Rodeo Lesson

I went to the 69th Annual New Windsor Rodeo last night. I had a good time because, as many of you know, I love animals, I love clowns, I love pyrotechnics, I love food (though I think the onion rings and funnel cake took three years off my life), I love hats*, and I love berries.

What do berries have to do with the rodeo? Not much, but they are a segue into my lesson this week. The New Windsor Rodeo had a carnival ride called the Berry-Go-Round. Cute name. To quote Butler Amusements, the Berry-go-Round is a ride where you ride inside one of four strawberry-shaped enclosures which "revolve independently while the entire ride slowly turns in a gentle circle." There are many different variations of this type of epicyclic ride, varying according to the shape of the object you ride in, and the speed of the ride (not always "gentle"). I remember as a kid riding in one with giant teacups to ride in. In that ride, you could control the speed and direction of rotation of your individual teacup, while the entire ride continued its circle independently. At the time, we were told by the carny that if you turned your teacup in the same direction as the main rotation, it would make the ride more exciting, but if you rotated the opposite way, it would help cancel out the rotation effect, and it would be less exciting.

I was young at the time,and didn't know physics, but when I thought about it again last night at the rodeo, I decided that the carny was wrong. If you think about adding the velocities due to the two separate rotational motions, then the direction of rotation will matter. But it isn't speed that you feel** making a ride exciting, it is acceleration. The acceleration due to each separate rotational motion is toward the center of rotation - no matter which way you rotate - so the direction of rotation would not matter.

To convince myself of this, I went back to first principles. The acceleration is the second derivative of the position. If we take the center of the ride as our origin, the rider's position vector can be taken as the sum of the vector pointing from the center of the ride to the center of the berry plus the vector pointing from the center of the berry to the rider's position within the berry. Each of these motions will be assumed to be uniform circular motion (although varying the speed of your berry could add interesting complexity to this problem. Left as an exercise for the reader :).
r = rcos(+wt+p)i + rsin(+wt+p)j + Rcos(-Wt+P)i + Rsin(-Wt+P)j

I have chosen the following variable names: r=radius of rotation of rider in the berry, w=angular speed of rotation of the berry (radians per unit time), and p= the phase of the berry rotation, which could be taken to be zero, if we start the time clock when the rider is exactly in the x-direction from the center of the berry. The capital letter variables are the same, except referring to the motion of the entire ride rotating in its gentle circle. This rotation is assumed to be clockwise (negative), as I think I remember it rotating that way last night, but the rotation of the berry can be positive or negative - rider's choice.

Note that at any time when +wt+p = Wt+P =0, the position is R+r in the x-direction. The rider is as far from the center of the ride as they will ever be. When +wt+p=pi, but Wt+P =0, the position is R-r in the x-direction, and the rider is as close to the center of the ride as they will ever be.

The first derivative of this position vector is the velocity.
v = -/+wrsin(+wt+p)i +/- wrcos(+wt+p)j + WRsin(-Wt+P)i - WRcos(-Wt+P)j

If the berry is rotating in the same sense (clockwise or negative) as the ride, then the speed is fastest (v=WR+wr) when the rider is farthest from the center of rotation, and slowest (v=WR-wr) when the rider is closest to the center of rotation.

Just the opposite is true if the berry rotates counterclockwise. In that case, the speed is slowest (v=WR-wr) when the rider is farthest from the center of rotation, and fastest (v=WR+wr) when the rider is closest to the center of rotation. Note that the maximum speed is the same in either case, it just occurs at different points in the motion.

But what about acceleration, which is what you feel in a ride? Taking another derivative yields
a = -w²rcos(+wt+p)i -w²rsin(+wt+p)j -W²Rcos(-Wt+P)i -W²Rsin(-Wt+P)j

When the rider is farthest from the center of the ride, the magnitude of their acceleration is w²r+W²R. It is the same no matter which direction the berry rotates. Likewise, the acceleration has a smaller magnitude (W²R-w²r) when the rider is nearest the center of the ride.

So, what is the conclusion? Is the ride no more or less exciting depending on the direction that your berry rotates? The acceleration is the same and that is what gives you the feeling of motion.

However, we have other senses that contribute to the excitement of the ride. You will only be moving fastest at the same time that you are accelerating the most if the berry and the ride are rotating in the same sense. You can see your friends whizzing by you while you feel the effects of large acceleration at the same time. This could add to the excitement of the ride.

I'd be interested in comment from anyone who has ridden such a ride more recently than I have.

_{*My fave was a purple mesh cowboy hat with a silver band, white silk flowers all around the band, and two dozen ribbons in various shades of pink and purple hanging down the back!
**That's special relativity}