Clay-stick inertia & energy problem

AI Thread Summary
A thin stick of mass 2.8 kg and length 2.2 m is hinged at the top, and a clay piece of mass 0.8 kg strikes it at a distance of 1.65 m from the hinge. The discussion revolves around calculating the ratio of final mechanical energy to initial mechanical energy after the collision. Participants note that angular momentum conservation applies, but mechanical energy is not conserved due to the inelastic nature of the collision. One contributor suggests that if treated as a single system without friction, the ratio could be 1:1, while another clarifies that initial mechanical energy is calculated before the collision, indicating that energy is lost. The conversation highlights the complexities of energy conservation in inelastic collisions.
MickeyBlue
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Homework Statement


A thin stick of mass M = 2.8 kg and length L = 2.2 m is hinged at the top. A piece of clay, mass m = 0.8 kg and velocity V = 2.7 m/s hits the stick a distance x = 1.65 m from the hinge and sticks to it.

Q2: What is the ratio of the final mechanical energy to the initial mechanical energy?

Homework Equations


1. Lf = Li (conservation of angular momentum)
2. Is = ⅓ML2
3. Ic = mx2
4. Krot = ½Iω2
5. Kl =½mVt2

The Attempt at a Solution


Clay-stick problem.jpg

I can't find the problem with my working in part 2; wondering if someone can point me in the right direction please. I calculated the angular velocity as 0.53 rad/s.
 
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MickeyBlue said:

Homework Statement


A thin stick of mass M = 2.8 kg and length L = 2.2 m is hinged at the top. A piece of clay, mass m = 0.8 kg and velocity V = 2.7 m/s hits the stick a distance x = 1.65 m from the hinge and sticks to it.

Q2: What is the ratio of the final mechanical energy to the initial mechanical energy?

Homework Equations


1. Lf = Li (conservation of angular momentum)
2. Is = ⅓ML2
3. Ic = mx2
4. Krot = ½Iω2
5. Kl =½mVt2

The Attempt at a Solution


View attachment 106431
I can't find the problem with my working in part 2; wondering if someone can point me in the right direction please. I calculated the angular velocity as 0.53 rad/s.
I get almost the same, but you might have some rounding error. Keep another digit of precision all the way through. I get .325.
I note that it asks for the ratio, so technically the answer should be of the form 0.32:1. Do you know what the official answer is?
 
No, I don't have an official answer yet. I thought the same about the ratio, but my online submission doesn't recognise colons.
 
I'd like to follow up from my previous answer: if the clay and the stick are taken as one system, and we assume no friction about the pivot, is the case not that mechanical energy is conserved? That would make the ratio 1:1 (or just 1).

I'm not 100% sure on this.
 
MickeyBlue said:
I'd like to follow up from my previous answer: if the clay and the stick are taken as one system, and we assume no friction about the pivot, is the case not that mechanical energy is conserved? That would make the ratio 1:1 (or just 1).

I'm not 100% sure on this.
The initial mechanical energy must be that before the collision. Since the bodies coalesce, it cannot be conserved.
 
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