Impulse and distance from the axis of rotation

In summary, the speaker is struggling with modeling the linear collision of a bat and ball using the conservation of angular momentum. They have had difficulty finding information on the reverse scenario and are unsure of their calculations. They also mention difficulties relating torques and forces and ask about the optimal place for the bat to hit the ball in terms of torques and forces. The collision is not perfectly inelastic and the ball is stationary before and after the collision.
  • #1
crudux_cruo
23
11
I'm trying to model the linear collision of a bat and a ball using the conservation of angular momentum. The ball is a point particle with at rest wrt the axis of rotation, and the bat is being treated as a rod of negligible radius. I have had to work through several problems involving a ball with a given linear momentum hitting a pole and finding the resulting angular momentum of the system, but I can't find any information on the reverse.

The red dot marks the axis of rotation

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My intuition says that the farther away from the handle that the bat hits the ball, the more force will be imparted on the ball. I cannot find anything to support this, and when I try the math myself I get very confusing results.

For reference I assume the bat has zero momentum after the collision, and so all momentum is transferred to the ball. If the ball (the yellow circle) is here, it has a given moment of inertia.
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However if the ball is closer to the handle, it would have less rotational inertia. This would mean that it would have greater angular velocity. Converting it to linear velocity shows the same trend.
proxy-image (1) copy 2.jpeg

I am almost positive that I am making a fundamental mistake on what I am sure is a painfully simple problem, but I cannot make sense of it. I am alright with challenging my intuition, but I'm not sure if I am going about it the right way.
 
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  • #2
What exactly are you trying to calculate? What are the initial conditions? Can the ball rotate? Is the collision elastic?

crudux_cruo said:
have had to work through several problems involving a ball with a given linear momentum hitting a pole and finding the resulting angular momentum of the system, but I can't find any information on the reverse.
Start at the bottom line and read upwards! I don't understand this issue...an equation is an equation.
 
  • #3
hutchphd said:
What exactly are you trying to calculate? What are the initial conditions? Can the ball rotate? Is the collision elastic?Start at the bottom line and read upwards! I don't understand this issue...an equation is an equation.
I apologize for the low quality post. I will figure it out another way.
 
  • #5
I formulated the problem using conservation of angular momentum and treated the system as objects with moment of inertia as stated above.

I am basing my logic off the logic that was used in a textbook. Angular collisions were never described in any detail in that book, and my ability to parse things together secondhand has not filled whatever gap there seems to be.

I am not sure how else to reformulate my question, and the fact that this is a consistent problem when posting on here tells me that my posts are low quality.
 
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  • #6
I'm happy to help but that is your call...
 
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  • #7
I genuinely appreciate it, and I apologize that I cannot communicate more effectively otherwise I would consider the offer. I hope that you enjoy the rest of your day.
 
  • #8
hutchphd said:
I'm happy to help but that is your call...
I realize that my responses were flustered and needlessly rude, so I apologize. I am going to take you up on your offer if you are still willing, but first I am going to figure out how to best reexpress my question. I think I am making the mistake of talking strictly about my intuition and being very vague about the actual physics.

Again, sorry for the earlier attitude. I'd hate to respond to your genuine attempt at help by acting chastised.
 
  • #9
hutchphd said:
I'm happy to help but that is your call...
I am having trouble relating torques and forces. I can convert a force to a torque, but I am having difficulty conceptualizing how I can do the reverse (in terms of when it would be appropriate to actually do that).

I asked about the baseball bat hypothetical because I was curious if there was an optimum place for the bat to hit the ball. Purely idealized hypothetical, I am speaking purely and entirely in terms of torques and forces.

Does the bat apply a greater force to the ball the further away the collision occurs from the axis of rotation (in this case, the handle)?

It's not a perfectly inelastic collision, the ball is stationary and the bat has a given angular momentum. The collision occurs and the bat is no longer moving and the ball must now have that same angular momentum. Doing the math and treating the ball as a point particle, the ball seems to have less moment of inertia the closer it is to the handle and thus more angular velocity. The ball is not rotating upon itself.

I have struggled thinking about how I can best phrase this question, so if I am still not clear please tell me and I will reiterate where I may be failing. I am trying to use the hypothetical in bold to arrive at an answer for my question that's italicized. Instead I am confusing myself and my intuition.

I hope I make a little more sense now, but if I am just repeating earlier points let me know and I will happily try again.
 
  • #10
crudux_cruo said:
I am having trouble relating torques and forces. I can convert a force to a torque, but I am having difficulty conceptualizing how I can do the reverse (in terms of when it would be appropriate to actually do that).
To my knowledge, I don't think you can find the force from the torque without specifying a direction for the force. You can use vector product to calculate torque ##\mathbf{\mathcal T}=\mathbf r\times\mathbf F##. However, its reverse operation will not give you a definitive direction for the force. Therefore it is impossible to infer both the direction or the angle the force vector makes with the arm vector and the magnitude of the force vector from the direction and the length of the torque created.

To learn more: https://math.stackexchange.com/questions/32600/whats-the-opposite-of-a-cross-product
 

FAQ: Impulse and distance from the axis of rotation

1. What is impulse and how is it related to distance from the axis of rotation?

Impulse is a measure of the change in momentum of an object and is calculated by multiplying the force applied to the object by the time it is applied. The distance from the axis of rotation is important because it determines the lever arm, which affects the torque and therefore the impulse produced.

2. How does increasing the distance from the axis of rotation affect the impulse?

Increasing the distance from the axis of rotation increases the lever arm, which in turn increases the torque and therefore the impulse. This is because a longer lever arm allows for a greater perpendicular distance between the applied force and the axis of rotation, resulting in a greater torque.

3. What is the relationship between impulse and angular velocity?

Impulse and angular velocity are directly proportional. This means that as the impulse increases, the angular velocity also increases. This is because the impulse causes a change in the object's angular momentum, resulting in a change in its angular velocity.

4. How does the direction of the applied force affect the impulse?

The direction of the applied force relative to the direction of motion affects the impulse. If the force is applied in the same direction as the motion, it will increase the object's momentum and therefore its impulse. However, if the force is applied in the opposite direction, it will decrease the object's momentum and impulse.

5. Can the impulse be negative?

Yes, the impulse can be negative. This occurs when the force is applied in the opposite direction of the object's motion, resulting in a decrease in its momentum. This can also happen if the force is applied at an angle, causing a decrease in the object's angular momentum and therefore a negative impulse.

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