---
title: 'How Backspin Actually Works - in Super Slow Motion'
source: 'https://youtube.com/watch?v=hgcEIKHCI00'
video_id: 'hgcEIKHCI00'
date: 2026-07-28
duration_sec: 949
---

# How Backspin Actually Works - in Super Slow Motion

> Source: [How Backspin Actually Works - in Super Slow Motion](https://youtube.com/watch?v=hgcEIKHCI00)

## Summary

This video explores the physics of spinning ball collisions, particularly backspin, using slow-motion footage. It explains how the horizontal coefficient of restitution (HCR) provides a framework for predicting ball behavior, but reveals that HCR is not a constant and depends on collision parameters like angle, velocity, and spin. The host demonstrates experiments with various balls and surfaces to illustrate the concepts.

### Key Points

- **Simplifying Non-Spin Collisions** [00:35] — For collisions without spin, physicists ignore force-time details and focus on impulse, the area under the force-time graph, which equals change in momentum.
- **Coefficient of Restitution** [03:30] — The coefficient of restitution (COR) quantifies elasticity, ranging from 1 (perfectly elastic) to 0 (perfectly inelastic), simplifying energy loss analysis.
- **Friction in Spin Collisions** [04:24] — When a spinning ball hits a surface, friction acts opposite to the contact point's motion, altering horizontal velocity and applying torque that reduces spin.
- **Horizontal Coefficient of Restitution** [05:45] — The HCR describes the ratio of horizontal velocity of the contact point after and before collision. It can be negative, zero, or positive.
- **Static Friction Dominance** [07:03] — In many collisions, static friction eventually engages, causing the ball to roll (HCR=0) as the contact point becomes stationary relative to the surface.
- **Equations for Spin and Velocity** [08:34] — Using HCR, one can write equations for post-collision horizontal velocity and spin, though these require measured HCR values specific to the collision.
- **HCR Variability** [11:43] — Unlike vertical COR, HCR is not constant; it depends on collision parameters such as angle, speed, and incoming spin, limiting its predictive power.
- **Real-World Example** [13:02] — In the original ball-under-table experiment, the assumption of rolling during collision is invalid because spin changes direction, indicating HCR near 1.
- **Angular Momentum Clarification** [14:09] — The host corrects a previous error: angular momentum and linear momentum are independent; one is not converted to the other.

### Conclusion

While the horizontal coefficient of restitution provides a useful framework for analyzing spinning collisions, it is not a constant like its vertical counterpart, making predictions less straightforward. The video emphasizes the importance of understanding the role of friction and collision time in determining ball behavior.

