Center of percussion of a baseball bat

[CricK0es]

Homework Statement

A baseball rests on a frictionless, horizontal surface. The bat has a length of 0.900m, a mas of 0.800kg, and its center of mass is 0.600m from the handle end of the bat (see figure below). The moment of inertia of the bat about its center of mass is 0.0530 kg.m^2. The bat is struck by a baseball traveling perpendicular to the bat. The impact applies an impulse

$$J= \int_{t1}^{t2} F dt$$

at a distance x from the handle end of the bat. What must x be so that the handle end of the bat remains at rest as the bat begins to move? (Hint: consider the motion of the center of mass and the rotation about the center of mass. Finc x so that these two motions combine to give v=0 for the end of the bat just after the collision. Also, remember that . The point on the bat you have located is called the center of percussion. Hitting a
pitched ball at the center of percussion of the bat minimizes the "sting" the batter experiences on the hands.

Homework Equations

$$J= \int_{t1}^{t2} F dt$$

Standard equation for torque[/B]

The Attempt at a Solution

This involves both translation and rotational steps. In order to make v = 0, the bat must move in such a way, so as to counteract the rotation. This isn't a homework question, just one I've found in a book and I'm unsure how to proceed. I would appreciate someone explaining how the answer of 0.710m is found. I have a class test in a few days and I want to try and cover as many of my bases as possible. Many thanks[/B]

 

[Doc Al]

First things first: Given the impulse, how can you find the resulting speed of the center of mass and the rotational speed about the center of mass?

 

[CricK0es]

Integral of the force= the impulse, will be equal to m(V₂-V₁) with V₁ being equal to zero (Centre of mass). The rotational speed I'm not too sure... Could we divide the impulse by Δt and multiply by the distance x, to get torque. Then integrate that between t2 and t1 to get angular momentum, for which, L = I . ω ?

 

[Doc Al]

Integral of the force= the impulse, will be equal to m(V2-V1) with V1 being equal to zero (Centre of mass).

Good.

The rotational speed I'm not too sure... Could we divide the impulse by Δt and multiply by the distance x, to get torque. Then integrate that between t2 and t1 to get angular momentum, for which, L = I . ω ?

Exactly. Make sure you are taking the torque about the center of mass.

Once you have those squared away, what do you think the criteria would be for the handle end of the bat to be not moving after the collision? (You'll need to combine those two motions.)

 

[CricK0es]

The magnitude of Vcm and Vtan (Velocity of centre of mass/tangential) will need to be the same at that point, but in opposite directions, or is that an over simplification? I understand that the two velocities cancel one another out... But how would I go about calculating that?

 

[Doc Al]

The magnitude of Vcm and Vtan (Velocity of centre of mass/tangential) will need to be the same at that point, but in opposite directions, or is that an over simplification?

Good. That's all there is to it.

I understand that the two velocities cancel one another out... But how would I go about calculating that?

How would you calculate Vtan? (Use ω.)

 

[CricK0es]

So it's just using v=rω? Okay, I'll have a run through it tomorrow and see if I have any more luck.Thanks

 

[Doc Al]

So it's just using v=rω?

Yep.

Okay, I'll have a run through it tomorrow and see if I have any more luck.

Good!

 

[CricK0es]

I know this is from ages ago, but I only just got around to doing it. I got it working with x = 0.710m! So thank you. I really appreciate it

 

[nathane14dur]

how does the part where we integrate torque with respect to time work without time values?

 

[haruspex]

how does the part where we integrate torque with respect to time work without time values?

If you study the thread you will see that no integral was performed. It suffices to consider the impulse, J, as a whole. It supplies both a change in linear momentum and a change in angular momentum.