Disambiguation of my physics homework

In summary, the conversation discusses the concept of angular momentum and how it relates to a ball sliding and rotating on a surface. It explores the idea of net torque being zero and the implications on the ball's angular momentum. It also discusses the initial and final angular momentum about a point on the surface and how it is affected by friction and the ball's mass and radius. The conversation also raises questions about how a translating object starts to rotate and if angular and linear momentum can be equated.
  • #1
Whitebread1
28
0
This is driving me crazy. I need to make sure I'm reading my physics homework correctly.

Alright, a uniform, solid ball is pushed with initial velocity V(0) so that it slides. As the ball moves, kenetic friction between the ball and the surface it is traveling on causes the ball to rotate until it reaches some constant speed V(f). I need to find V(f) in terms of V(0) and the question walks through different ways to find it.

Part B: Show that the net torque on the ball is zero about any point O along the ball's path on the surface. What does this imply about the Ball's total angular momentum about O?

This question is attempting to show that the torque about any point on teh surface of the ball is 0, correct? Being tha the torque is 0 about such points, the angular momentum about such points would be constant.

Part C: Write expressions for the ball's initial and final angular momentum about point O. Equate these to find the ball's final linear speed V(f). Does yoru result depend on the ball's mass M and/or radius R?

If point O is on the surface of the ball, then the torque about O is 0 as we established in Part B. How the hell can you have a initial and final angular momentum about O if the torque about O is 0?! I must be misreading this part and its driving me crazy.
 
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  • #2
Before the ball starts to rotate, its center of mass is moving parallel to the table surface. The ball has the same angular momentum about any point on the table that it will touch as it moves in a straight line. Look at the fundamental definition of angular momentum.
 
  • #3
Whitebread1 said:
kenetic friction between the ball and the surface it is traveling on causes the ball to rotate until it reaches some constant speed V(f).

I'm wondering if the key is here -- "UNTIL". It rotates until at some point they then assume that it just slides?

Maybe it begins to rotate, then the rotation slowly dies away... as the friction decreases as the ball speeds up... Maybe this is why they set them equal?
 
  • #4
OlderDan said:
Before the ball starts to rotate, its center of mass is moving parallel to the table surface. The ball has the same angular momentum about any point on the table that it will touch as it moves in a straight line. Look at the fundamental definition of angular momentum.
Right, so I'm wondering how the next part can ask for an initial and a final angular velocity about point O, when its well known that the angular momentum about O is always the same.

My problem is basically where they come up with this initial and final angular momentum when point O is located on the surface the ball is sliding or rotating on.
 
  • #5
physics girl phd said:
I'm wondering if the key is here -- "UNTIL". It rotates until at some point they then assume that it just slides?

Maybe it begins to rotate, then the rotation slowly dies away... as the friction decreases as the ball speeds up... Maybe this is why they set them equal?
It slides first, then rotates.
 
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  • #6
Whitebread1 said:
Right, so I'm wondering how the next part can ask for an initial and a final angular velocity about point O, when its well known that the angular momentum about O is always the same.

My problem is basically where they come up with this initial and final angular momentum when point O is located on the surface the ball is sliding or rotating on.
The initial angular momentum is all from translation of the center of mass. The final angular momentum about a point O that is instantaneously under the rolling ball has two contributions.
 
  • #7
OlderDan said:
The initial angular momentum is all from translation of the center of mass. The final angular momentum about a point O that is instantaneously under the rolling ball has two contributions.
O, it would be the lower angular momentum due to the translational velocity decreasing as well as the rotation of the ball.
 
  • #8
Whitebread1 said:
O, it would be the lower angular momentum due to the translational velocity decreasing as well as the rotation of the ball.
Thats it. Figure out the angular momentum of a ball rotating about a surface point in terms of the linear velocity of its center and you have all you need.
 
  • #9
This brings up a question about how a translating round object starts to rotate as it moves over a flat surface with no incline that has friction.

As the object moves initially, the point of the object in contact with the surface has a linear velocity equal to the velocity of the center of mass. AS the object begins to rotate, the linear velocity of the center of mass decreases and angular momentum increases. This means that part of the momentum that was once linear is now angular, right? Can angular and linear momentum be equated in such a fashion? And still will this hold true if the reference point is the center of mass of the object?
 
  • #10
Whitebread1 said:
This brings up a question about how a translating round object starts to rotate as it moves over a flat surface with no incline that has friction.

