How Does a Ferris Wheel Ride Reveal Physics Principles?

[clarineterr]

Homework Statement

A woman rides on a Ferris wheel of radius 16m that maintains the same speed throughout its motion. To better understand physics, she takes along a digital bathroom scale (with memory) and sits on it. When she gets off the ride, she uploads the scale readings to a computer and creates a graph of scale reading versus time. Note that the graph has a minimum value of 510 N and a maximum value of 666N. What is the woman's mass?

Homework Equations

a[tex]_{c}[/tex] = [tex]\frac{mv^{2}}{r}[/tex]

and Newtons 2nd law

The Attempt at a Solution

So at the top we have gravity acting on her and the normal force of her seat acting on her which we will call Nt.

so [tex]\frac{mv^{2}}{r}[/tex] = mg - Nt

Then at the bottom, where Nb is the normal force of her seat acting on her

[tex]\frac{mv^{2}}{r}[/tex] = Nb - mg

Then solving these equations I got 2mg = Nt + Nb

I really don't know if I am right about this...can someone please help me?

 

[tiny-tim]

Hi clarineterr!

So at the top we have gravity acting on her and the normal force of her seat acting on her which we will call Nt.

so [tex]\frac{mv^{2}}{r}[/tex] = mg - Nt

Then at the bottom, where Nb is the normal force of her seat acting on her

[tex]\frac{mv^{2}}{r}[/tex] = Nb - mg

Then solving these equations I got 2mg = Nt + Nb

I really don't know if I am right about this...can someone please help me?

Looks good to me!

 

[nlightner]

Your attempt at a solution is on the right track! However, there are a few things that need to be clarified. First, the equations you used are correct, but they are not the full picture. In addition to the centripetal acceleration equation and Newton's second law, we also need to consider the fact that the scale is recording a range of values, not just a single value. This means that the normal force on the woman is changing throughout the ride, and we need to take that into account.

Also, it's important to note that the normal force is not just the force of the seat pushing up on the woman. It also includes the force of gravity acting on the woman, since the scale is measuring the total force being exerted on her (including her own weight).

With that in mind, let's break down the problem into two parts: the top of the ride and the bottom of the ride.

At the top of the ride, the woman's weight is acting downward, and the normal force from the seat is acting upward. This means that the total force measured by the scale will be the sum of these two forces: Nt + mg. We can set this equal to the centripetal force (mv^2/r) and solve for the mass:

Nt + mg = mv^2/r

m = (Nt + mg)r/v^2

Similarly, at the bottom of the ride, the total force measured by the scale will be the sum of the normal force from the seat and the woman's weight: Nb + mg. Setting this equal to the centripetal force, we get:

Nb + mg = mv^2/r

m = (Nb + mg)r/v^2

Now, here's the key: we know that the minimum value recorded by the scale is 510 N, and the maximum value is 666 N. This means that at the top of the ride, the force measured by the scale is 510 N, and at the bottom of the ride, it is 666 N. So we can plug these values into our equations and solve for the mass:

At the top: 510 N = (Nt + mg)r/v^2

At the bottom: 666 N = (Nb + mg)r/v^2

Subtracting the two equations, we get:

666 N - 510 N = (Nb + mg)r/v^2 - (Nt + mg)r