Mass of a Person on a Ferris Wheel - Physics Homework

In summary: Friction is 100NThe force you apply is exactly opposed by the reaction force [excluding air resistance] - so the magnitude of the reaction force is exactly that required to exactly oppose the force you are applying.In summary, the normal reaction force is a reaction to the force of gravity, and its magnitude is dependent on the force applied by the object on which it is acting. It is larger at the bottom of the ferris wheel because the centripetal force required to maintain circular motion is larger at the bottom, thus requiring a larger reaction force to oppose it.
  • #1
Fleet
8
0

Homework Statement



I have to find the mass m of a person riding a ferris wheel sitting on a bath scale.
Given data:
R=16 m
Scale reading on top of ferris wheel = 510 N
Scale reading at bottom of the ferris wheel= 666 N

Homework Equations



[itex]F=ma[/itex]

For uniform circular motion we have:
[itex]a_{rad}=\frac{v^2}{R}[/itex]

The Attempt at a Solution



I know that three forces are acting on the person on both top and bottom of the wheel:
gravity Fg, the normal force Fn and the centripetal acceleration.

The reason that I'm stuck is that I can't understand this:

What is the resulting force? the centripetal acceleration?
I have read that the normal force in this situation is bigger at the bottom that on the top? Why is that?

If I answer these questions I can use Newtons 2nd law.

Thanks.
 
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  • #2
Fleet said:

Homework Statement



I have to find the mass m of a person riding a ferris wheel sitting on a bath scale.
Given data:
R=16 m
Scale reading on top of ferris wheel = 510 N
Scale reading at bottom of the ferris wheel= 666 N

Homework Equations



[itex]F=ma[/itex]

For uniform circular motion we have:
[itex]a_{rad}=\frac{v^2}{R}[/itex]

The Attempt at a Solution



I know that three forces are acting on the person on both top and bottom of the wheel:
gravity Fg, the normal force Fn and the centripetal acceleration.

The reason that I'm stuck is that I can't understand this:

What is the resulting force? the centripetal acceleration?
I have read that the normal force in this situation is bigger at the bottom that on the top? Why is that?

If I answer these questions I can use Newtons 2nd law.

Thanks.

Acceleration is not a force!

You should be referring to, and calculating, centripetal Force mv2/R

The force you are calling the Normal Force, is correctly named the Normal Reaction Force.

Being a Reaction Force it is as big as it needs to be, that is why it is bigger at the bottom of the loop than at the top.

The net force [the centripetal force] at the bottom is directed up. You have to add a large upward Reaction force to the weight force to get an answer that is directed up.
The net force [the centripetal force] at the top is directed down. You have to add a smaller upward Reaction force to the weight force to get an answer that is directed down.

and note: there are only two forces acting on the passenger. The weight force, due to gravity, and the Normal Reaction Force - due to the passenger contacting the seat.

Those two combine to give the net force in many situations - In this case it is the centripetal force.
 
  • #4
Thanks for the clarifications. The normal reaction force is directed upwards at both top and bottom, right? So Fn is less than gravity on the top and bigger than gravity at the bottom?

What exactly makes the normal reaction force change magnitude?
 
  • #5
Fleet said:
Thanks for the clarifications. The normal reaction force is directed upwards at both top and bottom, right? So Fn is less than gravity on the top and bigger than gravity at the bottom?

What exactly makes the normal reaction force change magnitude?

The reaction force is exactly that - a reaction.
Place you hand on a table - the table pushes back with a certain force.
Now push a little harder on the table - the table pushes back a little harder.
Now push on the table as hard as you can - the table pushes back as hard as you could.

place a 500kg mass on the table - and either the table will push back with sufficient force to support it, or the table will collapse.

Friction is a reaction force too. the uR calculation [coefficient of Friction x Reaction force] tells you the maximum possible friction force [analogous to the strength of the table above].

Suppose uR worked out to be 100N

if you push a mass on that surface with a force of 5N, friction will be 5N, and nothing moves.
Push with 20N, friction is 20N
Push with 80N, friction is 80N
Push with 100N, friction is 100N
Push with 120N and the mass moves.
 

1. What is the formula for calculating the mass of a person on a Ferris wheel?

The formula for calculating the mass of a person on a Ferris wheel is: m = (mv²)/r, where m is the mass of the person, v is the velocity of the Ferris wheel, and r is the radius of the Ferris wheel.

2. How does the mass of a person affect their experience on a Ferris wheel?

The mass of a person does not have a significant effect on their experience on a Ferris wheel. The main factor that affects their experience is the velocity of the Ferris wheel, as it determines the amount of centripetal force acting on the person.

3. Can the mass of a person change while they are on a Ferris wheel?

Technically, the mass of a person can change while they are on a Ferris wheel due to changes in their weight (which is affected by mass and gravity). However, for the purposes of this formula and calculation, we assume that the mass of the person remains constant.

4. How does the radius of the Ferris wheel affect the mass of a person on it?

The radius of the Ferris wheel does not directly affect the mass of a person on it. However, it does indirectly affect the velocity of the Ferris wheel, which in turn affects the amount of centripetal force acting on the person.

5. Why is it important to calculate the mass of a person on a Ferris wheel in physics?

Calculating the mass of a person on a Ferris wheel is important in physics because it helps us understand the relationship between centripetal force, velocity, and radius. It also allows us to make accurate predictions and calculations for other similar systems and situations.

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