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.

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