# Finding the period

## Homework Statement

The pendulum shown below right consists of a uniform disk with radius r = 20.0 cm and mass m = 1.2 kg attached to a uniform rod with length L = 40 cm and mass M = 0.8 kg. What is the period of the motion? T = 2∏√(I/mgL)

## The Attempt at a Solution

I dont know what to plug into what.

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Doc Al
Mentor
To start with, you need to calculate the rotational inertia (I) of the system about the axis of rotation. How would you do that?

Also, in your formula for T what does L represent? You'll have to find that as well.

To start with, you need to calculate the rotational inertia (I) of the system about the axis of rotation. How would you do that?

Also, in your formula for T what does L represent? You'll have to find that as well.

T = 2∏√(I/mgL)

L represents the distance to the center of mass of the system?
L = [(1.2*(0.4+0.6)/2)+(0.8*(0.4/2)]/(1.2+0.8)
= 0.38m?

using parallel axis theorem:

I = 1/2MR2 + ML2
= (0.5)(1.2)(0.2)2 + 0.8(0.2+0.4)2
=0.312?

Doc Al
Mentor
T = 2∏√(I/mgL)

L represents the distance to the center of mass of the system?
Right.
L = [(1.2*(0.4+0.6)/2)+(0.8*(0.4/2)]/(1.2+0.8)
= 0.38m?
What's the distance from the center of the disk to the axis?

using parallel axis theorem:

I = 1/2MR2 + ML2
= (0.5)(1.2)(0.2)2 + 0.8(0.2+0.4)2
=0.312?
Things are a bit mixed up here. Find the rotational inertia for each piece separately, then add them up. To find the I of the disk about the axis, you'll need the parallel axis theorem. What's the I of a thin rod about one end?

Right.

What's the distance from the center of the disk to the axis?

T = 2∏√(I/mgL)

L represents the distance to the center of mass of the system?
L = [(1.2*(0.4+0.6)/2)+(0.8*(0.4/2)]/(1.2+0.8)
= 0.38m?

the distance from the center of the disk to the pivot is 0.6m.

Things are a bit mixed up here. Find the rotational inertia for each piece separately, then add them up. To find the I of the disk about the axis, you'll need the parallel axis theorem. What's the I of a thin rod about one end?
L = [(1.2*0.6)+(0.8*(0.4/2)]/(1.2+0.8)
= 0.44m?

using parallel axis theorem:

I of disk = 1/2MR2 + ML2= 1/2(1.2)(0.2)2 + (1.2)(0.6)2= 0.456?
I of thin rod = 1/3ML2 = 1/3(0.8)(0.4)2= 0.0427?

I total = 0.4987?

I have an exam in 45 minutes so i really need to get this cleared up before that.

Last edited:
Doc Al
Mentor
L = [(1.2*0.6)+(0.8*(0.4/2)]/(1.2+0.8)
= 0.44m?

using parallel axis theorem:

I of disk = 1/2MR2 + ML2= 1/2(1.2)(0.2)2 + (1.2)(0.6)2= 0.456?
I of thin rod = 1/3ML2 = 1/3(0.8)(0.4)2= 0.0427?

I total = 0.4987?
Looks good.

Right.

What's the distance from the center of the disk to the axis?

Things are a bit mixed up here. Find the rotational inertia for each piece separately, then add them up. To find the I of the disk about the axis, you'll need the parallel axis theorem. What's the I of a thin rod about one end?
Looks good.
wen i plug m into the formula, T = 2∏√(I/mgL)

do i use the mass of the entire system? so 1.2 + 0.8 = 2kg?

Doc Al
Mentor
wen i plug m into the formula, T = 2∏√(I/mgL)

do i use the mass of the entire system? so 1.2 + 0.8 = 2kg?
Yes.

For more insight into the physical pendulum and how that equation comes about, review this: Physical Pendulum

Yes.

For more insight into the physical pendulum and how that equation comes about, review this: Physical Pendulum
thanks so much for ur help! really appreicate it. glad there are people like you out there.

but last question, can you explain why it has a negative sign in

Torque = -mgLsinθ?

Doc Al
Mentor
but last question, can you explain why it has a negative sign in

Torque = -mgLsinθ?
Because the torque is in the direction of decreasing θ. (It's a restoring force, tending to bring the system back to the equilibrium point.)