If you're holding on to a playground spinning wheel and then let go, is your trajectory straight or curved?

Let's say you are on one of those playground spinning wheels and then you let go. I've heard that you will fly off tangentially to the wheel, but do you really fly off in a straight line, since you must have some angular momentum? Or do you fly in a straight line, but spin on your own axis? Ignore air resistance and gravity, and you are a spherical cow.

• Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. Commented Jun 19 at 17:01

A child on a playground wheel normally rotates once per revolution of the wheel, as seen from the ground. If the child lets go, they will move in a straight line, but will continue to rotate at the same rate.

Note that after the child lets go, it retains the same angular momentum relative to the axis of the wheel, even though it is now moving in a straight line. An object frequently has angular momentum relative to an axis even if it's not revolving around the axis. In particular, an object moving in a straight line has nonzero angular momentum relative to any axis that does not intersect with the line.

There is no intrinsic requirement that the child rotate in synchrony with the wheel. It is entirely possible for the child to be non-rotating relative to the ground, or to rotate at a different rate. In that case, whatever its rotation rate is, it will retain it when it leaves the wheel. (Of course, in practice, friction and normal forces from the wheel will quickly synchronize the rotation of objects on it to the wheel's rotation.)

"Newton's first law expresses the principle of inertia: the natural behavior of a body is to move in a straight line at constant speed." When the child is on the spinning wheel there is a centripetal force exerted on them that keeps them moving in a circular path. When the child lets go, there are no forces acting on the child (we are ignoring gravity) so the child does what any other body does according to Newtons first law, which is continue in a straight line at constant velocity. Angular momentum is conserved. As the child moves away from the wheel, their radius vector from the centre of rotation increases, but the component of their linear velocity that is tangential to their radius vector reduces and $$M\times R\times V$$ remains constant. Any spin angular momentum due to the rotation of the child about its own axis s also conserved and remains the same before and after. If the child was not rotating in the reference frame of the spinning wheel their spin angular velocity would be the same as the initial angular velocity of the wheel according to a non rotating inertial observer, so the child would be moving in a straight line but rotating about their own comoving axis.

Since we are just ignoring air resistance but not air friction, the Magnus Effect comes into play.

Since you asked how I would ride a spinning wheel, and I prefer to keep my own rotation synchronized to that of the spinning wheel, I'll continue to spin after the release.

The Magnus Effect would provide a force perpendicular to the (initially tangential) trajectory, so my center of mass would follow a curved line. The force is caused by differences in air pressure.

Note that I, the spherical cow, does not necessarily have a uniform distribution of mass inside my body. To an observer, it might there look like my body "wobbled" along the trajectory. This also complicates the calculation of the Magnus Force.

Note that I assume that my body surface is uniformly smooth (otherwise delays of boundary-layer separation will make this more complicated as we add Knuckleball Trajectories (which will become more dominant if the wheel is spinning rather slowly) and which appear random in both magnitude and direction).

I also assume that there are no magnetic fields, electric fields, wind (and pressure gradients other than caused by the Magnus Effect), photons (light/heat), adjacent objects (to eliminate the Casimir Effect). Oh, and also assuming that there are no collisions and temperature or humidity gradients. Let us also assume that the spherical cow does not leak milk, vomit, defecate, breathe and radiates heat uniformly. We also ignore all effects that the spinning wheel could have on the cow post-release.

By all practical means...

...yes, the trajectory can be regarded as a reasonable approximation to a tangential trajectory.

• "...the spherical cow, does not necessarily have a uniform distribution of mass..." Sounds like it's time for the CIPM to establish a committee to draw up a proposal for the international standard cow. Commented Jun 16 at 12:18
• "we are just ignoring air resistance but not air friction". There's a difference? Commented Jun 16 at 17:47
• @RonJohn I suppose air resistance could be limited to aerodynamic drag forces, which by definition operate along the direction of motion. The Magnus effect operates perpendicular to motion and is actually a type of lift - I agree it's splitting hairs, but I suppose you could say the Magnus effect is caused by a pressure differential created by air friction but not air resistance. In most cases, though, I'd agree air friction, air resistance, and drag are the same thing. Commented Jun 17 at 13:47
• @RonJohn if you go into detail then yes there can be differences .. an example would be induced drag, which is a drag also in an inviscid fluid (or a at least an inviscid mathematical model) Commented Jun 18 at 4:08

Yes, you will travel in a straight line when you leave the merry-go-round.

