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hutt132
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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 mrod = 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 Wg = -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
Wgravity = -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
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