Yes. I just posted about MythBusters. However, I thought this would be an appropriate time to also talk about the physics involved with the "shoot the merry go round" myth.
The basic idea of the myth was to test this scene from some movie where a guy shoots a merry go round to make it spin. I think the MythBusters did their standard fine job of testing this. But what about the physics? Diagram time.
After the bullet strikes the merry go round, let me assume that the bullet sticks (which isn't very likely) to the structure. This then causes the merry go round to spin with an angular speed ω about its axis. So then, what is the main physics principle that applies here? If you said "conservation of momentum", that would be an excellent answer. Excellent, but wrong. You can say momentum is conserved when there are no external forces on the system. In this case, the system would be the bullet plus the merry go round and there is an external force. No, not gravity (well, yes) but I was thinking of the axle. The merry go round can turn, but its center of mass can't move. When the bullet hits, the axle exerts a force to prevent the merry go round from moving so that momentum is not conserved. You could make momentum conserved, but you would have to include the Earth in the system also. You probably don't want to do that.
If momentum is not conserved, what can we do? We can use angular momentum. The angular momentum principle says:
This is very similar to the momentum principle - the change in momentum is equal to the net force. For the angular momentum principle, the change in angular momentum is equal to the net torque. If the system is the bullet and the merry go round, the net torque is zero. This means that the change in angular momentum is zero or that the angular momentum before is equal to the angular momentum afterwards. But what is angular momentum?
For a point mass, the angular momentum (about some point o) can be defined as the scalar (though it is really a vector):
If this point mass is moving in a straight line near some point o, then ro is the distance from the point o to the mass. You might find it surprising that the angular momentum of this object would be constant as it gets closer to the point o.
The easiest way to find the angular momentum of a point mass (like a shooting bullet) would be to use the perpendicular distance of the path of the bullet to the point about which you want the angular momentum.
For an extended object (like the merry go round), the angular momentum is (again - the scalar form):
Here, I is the moment of inertia for that object (or what I like to call the rotational mass). Basically, it depends on the mass of that object, the size, and how the mass is distributed about the axis of rotation. ω is the angular velocity of the object. If I assume the merry go round is like a cylinder, I can say:
Ok. I know that was short, but I wanted to get to the calculations. Let's get it going. Using my diagram from above, I can say the before and after angular momentum are:
What is the moment of inertia for the merry go round with a bullet stuck in it? Technically it would be:
Since the merry go round has a weight of around 500 pounds (at least that is what they said in the show) and the bullet has a mass of a few grams, the contribution of the bullet just doesn't matter. This means that the final angular velocity of the merry go round would be:
Data from MythBusters
Now for some estimated values. From the show, they shot several rounds at the merry go round. The 9 mm round was listed has having a kinetic energy of 383 foot-pounds and a velocity of 1300 feet per second (396 m/s). 383 foot pounds is the same as 519 Joules (you can do this conversion with google calculator). If the KE and the velocity are known, I can solve for the mass of the round:
Using this give a mass of 6.6 grams. Seems ok to me. What about the other values? For the merry go round, it looks like they used this 8 foot diameter one. That means that R is about 1.2 meters and the mass is about 227 kg. Sure, it isn't actually a cylinder, but it is close enough. For ri (the distance that the bullet hits the merry go round), I will use 1.1 meter.
That is all that I need to calculate the final angular velocity. Putting those values in, I get:
With that angular speed, it would take almost 6 minutes to make one revolution. Oh, and that is assuming there is no friction. What about with that 50 cal. sniper rifle thing? MythBusters list it as having a kinetic energy of 13,000 foot pounds (17,625 Joules) and a speed of 2900 ft/s (884 m/s). Using the same ideas as above, this means it has a mass of 0.045 kg. If it sticks to the merry go round (or at least stops when it hits) it would give a final angular speed of 0.27 rad/sec. This would take 23 seconds to make one rotation. Not too bad. Oh, this is without friction.
Oh, just for comparison - what about a person? Suppose a 65 kg person runs with a speed of 4 m/s and hits the merry go round (but doesn't jump on) and stops. Using the same expression above, this would give the merry go round an angular speed of 1.7 rad/s. Big difference.
Taking into account friction
For the first set of tests, the MythBusters used a seemingly standard merry go round. They pulled with a force gauge to get an estimate of the frictional force. When pulling near the edge, it took 8.6 pounds (38 Newtons) to turn. Let me assume this gives a friction-torque of 38 N * 1 meter = 38 N*m (about).
Suppose the 9 mm hit the merry go round. If I take the bullet + merry go round as the system, then the friction-torque would do work on it. I can write:
The work done by a torque is just the torque times the angle through which the thing rotates. The rotational kinetic energy is:
Putting this together, I get:
Using the torque above and the initial (which I called the final) angular velocity of 0.018 rad/sec, I get an angle of 7 x 10-4 radians or just 0.04 degrees. With a radius of 1.2 meters, this would be a displacement on the edge of 0.08 cm (which seems about what they showed on MythBusters).
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