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Centripetal Acceleration

HOME » CLASSROOM » CIRCUS PHYSICS » CENTRIPETAL ACCELERATION


Trick riders may just ride in circles, but that doesn’t mean they aren’t accelerating. Moving in a circle requires steady changes in direction. This is a form of acceleration directed inward, with a magnitude that depends on the velocity of the horse and rider and the radius of the circle. This is the same acceleration that we feel when going around bends in a car, and that holds us in our seat on the rollercoaster. In this unit, you’ll learn the basics of circular motion.

Watch the Video

Circus Physics: Centripetal Acceleration

Watch as trick riders jump on and off a horse at galloping speed! Along the way, you’ll learn about circular motion and why circus rings are as large as they are.


Questions to Consider While Watching the Video

  1. What force does the horse exert?
  2. What forces are acting on the horse?
  3. How does the size the ring affect the speed at which the horse can gallop?
  4. What kinds of acceleration do you see?

Digging Deeper

Here is a top-down view of the trick-rider's horse as it runs around a ring. R is the ring's radius, V is the horse's velocity around the ring. The blue arrow is the component of the horse's velocity in the East-West direction, the x-direction. The red arrow is the component in the North-South, or y-direction.

Centripedal acceleration diagram

As the horse runs around the ring, its velocity is always directed around the circle. As the horse turns, however, its x and y component velocities change. Changing velocity is also known as acceleration.

A = change in v / change in t

But now we have a conundrum, if the horse is accelerating, why doesn't it seem to be speeding up?

Notice that the horse's velocity arrow at the bottom of the ring is completely directed East (blue), with no North-South component (red). As the horse runs counter-clockwise, its East velocity decreases, while the North velocity increases. When the horse reaches the due East position, its velocity is entirely North, with no East-West component. The red arrow then decreases as the blue arrow increases again, this time headed West.

No matter where the horse is, one of its velocity components is decreasing, always being pulled back to zero in its respective direction. For this reason, we say the acceleration is directed toward the center of the circle. This is commonly known as centripetal acceleration.

We can find the magnitude of this acceleration with a simple formula:

A = V²/R

Recall from the Newton's Laws of Motion unit that a net acceleration implies a net force.

This acceleration is caused by whatever is keeping the object moving in a circle. In the horse's case, it's caused by hooves pushing against the ground. In the case of a rollercoaster, it's caused by the tracks.

As you go through a horizontal rollercoaster loop, the tracks push against the car, and by extension your bottom, to keep you going in a circle. Your body pushes back, according to Newton's 1st and 3rd Laws. The force of your body pushing back is what keeps you in your seat.

The acceleration required to pull this off must be equal to or greater than gravity. We call this a "g". A 1 g turn is then a turn in which you feel an amount of centripetal acceleration equal to the acceleration due to gravity.

Your Turn

Use the concepts and formulas from this unit to figure out the following:

How fast must a roller coaster go through a 25 meter-radius horizontal loop in order to maintain constant centripetal acceleration of 1 g?

Answer: ≈56 km/h

1 g is the acceleration due to gravity, 9.8 m/s/s. Use the centripetal acceleration equation a = v²/r, with a = 9.8 m/s/s and r = 25 m.

Solving for v gives:

v = √(25m x 9.8 m/s/s) ≈ 15.6 m/s

Converting this to km/h:

15.6 m/s x 60 s/min x 60 min/hr x 1 km/1000m ≈ 56 km/h)

Further Explanation

If you'd like to learn more about centripetal acceleration and trick riding, check out these links: