Merry Go Round (Centripetal Force)

In summary, the conversation discusses two identical twins, Jena and Jackie, riding on a frictionless merry-go-round in northern Minnesota. Jena, sitting at a distance of 1.80 m from the center, must hold on to a metal post with a force of 58.0 N to stay on the merry-go-round. Jackie, sitting at a distance of 3.67 m from the center, must hold on with a force of 117.6 N to prevent from falling off. If Jackie falls off, her horizontal velocity will be 3.8585 m/s. The conversation also discusses an attempted solution to the problem, which was incorrect due to not taking into account the angular velocity of the twins.
  • #1
Chandasouk
165
0

Homework Statement



Two identical twins, Jena and Jackie, are playing one December on a large merry-go-round (a disk mounted parallel to the ground on a vertical axle through its center) in their school playground in northern Minnesota. Each twin has a mass of 29.0 kg. The icy coating on the merry-go-round surface makes it frictionless. The merry-go-round revolves at a constant rate as the twins ride on it. Jena, sitting a distance 1.80 m from the center of the merry-go-round, must hold on to one of the metal posts attached to the merry-go-round with a horizontal force of 58.0 N to keep from sliding off. Jackie is sitting at the edge, a distance 3.67 m from the center.

With what horizontal force must Jackie hold on to keep from falling off?

If Jackie falls off, what will be her horizontal velocity when she becomes airborne?


I attempted this problem but got the wrong answer. What I did first was

58N = 29kg*V^2
=========
1.80

V = 1.89 m/s

Fnet = m*v^2
---------
R

Fnet = (29kg)(1.89m/s)^2
--------------------
3.67

Fnet = 28.4N


What did I do wrong?
 
Physics news on Phys.org
  • #2
1.89 m/s is not the velocity of Jack. Their angular velocities is the same.
And angular velocity ω = v/R.
So v1/R1 = v2/R2. So find v2.
 
  • #3
So

1.89m/s
-------- = V2/3.67m
1.80 m

V2 = 3.8585 m/s ?

Then

Fnet = (29kg)(3.8585m/s)^2/3.67m = 117.6N ?
 
  • #4
Yes. That is right.
 

FAQ: Merry Go Round (Centripetal Force)

What is a Merry Go Round?

A Merry Go Round, also known as a carousel, is a spinning amusement ride consisting of a rotating circular platform with seats for riders. It is commonly found in amusement parks and playgrounds.

How does a Merry Go Round work?

A Merry Go Round works by using centripetal force to keep the riders moving in a circular motion around the central axis. The riders are held in place by the seats and the force of gravity, while the movement is generated by the rotation of the platform.

What is centripetal force?

Centripetal force is the force that acts on an object moving in a circular path towards the center of the circle. In the case of a Merry Go Round, it is the force that keeps the riders moving in a circular motion towards the center of the ride.

Is centripetal force dangerous?

No, centripetal force is not dangerous as long as it is within safe limits. The force required to keep the riders on a Merry Go Round is relatively low and is not strong enough to cause harm. However, if the ride is not properly maintained or if riders are not following safety guidelines, accidents can occur.

Can you adjust the centripetal force on a Merry Go Round?

Yes, the centripetal force on a Merry Go Round can be adjusted by changing the speed of the rotation or the distance between the riders and the center of the ride. This can be done by the operator to increase or decrease the level of thrill for the riders.

Back
Top