\mathrm{\theta }}$………(i)

Moment of inertia of solid sphere, $I=\frac{2M{R}^2}{5}$……….(ii)

The period of oscillation,$T=2\pi \sqrt{\frac{\mathrm{I}}{k}}$……….(iii)

Calculation of period of oscillation

Using equation (i) and the given values, we get

$$\begin{gathered} k=\frac{0.20}{0.85} \\ =0.2352\mathrm{N}.\mathrm{m}/\mathrm{rad} \end{gathered}$$

Using equation (ii) and the given values, we get

$$\begin{gathered} I=\frac{2\left(95\right)\left(0.{45}^2\right)}{5} \\ =0.855\mathrm{kg}.{\mathrm{m}}^2 \end{gathered}$$

Now, using equation (iii), period of oscillation is given by:

$$\begin{gathered} \mathrm{T}=2\left(3.14\right)\sqrt{\frac{0.855}{0.2352}} \\ =11.97\sec \\ \approx 12\sec \end{gathered}$$

Hence, the value of period of oscillation is 12 sec.