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# Energy Conservation Rotational Motion watch

1. I've attempted this question so many times and my answer is always slightly off. I'm unsure what I'm doing wrong. Please could someone have a go?

I've attached a picture with the necessary information. The rod was held horizontally and released from rest. The question asks to find the linear velocity of A at the bottom of its path. According to the textbook, the correct answer is 1.46m/s. Thanks

2. (Original post by PhyM23)
I've attempted this question so many times and my answer is always slightly off. I'm unsure what I'm doing wrong. Please could someone have a go?

I've attached a picture with the necessary information. The rod was held horizontally and released from rest. The question asks to find the linear velocity of A at the bottom of its path. According to the textbook, the correct answer is 1.46m/s. Thanks
Is it possible that you post your working?
3. (Original post by Eimmanuel)
Is it possible that you post your working?
Of course. Here you go
Attached Images

4. (Original post by PhyM23)
Of course. Here you go
Can clarify about the length of the rod:
in the first post, the separation between the mass A and B is l but in the 3rd posting, the separation becomes 2l.
So which is the correct data?
5. (Original post by Eimmanuel)
Can clarify about the length of the rod:
in the first post, the separation between the mass A and B is l but in the 3rd posting, the separation becomes 2l.
So which is the correct data?
Yes my apologies. The length of the rod is 0.8m. I used different notation on different occasions which is where the confusion lies. So in the second case, 2l=0.8m so l=0.4m
6. (Original post by PhyM23)
Of course. Here you go
It seems that you are mixing two different speeds together: the speed of the rod and mass.

I suggest you work on the angular speed omega and use it to compute the linear speed of mass A.

I can get 1.46 m/s.
7. (Original post by Eimmanuel)
It seems that you are mixing two different speeds together: the speed of the rod and mass.

I suggest you work on the angular speed omega and use it to compute the linear speed of mass A.

I can get 1.46 m/s.
Thank you for having a go. I've tried that I believe but I think I'm confused which energies I should use. Please may you show your working?
8. (Original post by PhyM23)
Thank you for having a go. I've tried that I believe but I think I'm confused which energies I should use. Please may you show your working?
Change in PE = Change in KE

Change in PE =

Change in KE =

I am using the first post info.
m

You should be able to use the two equations to find and find the linear speed.
9. (Original post by Eimmanuel)
Change in PE = Change in KE

Change in PE =

Change in KE =

I am using the first post info.
m

You should be able to use the two equations to find and find the linear speed.
May I ask why you use the moment of inertia of the rod and not the whole system?
10. (Original post by PhyM23)
May I ask why you use the moment of inertia of the rod and not the whole system?
How do you find the moment of inertia of the whole system?
11. (Original post by Eimmanuel)
How do you find the moment of inertia of the whole system?
I did

Where is the distance from each mass to the centre.
12. (Original post by PhyM23)
I did

Where is the distance from each mass to the centre.
Then you are double counting the KE of A and B based on your working.
13. (Original post by Eimmanuel)
Then you are double counting the KE of A and B based on your working.
Ah yes so I am. Thank you very much for your help; I really appreciate it.

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Updated: January 5, 2017
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