Mass of a Person on a Ferris Wheel - Physics Homework

  • Thread starter Thread starter Fleet
  • Start date Start date
  • Tags Tags
    Physics Wheel
AI Thread Summary
To find the mass of a person on a Ferris wheel using scale readings, the normal force varies due to the effects of centripetal acceleration. At the top of the wheel, the normal force is less than the gravitational force, while at the bottom, it exceeds gravitational force. The net force acting on the person provides the necessary centripetal force for circular motion. Understanding that the normal reaction force adjusts based on the person's position on the wheel is crucial for solving the problem. This relationship between forces is key to applying Newton's second law effectively.
Fleet
Messages
7
Reaction score
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



F=ma

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

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.
 
Physics news on Phys.org
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



F=ma

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

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.
 
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?
 
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top