I bet you thought there was going to be an awesome video. Sorry, but there is not. We used to have a nice merry go round at our park, but it is now gone. Anyway, I can't remember where I saw this question. It seems someone was working on a homework question.
The answer is no. If you just step off, the merry go round will keep going the same speed (angular speed). But why? Let me start with a diagram showing you just before and right after you step off.
The key concept here is angular momentum. Angular momentum is a lot like normal linear momentum except that it is totally different. In the simple, algebra-based course model angular momentum can be described as:
A quick note: really these should be vectors. However, in an introductory course, these are often described as scalars. If the object is on a fixed rotation axes, this is ok. There, I feel better for saying that. So, what is the I term? That is usually called the 'moment of inertia'. Probably a better name would be 'rotational mass'. Just like momentum (regular linear kind) is the product of mass and velocity, the angular momentum is the product of the rotational mass and the rotational velocity. You see how nice that is?
Here is a great demo showing the difference between mass and rotational mass. Rotational mass depends not only on the mass, but where the mass is with respect to the axis of rotation. In this demo, the two sticks have the same mass but different rotational masses. You should try something like this yourself - it is quite easy to set up.
Wasn't this post supposed to be about merry go rounds? Oh, right. Let me get to the angular momentum principle. This is very similar to Newton's laws (again, not the best name). Look at these two expressions.
What is that funny look τ? That is the torque. I will just say that torque is like the rotational force (get it)? It is ok, the net torque on the merry go round is zero (which really should be a vector). This means that the angular momentum does not change. This is just like the case where the net force is zero and the momentum (linear) does not change.
Why is there no torque on the merry go round? There is no torque because you just stepped off. If you had jumped off, it could make a difference - unless you jumped off in a radial direction (this also would exert no torque). No torque = no CHANGE in angular momentum. The mass and the shape of the merry go round didn't change so I doesn't change. This leaves the angular speed ( ω ) remaining the same.
But wait! (I know what you are thinking) Doesn't this mean that the total angular momentum of the guy plus the merry go round decreased? The guy (or girl) is no longer rotating. Ah HA! There is the trick. When you (or whoever) steps off the merry go round, you still have angular momentum even though you are not moving in a circle. Really.
If you are moving in a straight line, you could think of this as a non-constant angular velocity. Also, you can think of the moment of inertia of the person as changing since the person is moving farther away from the point of rotation. Here is a diagram showing the person moving in a straight line after leaving the merry go round.
At the first position, the person has an angular velocity and moment of inertia of:
Quick note: the "2" subscript is there because this is after the person jumped off the merry go round. Ok, now what about the next position? For the angular velocity, the radius changes as well as the component of the velocity that is going perpendicular to this radius (that part that is like it is moving in a circle). For the moment of inertia, the distance changes. This gives:
Let me get rid of the θ and r3 terms where:
This gives an angular momentum of:
The same as before. So, although the person is moving in a straight line the angular momentum (about that rotation point) is constant. The total angular momentum of the person-merry go round system is constant. Nothing happens to the angular speed when the person steps off.
What if the merry go round is going super fast? Here is an example.
Why would you do that? Well, you don't have to 'step off' in this case. And...if you like to see a video analysis of this event, here you go.