You are absolutely right that angular momentum must be conserved in this situation. It's not immediately intuitive, but it turns out that an object moving in a straight line can contain angular momentum. If you run your thought experiment backward in time you can see how linear motion can create circular motion. To simplify things, suppose you've thrown a heavy ball at a merry go round's push bar. The ball travels in a straight line, but when it hits the bar, the merry-go-round begins to spin. The reverse is equally true.

Angular momentum ($$\vec{l}$$) can be defined mathematically as the vector product (aka cross product) of an object's position and its momentum. The position vector has to be defined relative to an axis of rotation. Sometimes that axis is obvious (the center of the merry go round, for example) and sometimes it's not, but the math works out in the end regardless of where we put the axis of rotation.

There are a bunch of ways of describing what a cross product is, but for this it is really only important to know that if the two vectors are parallel or antiparallel (pointing exactly in the same or opposite directions) then the cross product is zero. In my high quality diagram below, we have two balls, one red and one blue, with equal momenta $$\vec{p}$$ flying toward a merry-go-round, as seen from above. The red ball is lined up so that it will hit right at the edge of the merry-go-round, and the blue ball is lined up to hit the center post of the merry-go-round.

From life experience you probably know that the red ball that hits the edge of the merry-go-round will cause it to spin. The math says that in that case, $$\vec{l} = \vec{r} \times \vec{p}$$ gives a non-zero result as well (i.e. neither the position nor momentum vector is zero and they aren't parallel or antiparallel). This is true both when the ball is headed toward the merry-go-round (left side of my drawing) and at the moment it collides with the merry-go-round (right side). (The fact that angular momentum is conserved tells us that the two vector products should be equal, and it turns out that, not surprisingly, they are).

In the case of the blue ball, intuition tells us that when it hits the center post it will not cause the merry-go-round to spin. Mathematically, when the ball is incoming, the ball's angular momentum is zero because its linear momentum is antiparallel to its position vector. At the moment of impact, the position vector is reduced to zero, so the angular momentum remains zero. This time $$\vec{l} = \vec{r} \times \vec{p} = 0$$ and there is no rotation from the impact.

• Note that the reverse case is not equivalent to the forward case. A ball that leaves a merry-go-round has angular momentum around its own axis. A ball that hits a merry-go-round is assumed to not have angular momentum of its own before it hits. Commented Jun 17 at 14:03
• Indeed, but the question didn't ask about the rotation of the person coming off of the merry-go-round; it asked about the trajectory. I'm assuming point object people/balls/and cows. Commented Jun 18 at 1:32
• The OP does ask about the rotation of the person coming off of the merry-go-round. There are other differences as well: when the ball hits the merry-go-round, it applies a force to it, making it spin up. Your explanation is great to explain that phenomenon. But that is not what the OP asked. When a person let go from a merry-go-round, there is no force, and it does not change rotation speed. Commented Jun 18 at 12:51
• @fishinear Fair enough - I missed the, "but spin on your own axis." C'est la vie. Commented Jun 18 at 17:15

When you let go of the spinning playground wheel, you'll shoot off in a straight line tangent to the wheel's edge. At the same time, you'll start spinning in the same direction to how the wheel was spinning. It's all about conserving angular momentum and linear momentum. There are no tangential forces causing curved motion, and the conservation of angular momentum ensures any rotational motion you exhibit after release is independent of your linear trajectory.

• I think you mean that you keep spinning in the same direction as how you and the wheel were spinning, not the opposite direction. Commented Jun 17 at 13:57
• ah yes good spot Commented Jun 17 at 15:26