Answer to Question #164078 in Physics for Vhench fernandez

Question

2. A Christmas ball in a shape of a hollow sphere is hung from a tree by a piece

of thread attached to the surface of the ball. The mass and radius of the ball

are 0 1059 and 0.12m, respectively. What will be its period of oscillation when

slightly displaced from its equilibrium position?

(Hint: 1 mm)

Answer 1

Let's consider a hollow sphere to be a physical pendulum. Then, the period of the pendulum can be written as follows:

$$T=2\pi\sqrt{\dfrac{I}{Mgd}}=2\pi\sqrt{\dfrac{I}{MgR}},$$

here, $I$ is the moment of inertia of the pendulum, $d=R=0.12\ m$ is the distance from the axis to the center of gravity (which is equals to the radius of the sphere).

We can find the moment of inertia of the pendulum by adding the moment of inertia of the hollow sphere and the moment of inertia of the pivot:

$$I=I_{sphere}+I_{pivot},$$ $$I=\dfrac{2}{3}MR^2+MR^2=\dfrac{5}{3}MR^2.$$

Finally, substituting $I$ into the formula for the period of the physical pendulum, we get:

$$T=2\pi\sqrt{\dfrac{\dfrac{5}{3}MR^2}{MgR}}=2\pi\sqrt{\dfrac{5R}{3g}},$$ $$T=2\pi\sqrt{\dfrac{5\cdot0.12\ m}{3\cdot9.8\ \dfrac{m}{s^2}}}=0.9\ s.$$