Solve Ball Torque Problem: 5.80 kg and 8.50 kg Balls

[irun4edmund]

Homework Statement

A 5.80 kg ball is dropped from a height of 13.5 m above one end of a uniform bar that pivots at its center. The bar has mass 8.50 kg and is 5.40 m in length. At the other end of the bar sits another 4.00 kg ball, unattached to the bar. The dropped ball sticks to the bar after the collision. How high will the other ball go after the collision?

Homework Equations

U = mgh
torque = Iα

The Attempt at a Solution

I set my equation up like this m₁gh₁ - Iα = m₂gh₂
where everything is known except for h₂ (what I am trying to find) and α (angular acceleration).

Im stuck because i have one equation with 2 unknowns, I am not sure what I am missing. Any help would be greatly appreciated.

 

[LowlyPion]

Homework Statement

A 5.80 kg ball is dropped from a height of 13.5 m above one end of a uniform bar that pivots at its center. The bar has mass 8.50 kg and is 5.40 m in length. At the other end of the bar sits another 4.00 kg ball, unattached to the bar. The dropped ball sticks to the bar after the collision. How high will the other ball go after the collision?

Homework Equations

U = mgh
torque = Iα

The Attempt at a Solution

I set my equation up like this m₁gh₁ - Iα = m₂gh₂
where everything is known except for h₂ (what I am trying to find) and α (angular acceleration).

Im stuck because i have one equation with 2 unknowns, I am not sure what I am missing. Any help would be greatly appreciated.

Is there a picture with this problem?

Is the bar horizontal, or elevated toward the end the mass starts on? Or is the second mass constrained to only go straight up?

 

[irun4edmund]

there was no picture to go with the problem. I wrote word for word from my physics book. I assume the bar is horizontal and that ball travels strait up, but the problem does not state that.

 

[LowlyPion]

there was no picture to go with the problem. I wrote word for word from my physics book. I assume the bar is horizontal and that ball travels strait up, but the problem does not state that.

Well let's assume that they mean for it to travel straight upward and that it will leave the bar at the same level as the first mass falls on it - that is with its ends level.

In this case then I would suggest that you look at the conservation of momentum.

You know that the initial mass m₁ and V₁ is all there is with everything else at rest.

After the process of attaching then the momentum should remain the same. So ...

m₁V₁ = I * w

where I is the combined moment of inertia and w is the resulting angular speed. And this speed w * r = V_{2 of the second mass}

The combined I or I_total = I_m1 + I_m2 + I_bar

I_m1 = m₁ * (L/2)²
I_m2 = m₂ * (L/2)²
I_bar = ... ? Figure the moment of inertia for the bar of length L.

Then you should be able to solve ... at least for the assumptions I would suggest.

 

[irun4edmund]

I calculated the velocity of the first ball to be v = (2gh)^1/2 or 16.27

Then the moment of inertia for the system to be I = m₁(L/2) + m₂(L/2) + (1/12)m_barL² or 57.12

Using these numbers i got w = 2.00 and v₂ = 10.8 and found that the ball goes 1.48 m high... but this isn't the right answer.

 

[LowlyPion]

I calculated the velocity of the first ball to be v = (2gh)^1/2 or 16.27

Then the moment of inertia for the system to be I = m₁(L/2) + m₂(L/2) + (1/12)m_barL² or 57.12

Using these numbers i got w = 2.00 and v₂ = 10.8 and found that the ball goes 1.48 m high... but this isn't the right answer.

My mistake, the moments of the masses are given by.

I = m*r² = m*L²/4

I just scribbled out the torque.

 

[irun4edmund]

ok, i corrected the moment of inertia, but i still didn't get the right answer... I took another look at linear momentum and figured this thing out.

I*w = m₁v₁r not where r = L/2
I*w = m₁v₁ like previously stated.

Thanks for your help. problem solved.