Center of Percussion. A baseball hat rests on a...

Center of Percussion. A baseball hat rests on a frictionless, horizontal surface. The bat has a length of $0.900 \mathrm{m},$ a mass of 0.800 $\mathrm{kg}$ and its center of mass is 0.600 $\mathrm{m}$ from the handle end of the hat (Fig. 10.64$)$ . The moment of inertia of the hat about its center of mass is 0.0530 $\mathrm{kg} \cdot \mathrm{m}^{2} .$ The bat is struck by a baseball traveling perpendicular to the bat. The impact applies an impulse $J=\int_{t_{1}}^{2} F d t$ at a point 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. Find $x$ so that these two motions combine to give $v=0$ for the end of the bat just after the collision. Also, note that integration of $\mathrm{Eq} .(10.29)$ gives $\Delta \boldsymbol{L}=\int_{t_{1}}^{t2}(\boldsymbol{\Sigma} \tau) d t$ (see Problem 10.88$) .1$ The point on the hat you have located is called the center of percussion. Hitting a pitched ball at the center of percussion of the bat minimees the "sting" the batter experiences on the hands.

Key Concepts

Center of Percussion

The center of percussion is the point on a swinging object, such as a bat or racket, where an impact causes rotation about the object's center of mass with minimal or zero reaction (i.e., translational acceleration) at the pivot or handle. This concept is important for optimizing impact locations to reduce shock felt by the user while transferring maximum energy to the object or projectile.

Impulse

Impulse is defined as the integral of force with respect to time and represents the change in momentum of an object due to a force acting over a time interval. It plays a key role in collision and impact problems by linking the applied force to the linear momentum of the object and, in rotational dynamics, to the angular momentum around a chosen axis.

Angular Momentum and Rotational Dynamics

Angular momentum is the rotational equivalent of linear momentum and describes the quantity of rotational motion an object possesses. In the context of an impulsive impact, both the impulse and the location of its application relative to the center of mass determine the resulting angular momentum. The added rotation, together with translation, defines the overall motion of a rigid body after the collision.

Center of Mass and Translational Motion

The center of mass is the point at which the mass of a body can be considered to be concentrated for the purposes of calculating translational motion. An impulse applied to a body not only initiates rotation but also translates its center of mass. Understanding the motion of the center of mass is crucial when relating the applied impulse to both the linear and rotational responses of the object.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotational motion about a given axis. It depends on the mass distribution relative to the axis of rotation. In impact problems, the moment of inertia about the center of mass helps determine the angular acceleration produced by an applied torque, which in turn influences the overall motion of the object after collision.

Transcript

00:01 Okay, so in this question, we're dealing with a baseball bat.

00:04 The bat has a length l of 0 .9000 meters.

00:10 It has a mass m of 0 .80.

00:17 I'll rewrite that.

00:18 0 .800 kilograms.

00:22 And its center of mass, which i'm going to call xcm, is 0 .0 .0 .0 .0.

00:33 600 meters from the handle end of the bat and we're also told that the moment of inertia of the bat around its center of mass is 0 .05 30 kilograms meters squared so for our x coordinate system i'll be calling the handle end of the bat are zero point with this information, we're supposed to analyze a collision.

01:04 A baseball hits the bat, and the bat is sitting horizontally on a frictionless surface.

01:13 So it's at rest, and there's no friction.

01:16 And a baseball hits the bat at a point that is a distance x from the handle end of the bat.

01:27 And we're supposed to find what this x value has to be in order for the handle not to move.

01:36 It can rotate, but it can't move translationally.

01:39 So as the bat rotates and starts to roll a little bit and moves translationally, the handle has to stay in place and only rotate.

01:51 So i'm going to draw a quick picture of this situation.

01:57 So we have our handle of the bat and then here is a very bad drawing of a bat.

02:06 So i'm going to say that the center of mass is here.

02:09 And then the ball, the baseball comes and hits it around this area.

02:17 And the reason that i said it was to the right of the center of mass instead of towards the handle end is because if it was towards the handle end, well, the we know that the bat is going to start moving downwards because it gets some added momentum going down.

02:34 But if it was hit on the left side of the center of mass, it would also start to rotate counterclockwise, which would mean the handle would start to move downward and then also rotate counterclockwise, which would make it go even further.

02:50 And the point of this question is that the handle does not move.

02:54 So in order for that to happen, we have to have downward.

02:57 Velocity if the ball hits it from this direction, but then the rotation of the bat has to cancel out that downward movement in the handle.

03:07 So the bat gets some downward velocity and then starts to rotate clockwise, which means that on this end, the linear velocity is here, v, going downward, but then the rotational velocity here, omega, goes counterclockwise.

03:26 And then the rotation and the linear movement balance out.

03:33 So what we need for that to happen is that v, the downward velocity overall of the bat, is equal to omega the rotational velocity of the bat times xcm.

03:52 We're saying that the translational speed of the rotational speed of the rotation.

03:57 At the handle end is going to be the rotational velocity of the bat, the bat, times the distance from the center of mass to the handle, since that will be our lever arm.

04:11 And what we're saying is that the translational velocity here of the rotation of the bat at the handle is equal magnitude to the linear velocity of the bat downward.

04:28 That will make it so that the handle of the bat only rotates but doesn't move, it doesn't change its position.

04:36 So that's something that we need to note, and we'll use that substitution later.

04:40 We're also given by the problem that the impulse j is equal to the integral from one point in time to another of the force times d t, and that the change in angular momentum is equal to the integral from one point in time to another of the net torque times d t so while these formulas independently aren't very helpful because we don't have any time measurements used in this equation we can also say that we can use this to say that impulse is going to be equal to the change in momentum and so we also have next to that the change in angular momentum.

05:36 And we've been able to connect linear momentum to angular momentum with the equation l is equal to r cross p, which when r and p, the lever arm and the momentum vector are perpendicular to each other, then the magnitude is just r times p...

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