Ball rotating on axle which is rotating itself

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Homework Help Overview

The problem involves a ball of mass m attached to a rotating axle, with the task of determining the linear and angular momentum of the ball. The context includes specific values for mass, rod length, and angular velocity, as well as a diagram provided by the original poster to illustrate the setup.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculation of linear momentum using the formula p=mv and the relationship between linear velocity and angular velocity. There is confusion regarding the calculation of angular momentum about point A, particularly due to the rotation of the axle. Some participants clarify that point A is a reference point and does not itself have rotation.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the rotating axle on the calculations. There is an acknowledgment of confusion regarding the setup, and some guidance has been offered about the nature of point A as a reference point.

Contextual Notes

Participants note that the problem may be interpreted differently, with some suggesting that there is only one ball involved, while others clarify the role of point A in the context of the rotating system.

NathanLeduc1
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Homework Statement


A ball of mass m is attached via a rod of length x to an axle that rotates with angular velocity ω. You can consider the ball to be a point mass.
m = 5 kg, x = 0.3 m, y = 0.4 m, ω= 30 rad/s

(a) What is the linear momentum (direction and magnitude) of the ball?
(b) What is the angular momentum (direction and magnitude) of the ball about point A?

I've included a diagram I made.The top ball has mass m = 5 kg and the bottom ball A is rotating with angular velocity ω. Hopefully that diagram makes sense...

Homework Equations


I = mr2
L=Iω
p=mv
v=rω


The Attempt at a Solution


(a) p=mv
v=rω=0.3m*30rad/s=9m/s
p=5kg*9m/s=45kgm/s clockwise
(b) L=Iω
L=mr2ω
This is where I am confused. It would be easy to calculate if A wasn't rotating but how do I . calculate the angular momentum of the ball about point A given the fact that A is rotating?
 

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NathanLeduc1 said:

Homework Statement


A ball of mass m is attached via a rod of length x to an axle that rotates with angular velocity ω. You can consider the ball to be a point mass.
m = 5 kg, x = 0.3 m, y = 0.4 m, ω= 30 rad/s

(a) What is the linear momentum (direction and magnitude) of the ball?
(b) What is the angular momentum (direction and magnitude) of the ball about point A?

I've included a diagram I made.The top ball has mass m = 5 kg and the bottom ball A is rotating with angular velocity ω. Hopefully that diagram makes sense...

Homework Equations


I = mr2
L=Iω
p=mv
v=rω


The Attempt at a Solution


(a) p=mv
v=rω=0.3m*30rad/s=9m/s
p=5kg*9m/s=45kgm/s clockwise
(b) L=Iω
L=mr2ω
This is where I am confused. It would be easy to calculate if A wasn't rotating but how do I . calculate the angular momentum of the ball about point A given the fact that A is rotating?
As I read this problem, there is only one ball. No ball at point A.

attachment.php?attachmentid=58379&d=1367386096.jpg
 
Ah, you again. Thanks for the help! :)

I wish I could show you the diagram on my paper... it demonstrates the problem a lot better. The axle itself is rotating with angular velocity ω. There might not necessarily be a ball at the end of the axle but the point at the end of the axle is labeled A.
 
NathanLeduc1 said:
Ah, you again. Thanks for the help! :)

I wish I could show you the diagram on my paper... it demonstrates the problem a lot better. The axle itself is rotating with angular velocity ω. There might not necessarily be a ball at the end of the axle but the point at the end of the axle is labeled A.
Do this mean you still haven't solved the problem ?
 
A is just a point in space taken as a reference. The axle rotates about there, but points don't have rotation.
 

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