# Lecture 20.10 – The Sweet Spot of a Baseball Bat

PETER SAETA: OK. So allegedly, we have

here a rod suspended from something that can actually move. And I’m going to thump it

at different places along, and we need to watch which way

does the top move in response. So what should happen if I hit

the rod in its center of mass? STUDENT: [INAUDIBLE] PETER SAETA: So what should

you be seeing up here if I hit the rod this way at the center of mass? Should go that way, right? The whole thing should translate. Let’s see if we can get this to work. STUDENT: Woah. PETER SAETA: Did it? STUDENT: Yeah. PETER SAETA: We’ll say yes. What if I hit it at the top? STUDENT: [INAUDIBLE] PETER SAETA: Goes the way I hit it. What if I hit it at the bottom? STUDENT: [INAUDIBLE] PETER SAETA: All the way the bottom. All right, let’s see. I’m not watching because I’m hitting. So what did you see? Remember, believing is seeing. STUDENT: [INAUDIBLE] PETER SAETA: Watch it again. STUDENT: It went back. PETER SAETA: It went back, right? It went this way. [WHISTLES] OK? That was the way it went initially. So somewhere in the middle,

there’s got to be a point where it doesn’t move either direction, since

I hit high enough, it goes that way. I hit it low enough,

it recoils this way. Somewhere in the middle,

it should stay at rest. So let me try where there’s

this suspicious red line. Unpersuaded. Try it again. STUDENT: [INAUDIBLE] PETER SAETA: Remember,

Ian, believing is seeing. One more time, Ian. You will see it at rest. Did you see it, Ian? STUDENT: A little bit. PETER SAETA: A little bit. OK, I’ll take that as a moral victory. STUDENT: You will see it. PETER SAETA: You will see

what I tell you you will see. Nope, wrong. That one. OK, so I we’d like to

understand how to calculate where we would hit the rod, which

we’ll think of as a baseball bat, only a very poorly manufactured one. So it’s uniform. And we will hit it at some

distance, y, below the top. And what we would like

is to figure out, how far down do we need to hit it so that the

top doesn’t want to move left or right? That spot is called the sweet spot. And if you’re playing baseball

and you hit the ball with the bat at the sweet spot of the bat,

then the end that’s in your hands doesn’t need to recoil

one way or the other. And so you don’t get your palm stung. But if you hit it away

from the sweet spot, then your hands will have to exert

a big force on the end of the bat, or you’re going to drop the bat. STUDENT: Which, in baseball,

is perfectly OK to– PETER SAETA: Which– STUDENT: [INAUDIBLE] [LAUGHTER] PETER SAETA: I make no

justification for baseball. OK? I have no rationale for it. So the question is, how would we–

well, what do we want– what does that mean that the top is

not moving instantaneously? How are we going to describe that? That seems to me like

a geometric condition. In fact, isn’t this a lot like

something we already worked on? Rolling without slipping? How come? STUDENT: Because the top part

has angular velocity such that it counteracts the linear velocity. PETER SAETA: Oh, OK. So his argument is– you’re thinking

with respect to the center of mass? STUDENT: Something like that. PETER SAETA: OK, if we thump it at

the center of mass, the center of mass moves to the right. How much is it going to want to rotate? STUDENT: None. PETER SAETA: None, because you

thumped it at the center of mass. So you thumped it and produced no

torque about the center of mass. So no angular momentum about the

center of mass, so therefore, it’s just the center

of mass translation. That would not be consistent

with the top being fixed. That would be consistent

with the top moving to the right at the

center of mass velocity. So hitting it in the

middle is not a winner. It’s going to have to

be below the midpoint. But if it’s in rigid rotation about

that endpoint, that’s what we want. Right? Then this point would stay at rest. The center of mass

would move at its speed. And the bottom would have to move

twice as fast as the center of mass in order for everything

to work out so that it rotates as a rigid body around

a fixed point at the end. Is that right? Because if you think about the

rotation– if that end were fixed, then the farther you are away,

the faster you need to be going. Right? And in fact, if we call this

distance y, then the speed at y, right after being thumped, should

have what dependence on omega? STUDENT: [INAUDIBLE] PETER SAETA: Omega times y. So we’re going to thump it

at some place down here. And that will change its momentum. Suppose that the change

in momentum is delta p. And we might call this

direction the x direction. So it’s delta p in the x direction. How fast is the center

of mass going to move? If we thumped it, and

so we’ve transferred a momentum delta p to the rod. So delta p transferred from– what is

it that’s coming in?– from the bullet, the bat, the ball– from

the ball to the bat. OK? So if the bat now has

momentum, delta p, then [? VCM ?] must be delta

p divided by M. Right? OK, if the top is going to be

fixed, then we need this to be true. Right? So if the top is momentarily

fixed or at rest, then [? VCM ?] must be

equal to omega times what? STUDENT: d/2. PETER SAETA: d/2. OK, so far, have I worried

about angular momentum? No. So now, we have to worry

about angular momentum, which means we get to pick a reference point. Uh-oh. One more attempt at

picking a reference point. Any ideas? STUDENT: [INAUDIBLE] STUDENT: Where the ball hits. PETER SAETA: Where the ball hits. Why is that convenient? STUDENT: Because the

angular momentum is 0 again. PETER SAETA: Angular

momentum is 0 again. So let’s pick point where

ball hits bat as reference. OK? Then, L0 is 0. Therefore, L final is 0. So we just have to express L final. L final has an orbital part,

R cross P plus I prime omega. See if you can write it

down, the two terms, using y as the distance below the

top where we thumped the rod. You’re going to need y

minus something, I suspect. STUDENT: [INAUDIBLE] PETER SAETA: So the distance

between– OK, we’re using this point, but the center of mass is up there. So the center of mass is

translating to the right. And we already figured out VCM. So this is d/2 times what’s this? Delta p, which would

be MV, center of mass. And what direction is that? STUDENT: [INAUDIBLE] PETER SAETA: So this our

reference, and that’s the momentum, R cross P. Is that not into the board? STUDENT: [INAUDIBLE] PETER SAETA: Which way is it? Is this the reference we’re using? Yeah, this is where

he said to use, right? So R cross P– OK, R cross P, that’s in. What’s the moment of inertia of

the rod about its center of mass? 112. And which way is that? That’s going to be out,

which is convenient since we have to add these things up and get 0. OK. So that says, y minus d/2

times M times V center of mass is equal to 1/12 Md squared

omega, except that omega is 2 V center of mass over d. So y minus d/2 M V center

of mass is equal to 1/6 M V center of mass times d. So wipe these things out and

throw d/2 to the other side. And we get y is equal to d times

1/2 plus 1/6, which is 2/3. So if we hit the rod

2/3 of the way down, the top will stay at rest

at least momentarily. Eventually, of course,

it has to slide off. But momentarily, during the collision,

there’s no impulse at the top.