Answer to Question #262076 in Mechanics | Relativity for lowrayne

Question

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

tread attached to the surface of the ball. The mass and radius of the ball are

0.105 kg and 0.12 m, respectively. What will be its period of oscillation when

slightly displaced from its equilibrium position? Hint:

𝐼 = 5

3 𝑚𝑟2

Answer 1

Explanations & Calculations

• Once given a small shake, the sphere starts oscillating about the hanging point at the tree.
• This set up makes up a compound pendulum whose periodic time is given by $\small T= 2\pi\sqrt{\frac{I}{mgl}}$ .
• $\small I$ is the moment of inertia about the axis of rotation. Since the thread here is just a spacer there is no contribution from it to the system.
• The total $\small I$ is the m.o.i about the axis of rotation.
• Distance should be measured up to the centre of mass of the sphere from the point of rotation.
• If the length of the thread is $\small L$, then the m.o.i about the hanging point will be

$\qquad\qquad \begin{aligned} \small I&=\small \frac{2}{3}mr^2+ m(L+r)^2\\ \end{aligned}$

• Then the period will be,

$\qquad\qquad \begin{aligned} \small T&=\small 2\pi\sqrt{\frac{\frac{2}{3}mr^2+m(L+r)^2}{mg(L+r)}} \end{aligned}$

• One needs to know the length of the tread to work this out.

• But the given m.o.i is just something about the point the tread is attached to the sphere where the m.o.i is calculated to be

$\qquad\qquad\small \frac{2}{3}mr^2+mr^2=\frac{5}{3}mr^2$

• If you use it in the equation above you get a period to be $\small 0.9\,s$ which seems less accurate to me.