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Could somebody please explain this problem to me?(torque/angular momentum)

  • Thread starter nlsherrill
  • Start date
1. The problem statement, all variables and given/known data

Okay I just need someone to explain to me 2 things in the solution to a problem.

Here is a link to the solution manual for the book I am using for my University Physics 1 course.

http://oak.cats.ohiou.edu/~tt106402/work/phys252/Physics%20Notes/ch10.pdf [Broken]


if you search (crtl F) type in "A uniform rod of length L1 and mass", and it will take you to problem 55.

I must also note that the solution in the back of my book has L2/L1=0.36, and this solution manual online has it = 0.88, so I hope the one online is wrong, because as you will read below, I don't see the connections.

Heres a picture of the problem...its the only one I could find but you have to have cramster.

http://www.cramster.com/answers-nov-09/physics/angular-momentum-uniform-rod-length-l1-mass-08-kg-attached_705957.aspx

2. Relevant equations



3. The attempt at a solution

I honestly don't understand part 1. It seems to indicate that the gravitational torque of the rod = the rotational kinetic energy right before impact. Why is this? I understand that when the bar is directly below the pivot, the torque=0, so is it at that exact point that ALL of the gravitational torque gets "converted" into rotational kinetic energy? I just haven't seen this in a problem before, and there is nowhere in my book that I have found a direct relationship between torque and kinetic energy. Clearly I must be missing something.

I understand part 2, that angular momentum right before and after impact is conserved(because torque=0).

I also don't understand part 3. It seems to have "plugged in" the angular velocity solved for in part 1 into part 2, and used that to = something.

If any advice can be given on this problem it would be GREATLY appreciated, as my final exam is tomorrow..

Lastly, I don't know why alpha=m/M is used.
 
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okay I understand step 1 now...I figured mechanical energy would not be conserved considering there was a pivot which sometimes have friction. As for the rest, I have seen 2 different solutions and they all have different values for L2/L1.
 

Doc Al

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I honestly don't understand part 1. It seems to indicate that the gravitational torque of the rod = the rotational kinetic energy right before impact. Why is this? I understand that when the bar is directly below the pivot, the torque=0, so is it at that exact point that ALL of the gravitational torque gets "converted" into rotational kinetic energy? I just haven't seen this in a problem before, and there is nowhere in my book that I have found a direct relationship between torque and kinetic energy. Clearly I must be missing something.
Realize that you haven't given the figure, which is needed to describe the problem. Nonetheless, part 1 has nothing to do with torque. It's applying energy conservation to find the angular speed just before impact.

I understand part 2, that angular momentum right before and after impact is conserved(because torque=0).
OK.

I also don't understand part 3. It seems to have "plugged in" the angular velocity solved for in part 1 into part 2, and used that to = something.
Looks like energy conservation applied after the collision.

If any advice can be given on this problem it would be GREATLY appreciated, as my final exam is tomorrow..

Lastly, I don't know why alpha=m/M is used.
Why not? It's just a constant.
 
Realize that you haven't given the figure, which is needed to describe the problem. Nonetheless, part 1 has nothing to do with torque. It's applying energy conservation to find the angular speed just before impact.


OK.


Looks like energy conservation applied after the collision.


Why not? It's just a constant.


Thank you for your assistance Doc Al. Can you see the cramster solution? That one seems to make much more sense, and the answer is much different. Can you tell me which one is right(if either are)? By the way I am just doing practice problems to prepare for the final, so an answer means nothing compared to understanding it. I just say this because I want to emphasize that I am not just trying to get an answer out of someone for credit in the course.

But doesn't alpha represent the angular acceleration? How do you know that the ratio of the masses=the angular accel?
 

Doc Al

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Thank you for your assistance Doc Al. Can you see the cramster solution? That one seems to make much more sense, and the answer is much different. Can you tell me which one is right(if either are)? By the way I am just doing practice problems to prepare for the final, so an answer means nothing compared to understanding it. I just say this because I want to emphasize that I am not just trying to get an answer out of someone for credit in the course.
I only have access to what you linked. Can you post the problem itself with the diagram? If you want me to look at some other solution, post it.

But doesn't alpha represent the angular acceleration?
Not here. Here it's just being used as a constant. But alpha is often used to represent angular acceleration, so that might be a bit confusing. (But angular acceleration is irrelevant for this problem.)
How do you know that the ratio of the masses=the angular accel?
It's not. You are misinterpreting the meaning of alpha. They are just using alpha and beta to simplify the writing of the equations.
 
I only have access to what you linked. Can you post the problem itself with the diagram? If you want me to look at some other solution, post it.


Not here. Here it's just being used as a constant. But alpha is often used to represent angular acceleration, so that might be a bit confusing. (But angular acceleration is irrelevant for this problem.)

It's not. You are misinterpreting the meaning of alpha. They are just using alpha and beta to simplify the writing of the equations.
http://www.cramster.com/answers-nov-09/physics/angular-momentum-uniform-rod-length-l1-mass-08-kg-attached_705957.aspx

I just tried the link and it didn't work the first time, but did the second. I believe this may indicated that you have to be signed in. The problem number is 67 and its from chapter 10 of Physics for Scientists and Engineers (6th) by Tipler and Mosca
 
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Doc Al

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Note that the problem in cramster has different masses than in your solution manual, so you can't just compare answers. (I can't see the cramster solution.)
 
Note that the problem in cramster has different masses than in your solution manual, so you can't just compare answers. (I can't see the cramster solution.)
oh geeze your right. Here's the cramster solution:

Answer Details
M = 0.8 kg, m =0.41 kg, θmax= 60°, find L2/L1

moment of inertia of the rod I = (1/3)ML1^2

the rod rotates down, mechanical energy conservation:

Mg(L1/2) =(1/2)Iω^2 (1)

inelastic collision, angular momentum conservation:

Iω = (I +mL2^2)ω' (2)

rod and particle rotate up, mechanical energy conservation:

(I + mL2^2)(1/2)ω'^2 =((1/2)MgL1 + mgL2)*(1 -cosθmax) (3)

eliminate ω and ω' from the above 3 equations to get L2/L1 ˜ 0.47
 

Doc Al

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The cramster solution looks OK to me. (I didn't work out the arithmetic though.) If that's easier to understand than the solution in your solution manual, just apply the same method with the other values for the masses and compare. That way you'll confirm your understanding.
 

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