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If a tennis ball and a bowling ball are dropped of a rooftop, they hit the ground at the same time. But if they are rolled down a slope, the bowling ball rolls faster. Why?

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closed as off-topic by David Hammen, Jon Custer, Bill N, Yashas, Tobias Kienzler Jul 30 '17 at 18:37

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    $\begingroup$ What's your evidence that the two drop through air at exactly the same rate? Regarding the slope, I'd expect the furry tennis ball to make greater surface contact & therefore have greater resistance. $\endgroup$ – Chappo Jul 28 '17 at 12:40
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    $\begingroup$ @Chappo assuming negligible air resistance. Honestly, that's just a complication that detracts from the root physics concept that the question is getting at $\endgroup$ – Jim Jul 28 '17 at 12:48
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    $\begingroup$ Also apart from the discussion of moment of inertia in the answer below, keep in mind that a soft tennis ball will experience much more kinetic energy loss during rolling simply from being soft (this is a part of so-called rolling friction) $\endgroup$ – Steeven Jul 28 '17 at 12:55
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    $\begingroup$ @Steeven Yet more details that are ultimately red herrings. Regardless how much rolling friction a tennis ball has, the reason it is slower when rolling is the moment of inertia. No need to over-complicate the issue; the gain of accuracy does not outweigh the loss of clarity $\endgroup$ – Jim Jul 28 '17 at 13:00
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    $\begingroup$ @all Why was this closed as HW-like? It asks about a physics concept. Yes, I can see this being a conceptual question on some student's assignment, but we can't disqualify valid questions because of that. Most of the questions on this site fall under that category $\endgroup$ – Jim Jul 31 '17 at 12:02
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The easy explanation is that the tennis ball is hollow.

When you merely drop the objects, they are subjected to the same acceleration - the aceleration due to gravity - and nothing else. Conservation of energy then says that their gravitational potential energy should be completely transformed into kinetic energy at the ground:

$$mg\Delta h=\frac{1}{2}mv^2\to v=\sqrt{2g\Delta h}$$

Since the initial heights $\Delta h$ are equal, they both have the same velocity as each other (though not constant in time) no matter how far they fall and, thus, hit at the same time.

However, when you roll them down the roof, the initial gravitational potential energy, $mg\Delta h$, is transformed not only into kinetic energy, but also into rotational energy. The rotational energy of something is $\frac{1}{2}I\omega^2$, where $I$ is the moment of inertia (the rotational equivalent of mass) and $\omega$ is the angular velocity ($\omega=v/r$; the velocity of the object divided by its radius).

This is all well and good, so the difference between the bowling ball and the tennis ball is now because the bowling ball is solid and the tennis ball is hollow. When just dropped, there is no difference. However, when rolling, the different distributions of mass affect the moments of inertia differently. A solid sphere has $I=\frac{2}{5}mr^2$, while a hollow sphere (I know the tennis ball is not perfectly hollow, but let's make this approximation, okay?) has $I=\frac{2}{3}mr^2$. What does this mean? Well, let's do the math (math is fun!).

For the bowling ball, we have: $$mgh=\frac{1}{2}\left(I\omega^2+mv^2\right)=\frac{1}{2}\left(\frac{2}{5}mr^2\cdot\frac{v^2}{r^2}+mv^2\right)\to v=\sqrt{\frac{10}{7}gh}$$

Whereas, for the tennis ball, we have: $$mgh=\frac{1}{2}\left(I\omega^2+mv^2\right)=\frac{1}{2}\left(\frac{2}{3}mr^2\cdot\frac{v^2}{r^2}+mv^2\right)\to v=\sqrt{\frac{6}{5}gh}$$

Notice that the mass of either ball is mostly irrelevant and that, since $\sqrt{\frac{10}{7}}>\sqrt{\frac{6}{5}}$, the forward velocity, $v$, of the bowling ball is greater than that of the tennis ball; just because one is hollow and one is solid.

It's also worth noting that the radius, as you may have concluded, does not ideally affect the forward velocity. This is something easily shown through the equations above as well as experimentally. Grab some solid spheres of different radii and roll them down an incline (I work in a physics teaching lab, so believe me when I say I've done this many times), you should see they hit the bottom at the same time. Yay! Physics is cool!

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  • $\begingroup$ Without trying to analyze the equations, is it true that if the tennis ball was solid and also the same mass that it would still roll slower because of the smaller radius? Thanks. $\endgroup$ – Brad S Jul 28 '17 at 12:59
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    $\begingroup$ @BradS No. You don't really need to analyze the equations, the final expression for the velocity of each ball only includes $g$, the gravitational acceleration, and the initial height, $h$. In an ideal situation, the radii of the balls make no more difference than their mass $\endgroup$ – Jim Jul 28 '17 at 13:02
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    $\begingroup$ @BradS $I$ does have a factor of $r^2$ in it, so you'd think there is some effect from the radius. But remember that $\omega$, the angular velocity, is just the forward velocity, $v$, divided by $r$; $\omega=v/r$. So when we have $I\omega^2$ in our rotational energy, the $r^2$ from $I$ gets cancelled out by the division by $r^2$ from $\omega^2$ $\endgroup$ – Jim Jul 28 '17 at 13:22
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    $\begingroup$ Most modern bowling balls are solid-ish (ignoring the holes), but they're not uniform density. Indeed, to make them roll faster they have high-density cores and low-density "covers", so the moment of inertia is even less than what you figure. Doesn't affect the validity of the answer, but it's an interesting fact. $\endgroup$ – hobbs Jul 28 '17 at 18:46
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    $\begingroup$ @hobbs It's very logical when you think of the function of a tennis ball vs. a bowling ball. A bowling ball is supposed to roll well. They make it hollow and heavy centered on purpose. A tennis ball is supposed to bounce well, so it has an appreciable amount of friction and a hollow compressible core. Tennis balls are poor rollers by design. $\endgroup$ – JMac Jul 29 '17 at 14:30
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I haven't tried this experiment but the first two factors that spring to mind are:

  1. Rolling Friction The bowling ball is hard and smooth while the tennis ball is fuzzy and softer. This would lead to a larger coefficient of rolling friction for the tennis ball.
  2. Distribution of Mass The tennis ball is hollow while the bowling ball is solid. This means that gram for gram the tennis ball will have a higher moment of inertia which means it takes a greater torque to get it rotate. See Jim's answer for much more detail. This is a variation of the classic classroom demo of a solid disc and a ring of equal mass and radius rolling down a ramp - the disk wins.
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Ignoring air resistance and other frictional effects other than those causing the objects to roll, the difference is due to the distribution of mass about the axis of rotation and not the actual mass of the two objects.

The moment of inertia of a body is a measure of the resistance the body to undergo an angular acceleration and the moment of inertia of a solid sphere (bowling ball) is proportionately smaller than that of a hollow sphere (tennis ball) by a factor of about $\frac 3 5$.
This means that the angular acceleration and so also the translational of the bowling ball is greater than that of the tennis ball.

Put another way, as the bowling ball rolls down a slope proportionately more of the gravitational potential it loses goes into translational kinetic energy and less into the rotational kinetic energy as compared with the energy transfers to a tennis ball.

The derivation for the acceleration of a solid sphere rolling down a slope is shown to be independent of the mass here and you can adapt the derivation to shown that the acceleration of the bowling ball is greater than that for a tennis ball.

When he objects fall again their accelerations are independent of mass and since all the loss in gravitational potential energy goes into only translational kinetic energy so the bodies accelerate at the same rate and reach the ground at the same time.

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