Normal Force at the bottom of a Ferris Wheel

In summary, the student of weight 678 N experiences a normal force of 565 N at the highest point on a rotating Ferris wheel. To find the normal force at the lowest point, the formula N = mv^2/r - mg is used. When the wheel's speed is doubled, the normal force at the highest point becomes N = mv^2/r - mg, while the normal force at the lowest point becomes N = mv^2/r + mg. By examining these expressions, it can be concluded that the normal force at the lowest point is greater than the normal force at the highest point.
  • #1
AnkhUNC
91
0

Homework Statement



A student of weight 678 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude of the normal force N on the student from the seat is 565 N. (a) What is the magnitude of N at the lowest point? If the wheel's speed is doubled, what is the magnitude FN at the (b) highest and (c) lowest point?

Homework Equations





The Attempt at a Solution



So M = 678N, NTop = 565N. Fc = mg - Ntop = 6079.4
So Nbottom = Nbottom - mg = 6079.4 which leads Nbottom to = 12723.8 but this is incorrect. Where am I going wrong?
 
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  • #2
At the top, normal force, weight, and acceleration all point down:
N + mg = mv^2/r; so N = mv^2/r - mg

At the bottom, normal force and acceleration point up, but weight points down:
N - mg = mv^2/r; so N = mv^2/r + mg
 
  • #3
I really don't need all that though do I? If I do how am I going to solved for v^2 or r? I only have one equation and two unknowns. At best I'd have Ntop+Nbottom = mv^2/r.
 
  • #4
AnkhUNC said:
I really don't need all that though do I?
Yep. It's the easy way!
If I do how am I going to solved for v^2 or r?
No need to solve for those.

Examining those expressions for N, how does Nbottom compare to Ntop? (Hint: What's Nbottom - Ntop?)
 

Related to Normal Force at the bottom of a Ferris Wheel

What is the normal force at the bottom of a Ferris Wheel?

The normal force at the bottom of a Ferris Wheel is the force exerted by the seat of the Ferris Wheel on the rider sitting at the bottom. It is equal to the rider's weight and acts in an upward direction.

How does the normal force at the bottom of a Ferris Wheel change as the wheel rotates?

The normal force at the bottom of a Ferris Wheel remains constant as the wheel rotates, as long as the speed and direction of the wheel's rotation are constant. This is because the force of gravity and the normal force are always balanced at the bottom of the wheel.

What factors affect the normal force at the bottom of a Ferris Wheel?

The normal force at the bottom of a Ferris Wheel is affected by the rider's weight, the speed and direction of the wheel's rotation, and the design of the wheel itself. A larger wheel or a faster rotation will result in a greater normal force.

Is the normal force at the bottom of a Ferris Wheel always equal to the weight of the rider?

Yes, the normal force at the bottom of a Ferris Wheel is always equal to the weight of the rider. This is because at the bottom of the wheel, the force of gravity and the normal force are equal and opposite, resulting in a balanced force.

Why is the normal force at the bottom of a Ferris Wheel important?

The normal force at the bottom of a Ferris Wheel is important because it ensures the safety and stability of the rider. Without the normal force, the rider would not stay in their seat and could potentially fall out of the wheel.

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