# Test Question

1. Oct 19, 2007

### bob1182006

Not original #'s but same questions:
1. The problem statement, all variables and given/known data
A 750N child rides a ferris wheel that is moving at a constant velocity. At the highest point the child has an apparent weight of 650N.
(a) what is the acceleration of the wheel?
(b) what is the apparent weight of the child at the lowest point?

2. Relevant equations
F=ma
a=v^2/R

3. The attempt at a solution
Since the wheel is accelerating it's not an inertial frame so there's a pseudoforce (centrifugal?) acting on the child outward while the acceleration of the wheel is pointing inward to the center.

W=750N, W=mg, mg=750N, m=750N/g~76.4kg
(a).
at the top of the ferris wheel the child experiences the downward force of the acceleration and gravity. the centrifugal force (C) is upward:
C-mg=m(-a), C-mg=-ma
at the top -ma = apparent weight:
C-mg=650N, C=650N+mg=650N+750N=1400N

C/m-g=-a
g-C/m=a
9.81-(1400)/76.4kg=9.81-18.3=-8.49 m/s^2

(b).
at the bottom acceleration is up, gravity downward, as well as C.
-C-mg=ma
-1400N-750N=-2150N apparent weight, -gives direction down so the child thinks he weighs 2150N.

I'm not even sure if this problem has a numerical solution (no R, v given)

2. Oct 19, 2007

### Staff: Mentor

Got to be careful here. Centrifugal force only appears if you insist on analyzing this from the noninertial frame of the rotating wheel, which is not necessary. Is that how your instructor wants you to solve such problems? (I would not recommend it.)

In any case, if you do choose to use pseudoforces and noninertial frames, do it right: The acceleration is zero in such a frame.

Viewed from the noninertial frame there are three forces acting on the child:
> mg acting down
> normal force of the seat acting up (that's the apparent weight)
> the centrifugal pseudoforce acting up

These add up to zero.

Of course you always just use an inertial frame of reference, in which the child is centripetally accelerated. The only forces acting then are gravity and the normal force.

3. Oct 19, 2007

### bob1182006

I've never done a problem like this, I'm used to when the radius is given and such.

So to analyze it in an inertial frame the wheel is accelerating so for the top I would have:

N-mg=-ma
N=-ma+mg
N=-650+750=100 (is that right? guess i screwed up my signs plugging in 650 for (-ma) ><).

and then continue from there? That's what I did though, so I thought I was analyzing the system from a noninertial frame but I just got Centrifugal force fromm there and used it in a inertial frame....which may have lead to a somewhat correct answer but improper reasoning :/

4. Oct 19, 2007

### Staff: Mentor

This is correct, but solve for the acceleration not the normal force:
ma = mg - N
No, not right. The apparent weight, which equals the normal force, is given. For some reason you set "ma = 650". (If you knew that, you'd be done! a is what we're trying to solve for.)

5. Oct 19, 2007

### bob1182006

o right was thinking about part b...
-ma=-mg+N
a=-N/m+g, a=-650/74.6+9.81=1.1 m/s^2

then for part b:

N-mg=ma
N=ma+mg=m(a+g)=74.6(1.1+9.81)=810N

yeash I can't believe i messed up that badly on 1 question...At least I'm not alone with that, some people got a formula as the answer :/.