# Seesaw Problem

1. Jan 15, 2005

### laminatedevildoll

My main goal was the find the angular acceleration of the seesaw.

The swing bar of mass is pivoted at a different point, as shown in the figure.

Find the magnitude of the angular acceleration of the swing bar. Be sure to use the absolute value function in your answer, since no comparison of m1 ,m2, mbar and has been made.

So naturally, I tried to find the moment of inertia
The inertia of the bar is 1/12m_bar(L)^2
The inertia of the two masses are (m_1)*(l/3)^2+(m_2)*(2*l/3)^2

Adding together the total inertia is

(m_bar/12)*(l/3+2*l/3)^2+(m_1)*(l/3)^2+(m_2)*(2*l/3)^2

where l/2 and 2l/3 are the radiuses that are away from m1, m2, respectively.

But unfortunately, this is not the correct answer.

I also tried to find the net torque. And this should be in absolute value because m2 and m1 are not given.

So I thought it was abs(g*(m_1*l/3-m_2*2*l/3)) but it was wrong.

This problem is driving me crazy! I would appreciate any help.

Thank you

Last edited: Jan 16, 2005
2. Jan 16, 2005

### Hyperreality

Is this system in equilibrium, and was later than applied by a negligible external force? And can you please varify the parameters of this solution?

3. Jan 16, 2005

### laminatedevildoll

Seesaw

This is the actual problem (static equilibrium)
In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses m_1 and m_2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction.

You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m_1>m_2.
m_2 is on the right side, m_1 is on the left side.
There were other parts to this problem, which I figured out.

Part A
The seesaw is pivoted in the middle, and the mass of the swing bar is negligible
Acceleration due to gravity =(2*(m_1-m_2)*g)/((m_1+m_2)*l)

Part B
Now consider a similar situation, except that now the swing bar itself has mass m_bar.
So, since the inertia of the bar is 1/12*m_bar*l^2
Angular A = (g*(m_1-m_2))/(l/2*(m_bar/3+m_1+m_2))

Part C
This time, the swing bar of mass is pivoted at a different point

.....l/3..m_bar......2l/3............
<______*__________________>
m_1................................m_2

This, time m_1, m_2 is negligible, so I have to use the absolute value for the net torque.

I first tried to calculate Inertia

Inertia of m_bar
I=(1/12)*m_bar*(l/3+2l/3)^2
Inertia of m_1
I= (m_1)*(l/3)^2
Inertia of m_2
I=(m_2)*(2l/3)

So the total inertia would be everything above added together, but the answer is different. I just don't see the problem.

The torque which I got was abs(g*(m_1*l/3+m_2*2*l/3 +m_bar*l)) is also wrong.

Last edited: Jan 16, 2005
4. Jan 16, 2005

### laminatedevildoll

The torque,which I calculated g*(abs((l/6)*m_bar)-((l/3)*m_1)-(2*(l/3)*m_2)) is wrong!!! There is something wrong with the signs.

5. Jan 16, 2005

### laminatedevildoll

Okay, so I had a lightbulb moment. Since the bar isn't centered, I need to use the parallel axis theorem. I=I_cm+Md^2
I bar is l^2/12*(m_bar+m_bar/3)

So, the inertia might be l^2/12*(m_bar+m_bar/3)+(l^2/9*(m_1+4*m_2)) but not so sure...

In any case, I am still having trouble with the torque...

Last edited: Jan 16, 2005
6. Jan 16, 2005

### laminatedevildoll

angular acceleration = (g*(abs((l/6)*m_bar)-((l/3)*m_1)-(2*(l/3)*m_2)))/((l^2/12*(m_bar+m_bar/3))+(l^2/9*(m_1+4*m_2)))
but when I typed it in, it says that I need to check over my signs... I need to include the absolute value for torque because the masses are unknown. HELP!

Last edited: Jan 17, 2005