Radius of ball matters in fall speed

[nhmllr]

In physics class today, our teacher wanted to demonstrate that mass doesn't have an effect on fall speed. (The class was about gravitational potential energy becoming kinetic, mgh = 1/2 *mv^2 and all that)

Here was the set up: He had a small rail (two parallel thin metal poles) sloping down, becoming completely horizontal at the end, then leading off of a table. He would roll a dense metal spherical ball bearing down the rail, it would roll down the slope, roll off with a horizontal velocity and roll off of the table onto the floor. Where the ball would land, he had a piece of carbon paper over a piece of copy paper, such that the ball would leave an imprint right where it landed.

He rolled a small ball bearing and a larger metal ball (3 or 4 times the radius). But each time he tried, the metal ball would always travel 4 or 5 inches further than the small ball. He tried lining up the middles and being careful in general, but the large ball always traveled farther. The teacher was a little bit disturbed.

Air resistance and friction should be negligible in this example.

So... Why should the large ball go significantly farther?

I thought about this for a bit. I thought that maybe some of the energy in the larger ball was going into rotational kinetic energy. But as the velocity of the center of mass is equal to the tangential velocity of rotation, the kinetic energy lost to rotational kinetic energy should be proportional for both the large and small balls. In other words, I don't see how the radius would affect the velocity.

So what gives!?

Thanks.

 

[schaefera]

Clearly it must be friction somewhere in the system, because the velocity at the bottom is independent of mass as are the equations describing horizontal and vertical positions with respect to time.

 

[Drakkith]

I'd guess that the larger ball has more mass and takes longer to lose its energy to friction.

 

[nhmllr]

Clearly it must be friction somewhere in the system, because the velocity at the bottom is independent of mass as are the equations describing horizontal and vertical positions with respect to time.

Okay then. Why should friction affect the small ball so much substantially more?

 

[K^2]

Rotational KE actually does reduce speed off the ramp.PE -> KE

$\small mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

Now, for a solid sphere of uniform density, $\small I = \frac{2}{5}mR^2$. And for any rolling object $\small v = \omega R$.

$\small mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mR^2\right)\left(\frac{v}{R}\right)^2 = (\frac{1}{2} + \frac{1}{5})mv^2$

In other words, instead of $\small v=\sqrt{2gh}$ for point mass, for a solid sphere you get $\small v=\sqrt{\frac{10}{7}gh}$, which is significantly smaller.

So now that we've established that rotation does matter, it should be clear that varying moment of inertia will result in different speed off the ramp. A cylinder, for example, will roll off the ramp traveling slower than a sphere.

If the larger sphere had a heavier core and a lighter shell, it would readily explain the situation. If both spheres were perfectly uniform, then friction is all you have left. How large was the ball bearing?

Okay then. Why should friction affect the small ball so much substantially more?

Weight of the object increases as cube of the radius. Drag as square. So at any given velocity, weight/drag is higher for larger sphere, assuming the densities are equal. That means larger sphere will have higher terminal velocity.

Of course, with distances involved, the small bearing would have to be pretty tiny to make a difference that significant. That's why I ask you how big it actually is.

 

[nhmllr]

If the larger sphere had a heavier core and a lighter shell, it would readily explain the situation. If both spheres were perfectly uniform, then friction is all you have left. How large was the ball bearing?

Ah, very interesting question. It didn't cross my mind that the large ball bearing might not have uniform density. While if both balls were of uniform density in a frictionless environment, they really should land in the same place.

I don't have very accurate measurements of this ball. I was only observing the teacher handle the ball in the front of the class 16 hours ago. But from my memory I'd say it was about... 1.5 to 2 inches in diameter. Would a ball of that size typically be made out of different material?

 

[Nugatory]

Here was the set up: He had a small rail (two parallel thin metal poles) sloping down, becoming completely horizontal at the end, then leading off of a table.

With this style of ramp, the smaller ball leaves the table at a slightly lower height from the floor - both the center of mass and the bottom surface of the ball that's going to touch the carbon paper are lower. Therefore it spends less time in flight and covers a shorter horizontal distance before it touches the ground.

 

[nhmllr]

With this style of ramp, the smaller ball leaves the table at a slightly lower height from the floor - both the center of mass and the bottom surface of the ball that's going to touch the carbon paper are lower. Therefore it spends less time in flight and covers a shorter horizontal distance before it touches the ground.

Hmm... that seems negligible though.

However, the space between the poles might have an effect on the balls.

The large ball is essentially on a flat surface, while small ball has the two poles touching very close to the ball's axis.

Now, the rotational kinetic energy of ball is dependent on the tangential velocity of the outermost rotating particles.

When the ball is rolling on the ground, this tangential velocity is equal to the velocity at the center of mass.

However, in this rail case, what might be happening is that the tangential velocity is equal to the much more than the center of mass velocity, because the tangential velocity that is equal to the center of mass velocity is the velocity of the particles closer to the center.

In this way, the rotational energy will sap more energy from the kinetic energy when the ball is smaller on this ramp.

I will talk to the teacher tomorrow and perhaps we will conduct the experiment with a flat board instead of the rails and see what happens.

 

[K^2]

The large ball is essentially on a flat surface, while small ball has the two poles touching very close to the ball's axis.

Ha, ha, ha. I didn't even think about that. That explains it completely. This changes relationship in equations above from $\small v = \omega R$ to $\small v = \omega R'$ for some $\small R' < R$. If you track it through to the final velocity formula, this gives you a decrease in final velocity. The larger the disparity between $\small R$ and $\small R'$ the stronger the effect. And this is definitely going to be a strong enough effect to explain the difference.

I can't believe I didn't think of this right away. I was even picturing it correctly in my head with the two wires, and the ball rolling close to axis... Guh.

Mystery solved, people. Move along.