Work and Kinetic Energy of a Trebuchet

[hutt132]

Homework Statement

This time you construct a trebuchet. A counterweight (M = 550kg) is at one end a distance ℓ² = 1.5m away from the pivot. The child (m = 10kg) is a distance ℓ¹ = 3m away from the pivot. The mass of the uniform rod connecting the two is m^rod = 35kg. When released from the horizontal position, what will be the child’s velocity when she gets to the top?

There are two extra things to take into account: the rotational kinetic energy of the rod (breaking it up into two rods ℓ₁ and ℓ₂ might help here), and the gravitational work done on the rod. This will be W_g = -mgΔy as usual, but where Δy is the vertical displacement of the center of the rod.

Here's a picture of the problem:
http://img46.imageshack.us/img46/2929/v8xv.jpg

Homework Equations

Sum of Work = Change in Kinetic Energy
KE = 1/2 * m * V²
W_gravity = -m * g * Δy
ω = v / r

The Attempt at a Solution

This has to be solved without using potential energy.

The example the teacher gave in class was without the rod having a mass, but this time it has a mass so I'm not sure how to factor that in.

Here's the example from class with the massless rod:
http://img607.imageshack.us/img607/3025/vw4t.jpg

 

[CAF123]

The example the teacher gave in class was without the rod having a mass, but this time it has a mass so I'm not sure how to factor that in.

You can model the system as a uniform rod pivoting about it's centre of mass with two point masses at either end. You will need the moment of inertia of a point mass and that of the rod about an axis through it's centre of mass.

 

[hutt132]

You can model the system as a uniform rod pivoting about it's centre of mass with two point masses at either end. You will need the moment of inertia of a point mass and that of the rod about an axis through it's centre of mass.

I'm not sure how to do that, but I tried to follow the example the teacher gave in class and I got it filled out up to the point where I have to fill in the KE for the rod that I split into two.
Here's my work: http://img543.imageshack.us/img543/3576/4zm6.jpg

Could you please help me fill in the KE of the rod part?

 

[haruspex]

The worked example uses potential energy, but in the OP you say it is not to be used. Wg = -mgΔy is an example of using PE, not an alternative. Indeed, there's no point in calculating KE if you're not allowed to use PE.
Without PE, the only way I can think of is to develop the torque and angular acceleration equations. That will require some calculus.

 

[hutt132]

The worked example uses potential energy, but in the OP you say it is not to be used. Wg = -mgΔy is an example of using PE, not an alternative. Indeed, there's no point in calculating KE if you're not allowed to use PE.
Without PE, the only way I can think of is to develop the torque and angular acceleration equations. That will require some calculus.

Thanks for clearing that up. I see now that PE is used. I just can't figure out how to do the KE side of the equation for the rod.

 

[CAF123]

I'm not sure how to do that, but I tried to follow the example the teacher gave in class and I got it filled out up to the point where I have to fill in the KE for the rod that I split into two.
Here's my work: http://img543.imageshack.us/img543/3576/4zm6.jpg

Could you please help me fill in the KE of the rod part?

You forgot about the rotational kinetic energy contribution of the two masses. The moment of inertia of a point mass about an axis and a rod rotating about its C.O.M are results you can just use from a textbook or course notes.

I am not sure if the analysis of the gain in potential energy of each rod segment is correct since each mass element of the rod will not be at the same height.