## Transcript

it's hard to bounce a ball under a table. It always wants to come back to you because of backspin. And I got a few comments along the lines of, "Well, actually happening during that collision? How does backspin actually
change the direction of the ball? I want to see that collision close up in slow course I had to do that, and I discovered some really cool things along the way. Collisions that don't involve spin are easy to understand by making
these simplifying assumptions that just work. You don't need slow-mo to understand it. But when you try to do the same with spin collisions, you end look at collisions that don't involve spin first because I want to show you
how making one simplification allows you to explain the behavior of a huge range glance, it might seem quite complicated. ball will be after the collision, we need to work out the change in its
momentum. Momentum changes when a force is applied. The momentum change is larger when the force is larger and when the force is applied for a longer time. force and the duration of the collision. And if we multiply the two together, we
get the change in momentum, sometimes called the impulse. And you can see that it's the area under this force-time graph. But measuring the force and duration is going to be really difficult. And in reality, the force
collision. So, we need to do an integration to get the impulse, to get the area under the graph. We're screwed, basically. And this is where the physicists say is, "Look, don't worry about the details. Don't worry about how
about the details. Don't worry about how high the force gets, how long the collision lasts. It's the area that's important." Like, if you double the force but halve the duration of the collision, the area would still be the
same. And so, physicists say, "Like, if we can find another way to work out what the area under the graph should be, what the impulse should be, which tells us the change in momentum, then we don't need to worry about the shape of it,
because that won't affect the end result once the collision has finished. In fact, let's just assume that it's arbitrarily thin. Let's assume that the collision is instantaneous. So, what is this other way of working out the area
under the graph, the impulse? I'll get to that in a sec, but first a note about instant aneous collisions. Real collisions are never instantaneous. And hold on, if the collision has a zero second duration, then the force must be
infinite at that single moment in time. And that's true. And generally in anyway, how could you work out the area under a graph that's infinitely thin and infinitely tall? Zero times infinity is undefined, except in this case it is
actually defined, because it's the limit of a graph that has a fixed area there. So, back to the question, how do you physicists decide what the area impulse should be? Well, you start by making another assumption. You assume
that the collision is perfectly elastic. That means that the momentum of the ball as the momentum of the ball going into let's define down as the negative direction, and let's imagine we have a
ball with a momentum of minus one going into the collision. If it's perfectly elastic, it'll have a momentum of plus one coming out of the collision. So, then the change in momentum is plus two. The impulse is plus two. Now, not all
collisions are perfectly elastic. What's great is you can take all the complexity of like, you know, thinking about the material properties of the ball and the surface, like, um, how is energy dissipated, what's the damping like, and
you just roll it into a single number called the coefficient of restitution. coefficient of restitution of one. Perfectly inelastic collisions have a coefficient of restitution of zero. If a ball has half the speed coming out of
the collision than it had going in, then that's a coefficient of restitution of a half, and so on. So, again, the details don't matter. You just measure the coefficient of restitution for your two objects by observing a collision. And
typically it stays fairly constant over a wide range of velocities, so you can scenarios. But if you try and do the same thing with collisions involving spin, it goes wrong. Like, there doesn't seem to be a way to make a simplifying
assumption that nicely predicts the direction and spin of the ball after the that, here's a qualitative description of what's going on. When a spinning ball hits a surface, it experiences friction. Friction always acts in the opposite
direction to the direction of motion. So, here, where the surface of the ball is traveling to the right, it's going to experience friction pointing to the force to the left, its velocity increases in that direction. So, we see
velocity towards the left. But because the force is being applied off center ball, the ball also experiences a torque, a twisting force. This torque is
clockwise, which is opposite to the way the ball is spinning. So, we expect the decrease, which is what we see. But when you try to analyze it quantitatively, things go wrong. Like, it starts off well, and I really like this. This is
learned from making this video. What you normally think of is the coefficient of ratio of the momentum of the ball coming out of the collision to the momentum of That's actually just the vertical coefficient of restitution. There's also
a horizontal coefficient of restitution, and it has some weird properties. horizontal coefficient of restitution a lot, so I'll be abbreviating that to HCR be really clear about what the horizontal coefficient of restitution
is. Like, imagine this ball is coming in. It's a really oblique angle, so we component of the coefficient of restitution. And imagine for some reason this ball isn't able to spin. I don't know why or how that would be, but just
imagine it, right? So, the ball hits the surface. I can feel the resistance there. So, the ball is deforming elastically. It's storing elastic potential energy, and then it springs back, making the ball travel back in the
direction that it came. Right? So, if the collision had an HCR of one, it would bounce back at the same speed that it came in with, if it wasn't able to right? It's not about the center of mass
the ball touches the floor. And that's important because balls actually can spin, of course. So, with an HCR of one, it's that point on the ball that's speed it came in with. But, the ball itself is still traveling to the right.
The ball must be spinning like crazy for the bottom of the ball to be traveling backwards. Something unusual about the HCR is that it's very often really close to zero. And remember when the vertical coefficient of restitution is zero, that
corresponds to the ball stopping dead. So, why would the HCR be zero so often? Well, it's because static friction is stronger than kinetic friction. Take this collision, for example. This is a pool ball, and I'm kind of simulating it
bouncing off a pool table cushion here. Having some give in the cushions on a pool table extends the duration of the collision. And watch what happens. To begin with, the ball is sliding against the surface. Sliding surfaces experience
