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Bowling ball, incline and torque

  • Thread starter Niles
  • Start date
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1. Homework Statement

If we look at a bowling-ball rolling up an uncline without slipping, I have to find the acceleration of this. I was wondering:

May I choose by myself where to look at the torque? I mean, can I decide by myself whether I look at the center of the ball where the sum of the torque is the friction_force *radius or if I want to look at the point where the ball and the ground touch eachother and where gravity is the only force in the torque? I just use parallelaxis-theorem in this case.

I hope you understand.
 
318
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1. Homework Statement

If we look at a bowling-ball rolling up an uncline without slipping, I have to find the acceleration of this. I was wondering:

May I choose by myself where to look at the torque? I mean, can I decide by myself whether I look at the center of the ball where the sum of the torque is the friction_force *radius or if I want to look at the point where the ball and the ground touch eachother and where gravity is the only force in the torque? I just use parallelaxis-theorem in this case.

I hope you understand.

The one which you are telling about the center.That is correct but the one where you are considering the torque about the point of contact there you do have to consider mgsin(angle of inclination) which acts on the center of mass parallel to the inclined surface
 
1,868
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Cool, thanks.

I have two new questions. Take a look at the picture.

1) I want to find the distance using torque. I know that the sum of the torque must equal zero for the meter stick not to rotate. What is the first thing to do, if I want to approach this problem that way? (I know I can find the center of gravity, but this way of approach seems better in my case).

2) (Not related to the picture) If I have a stick of 1 m and a stone attached the one end, where do I have to put the origin to calculate the center of mass? Is it in the middle of the stick?
 

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318
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Cool, thanks.

I have two new questions. Take a look at the picture.

1) I want to find the distance using torque. I know that the sum of the torque must equal zero for the meter stick not to rotate. What is the first thing to do, if I want to approach this problem that way? (I know I can find the center of gravity, but this way of approach seems better in my case).

2) (Not related to the picture) If I have a stick of 1 m and a stone attached the one end, where do I have to put the origin to calculate the center of mass? Is it in the middle of the stick?
I cannot see the diagram so i am not able to answer the first.

For the Second i am not able to understand the question.
Do you mean to find out the center of mass of the system?
You can do it by integration or by just applying the equation


[tex]x_{cm} = \frac{m_{1}x_{1} + m_{2}x_{2}}{m_{1} + m_{2}}[/tex]

Where m1 is the mass of the stick and m2 is the mass of the stone and x1 and x2 are the x coordinates of the center of mass of the stick and the stone respectively.This will give you the x coordinate of the centre of mass of the system.
 
1,868
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I cannot see the diagram so i am not able to answer the first.

For the Second i am not able to understand the question.
Do you mean to find out the center of mass of the system?
You can do it by integration or by just applying the equation


[tex]x_{cm} = \frac{m_{1}x_{1} + m_{2}x_{2}}{m_{1} + m_{2}}[/tex]

Where m1 is the mass of the stick and m2 is the mass of the stone and x1 and x2 are the x coordinates of the center of mass of the stick and the stone respectively.This will give you the x coordinate of the centre of mass of the system.

Ok, thanks. Does it matter, where I put my origin and find the x's from there?
 
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All the coordinates are wrt to the origin.

I am not telling anything further.Its not that i am unwilling to explain it to you.But you will understand it better if you carry on this prob from the statement given above.

You will have to check your answer for various situations of the origin.
 

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