Question

[3 points] Question: Consider a pendulum made from a uniform, solid rod of mass M and length L attached to a hoop of mass M and radius R. Find the moment of inertia I of the pendulum about the pivot depicted at the left end of the solid rod.

Answer 1

Answer:

I=$\frac{4}{3}ML^2+2MR^2+2MRL$

Explanation:

We are given that

Mass of rod=M

Length of rod=L

Mass of hoop=M

Radius of hoop=R

We have to find the moment of inertia I of the pendulum about pivot depicted at the left end of the slid rod.

Moment of inertia of rod about center of mass=$\frac{1}{12}ML^2$

Moment of inertia of hoop about center of mass=$MR^2$

Moment of inertia of the pendulum about the pivot left end,I=$\frac{1}{12}ML^2+M(\frac{L}{2})^2+MR^2+M(L+R)^2$

Moment of inertia of the pendulum about the pivot left end,I=$\frac{1}{12}ML^2+\frac{1}{4}ML^2+MR^2+MR^2+ML^2+2MRL$

Moment of inertia of the pendulum about the pivot left end,I=$\frac{1+3+12}{12}ML^2+2MR^2+2MLR$

Moment of inertia  of the pendulum about the pivot left end,I=$\frac{16}{12}ML^2+2MR^2+2MRL$

Moment of inertia of the pendulum about the pivot left end,I=$\frac{4}{3}ML^2+2MR^2+2MRL$

Answer 2

Answer:

Explanation:

Rod of pendulum of mass M and length L.

Mass of hoop is M and radius of hoop is R.

Moment of inertia of the rod about its centre is

$I_{R}=\frac{ML^{2}}{12}$

Moment of inertia of the rod about the pivot point

Use the parallel axis theorem

$I=I_{R}+M\left (\frac{L}{2} \right )^{2}$

$I=\frac{ML^{2}}{12}+\frac{ML^{2}}{4}=\frac{ML^{2}}{3}$    .... (1)

Moment of inertia of the hoop about its centre is

$I_{H}=MR^{2}$

Moment of inertia of the hoop about the pivot point

Use the parallel axis theorem

$I'=I_{H}+M(R+L)^{2}$

$I'=M(2R^{2}+L^{2}+2RL)$    .... (2)

Total moment of inertia of the pendulum about the pivot is

$I=\frac{ML^{2}}{3}+M\left ( 2R^{2}+L^{2}+2RL \right )$        .... from (1) and (2)

$I=M\left ( 2R^{2}+\frac{4L^{2}}{3}+2RL \right )$