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

Step-by-step 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)=(ML^(2))/(12)`

Moment of inertia of the rod about the pivot point

Use the parallel axis theorem

`I=I_(R)+M\left ((L)/(2) \right )^(2)`

`I=(ML^(2))/(12)+(ML^(2))/(4)=(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=(ML^(2))/(3)+M\left ( 2R^(2)+L^(2)+2RL \right )` .... from (1) and (2)

`I=M\left ( 2R^(2)+(4L^(2))/(3)+2RL \right )`

Answer 2

I=
`(4)/(3)ML^2+2MR^2+2MRL`

Step-by-step 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=
`(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=
`(1)/(12)ML^2+M((L)/(2))^2+MR^2+M(L+R)^2`

Moment of inertia of the pendulum about the pivot left end,I=
`(1)/(12)ML^2+(1)/(4)ML^2+MR^2+MR^2+ML^2+2MRL`

Moment of inertia of the pendulum about the pivot left end,I=
`(1+3+12)/(12)ML^2+2MR^2+2MLR`

Moment of inertia of the pendulum about the pivot left end,I=
`(16)/(12)ML^2+2MR^2+2MRL`

Moment of inertia of the pendulum about the pivot left end,I=
`(4)/(3)ML^2+2MR^2+2MRL`