As the object moves initially, the point of the object in contact with the surface has a linear velocity equal to the velocity of the center of mass. AS the object begins to rotate, the linear velocity of the center of mass decreases and angular momentum increases. This means that part of the momentum that was once linear is now angular, right? Can angular and linear momentum be equated in such a fashion? And still will this hold true if the reference point is the center of mass of the object?
This problem can be done from another perspective. Friction is a constant retarding force that decelerates the center of mass. At the same time, friction exerts a constant torque about the center of mass that causes angular acceleration of the ball about its center. This continues for some time until the angular velocity and the linear velocity achieve the ratio needed for rolling. At that point slipping stops and the only friction is the very small rolling friction associated with deformation of the mat. During the deceleration phase work is being done by the friction, which reduces the energy of the ball, but it is a bit tricky to calculate the work because the ball gradually begins to rotate. The distance the force acts on the surface of the ball is less than the distance the center of the ball moves because of the rotation.

There is a connection between linear momentum and angular momentum

L = r x p

Notice that the angular momentum about the center of mass is not conserved in this process. There is a torque about the center of mass.
 
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  • #11
OlderDan said:
This problem can be done from another perspective. Friction is a constant retarding force that decelerates the center of mass. At the same time, friction exerts a constant torque about the center of mass that causes angular acceleration of the ball about its center. This continues for some time until the angular velocity and the linear velocity achieve the ratio needed for rolling. At that point slipping stops and the only friction is the very small rolling friction associated with deformation of the matt. During the deceleration phase work is being done by the friction, which reduces the energy of the ball, but it is a bit tricky to calculate the work because the ball gradually begins to rotate. The distance the force acts on the surface of the ball is less than the distance the center of the ball moves because of the rotation.

There is a connection between linear momentum and angular momentum

L = r x p

Notice that the angular momentum about the center of mass is not conserved in this process. There is a torque about the center of mass.
Thank you!
 
  • #12
physics girl phd said:
I'm wondering if the key is here -- "UNTIL". It rotates until at some point they then assume that it just slides?

Maybe it begins to rotate, then the rotation slowly dies away... as the friction decreases as the ball speeds up... Maybe this is why they set them equal?

Make some experiment. Kick a ball, or toss a marble or a cylinder-shaped something on a table - if you do it properly, so as you do not let it roll during you touch it and speed it up, it will slide for a short time, then it will roll till it stops at the end.

It happens because of kinetic friction: it works against sliding motion - translation- but accelerates rotation. When there is pure rolling, the point of the body in touch of the surface is in rest with respect to the surface, there is no work done by friction any more.

To solve the problem you need to know how the angular momentum with respect to a point is defined, and the conservation of angular momentum when the net torque with respect to that point is zero. Do not forget that the angular momentum for the in-plane motion of an extended body, with respect to certain point, is the sum of the angular momentum of the centre of mass and the angular momentum with respect to the centre of mass.

The other thing to take into account is how the translational speed of a rolling body (velocity of its centre of mass ) is related to the angular velocity w around the CM. You know, if not, you can try out that one roll of a cylindrical body will move the CM to a distance equal to the perimeter of the cylinder. The displacement is s=2pi*r, during the time of one roll, T, so v(translational) = s/T=2pi/T * r = w*r.

Can you proceed from here?

ehild
 

Related to Disambiguation of my physics homework

1. What is disambiguation?

Disambiguation is the process of removing ambiguity or uncertainty from a statement, word, or concept. In the context of physics homework, it refers to clarifying any confusion or ambiguity in the problem or its solution.

2. When should I use disambiguation in my physics homework?

You should use disambiguation whenever you encounter a problem or concept that is unclear or has multiple possible interpretations. It is especially important in physics, where precision and accuracy are crucial.

3. How can disambiguation help me in solving physics problems?

Disambiguation can help you identify and clarify any unclear or confusing aspects of a problem, allowing you to approach it with a clear understanding. This can lead to a more accurate and efficient solution.

4. What are some strategies for disambiguation in physics homework?

Some strategies for disambiguation in physics homework include carefully reading the problem statement, defining all variables and symbols, drawing diagrams or sketches, and seeking clarification from the instructor or classmates.

5. Can disambiguation affect my grade on physics homework?

Yes, disambiguation can have a significant impact on your grade in physics homework. A clear and accurate solution is necessary for a good grade, and disambiguation can help you achieve that.

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