kinetic friction that slows the rotation of the ball and gives it some horizontal velocity. Eventually, the two surfaces are moving slow enough relative to each other that static friction kicks in, and they grip each other. In other words,
the ball is now rolling, which is to say, the surface of the ball is not moving relative to the green baize. And if two surfaces aren't moving relative that the coefficient of restitution is zero. That doesn't mean that the center
horizontally. It means that the contacting surface isn't moving horizontally. It means that the horizontal velocity and the rate of spin are now linked. The horizontal velocity is the angular velocity times the radius
that, it seems weird that the HCR would ever be greater than zero. But in the same way that a non-spinning ball compresses on collision, which stores elastic potential energy then releases it for the bounce, when a spinning ball
hits a surface, it wants to keep spinning, but the friction is stopping it. Because it's elastic, though, it will twist up, storing elastic potential energy until it can't twist anymore before releasing that energy in an
opposite twist. So, the HCR can be anywhere between minus one and plus one. And brilliantly, we have a couple of equations that make use of it. The equation for the horizontal velocity after a collision looks like this. And
the spin after the collision looks like this. I won't go into the derivation, the description. The point is, we seem to have achieved for spin collisions what we did for non-spinning collisions, which is to hide all the details behind
because the vertical component of the collision is independent of all this description of everything that's going on without having to worry about the details. Like, you know, what is the actual force and duration of the
collision, or what is the actual torque changing the angular momentum of the coefficients. I mean, it's worth acknowledging that these equations aren't necessarily intuitive, but if you put in specific values for the HCR, like
one or zero, then intuitive results do come out. And look, I captured a bunch of these scenarios. Like here, the HCR is really close to one, I think, in which case we expect the ball to end up spinning in the opposite direction,
which is what we see here. Here, the HCR is less than one, and I believe that's because energy is being dissipated in the wobbliness of the ball. Here's a the wobbliness of the ball. Here's a much more extreme version of that.
like we discussed before. But what about this one? Look at this ball is traveling to the right before the collision, and it's still traveling to the right after the collision. So, this collision has a negative HCR, which
that it just can't happen with the vertical component of coefficient of restitution. It happens because the collision time isn't long enough for static friction to kick in and bring the HCR up to zero. That's also true for
obvious. The issue here is that the collision time is just insanely short. Quick tangent. Real-world collisions can get really close to instantaneous, or at least on like human perception time scales. And we know from the discussion
of impulse earlier that the force must be really high for short collisions, years ago in one of my very first videos. You can just about see in this SD footage that the force between these really stiff steel balls is enough to
bust a hole in this sheet of paper. End of tangent. So, it seems like we have some simple equations where we can plug in some measured values and then model and predict the behavior of spinning collisions. So, where's the problem
here? Well, the horizontal coefficient of restitution doesn't just depend on things like, you know, how squishy is the ball, how does the ball dissipate energy. It also depends on this relationship between
And all that depends heavily on the parameters of the collision, the angle and the velocity and the spin of the ball going into the collision. So, collisions that don't involve spin are easy to model and understand intuitively
assumption that the coefficient of restitution stays fairly constant over a wide range of collision velocities. But that's just not true for the horizontal coefficient of restitution. It's really just not a constant in any useful sense.
collisions. On the left, static friction kicks in, meaning after the collision, the ball is spinning at just the right speed so that the bottom surface is stationary in the horizontal axis. In other words, an HCR of zero. Whereas on
over and the HCR is negative. I achieved that difference by making the surface just as easily have been because the ball wasn't spinning as fast on the left or was coming down faster. But I had the
home. By the way, if you're wondering how I got the ball spinning so fast for those collisions, this is how. What's really nice about this analysis though, is that it has its uses. The paper linked in the description analyzes
tennis specifically. They found that within that specific analysis, static friction always kicked in at some point. And so the HCR was always somewhere between one and zero. And between those two extremes actually, the exit angle
didn't change that much. Weirdly, exit spin was harder to predict, but the apply this to the original ball under a table video and see that our assumption that the ball rolls during the collision isn't right. That's like saying that the
HCR is close to zero, but the spin changes direction, meaning the HCR is closer to one. So it's been really cool to see how making simplifying assumptions like we do in physics all the time has still been useful in this
scenario. But it's also cool to know that when my brain says, "Hold on, I'm in slow motion." There's actually a good reason for that. There was a big aha moment in this video for me. It was when I was seeing if I could analyze what was
going on in terms of the conservation of angular momentum, where I realized that the angular momentum of a spinning object is independent of where you choose the origin to be. So it makes sense to put the origin in the middle of
the object to make that calculation easier. And then you can move the origin somewhere else for a different calculation, and that angular momentum that you calculated earlier is the same as it was before. And if there's now
some translational motion, well, you just add the angular momentum from that calculated. But, just seeing how angular momentum behaves in this particular problem that I was working through really helped to solidify how it all
works and clear up a few misunderstandings that I had about it. should make, actually. In the ball under a table video, I said that at one point angular momentum is converted to linear momentum. That's not how it works.
independent of each other. Anyway, the point is I feel like I have a new skill now. And I get the same feeling with the sponsor of this video, Brilliant. Like, these aha! moments always come when I'm working through an actual problem, and
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