Calculating Apparent Weight on a Rotor Ride

In summary: So you have a value for v, and you know the radius of the rotor, so you can calculate the centripetal acceleration of the person. You can also calculate the normal force on the person. Since you know the coefficient of friction, you can determine the maximum force of static friction, which would keep the person in circular motion. Now you can compare the normal force with the maximum force of static friction. If the normal force is less than the maximum force of static friction, the person will not slip on the wall; if the normal force is greater than the maximum force of static friction, the person will slip on the wall.In summary, a 30-kg child on a rotor ride at a carnival with
  • #1
LesterTU
7
0

Homework Statement


A 30-kg child is on a rotor ride at a carnival. The radius of the rotor is 3 m and it completes 0.4 revolutions in one second. The coefficient of static friction is 0.6. What is the magnitude of the child's apparent weight?

rotor1.png

Homework Equations


ΣFx : FN = mv2/r
ΣFy : Ff = Fg

The Attempt at a Solution


First I need to find v:
0.4 revolutions per second is equivalent to one revolution per 2.5 s, so T = 2.5 s.
v = 2πr / T => v = 7.54 m/s.

Since the apparent weight is the magnitude of the resultant force exerted on a body by a supporting surface, the way I see it FN is the only force that fits this description so I figured
apparent weight = FN = mv2/r = 568.5 N.

However the solution given by the author says that
apparent weight = (FN2 + Fg2)1/2 = 640 N.

I've been trying to understand why mg plays a role in the child's apparent weight but I don't see how in this case
 
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  • #2
Hello and welcome to PF!

Are you sure that FN is the only force exerted on the person by the wall? Consider the forces shown in your figure.
 
  • #3
Thanks!

I guess the friction applies in this case but for some reason because of its direct relationship with the normal force I didn't consider it. It makes a lot of sense though now that I think about it
 
  • #4
OK, good.
 

1. What is apparent weight in relation to a rotor ride?

Apparent weight is the weight that an object appears to have when experiencing a non-uniform acceleration, such as on a rotor ride. It may differ from the object's actual weight due to the effects of centripetal force.

2. How does the rotation of a rotor ride affect apparent weight?

The rotation of a rotor ride causes the rider to experience a centripetal acceleration, which creates a force in the direction of the center of rotation. This force can make the rider feel heavier or lighter depending on their position and the speed of rotation.

3. Can the apparent weight on a rotor ride be negative?

Yes, the apparent weight on a rotor ride can be negative. This means that the rider will feel lighter than their actual weight due to the direction of the centripetal force being opposite to the direction of gravity.

4. How does the radius of rotation affect apparent weight on a rotor ride?

The radius of rotation has a direct impact on apparent weight on a rotor ride. The smaller the radius, the greater the centripetal force and therefore the greater the apparent weight. Conversely, a larger radius will result in a lower apparent weight.

5. How does the speed of rotation affect apparent weight on a rotor ride?

The speed of rotation also plays a role in determining the apparent weight on a rotor ride. As the speed increases, the centripetal force also increases, resulting in a higher apparent weight. The opposite is true for a decrease in speed.

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