Ferris wheel Rotational Kinematics Question (2024)

Homework Statement

The given information is:
A Ferris wheel with a radius of 9.0 m rotates at a constant rate, completing one revolution every 37 s.
Suppose the Ferris wheel begins to decelerate at the rate of 0.22 rad/s^2 when the passenger is at the top of the wheel.

(1) Find the magnitude of the passenger's acceleration at that time.
(2) Find the direction of the passenger's acceleration at that time.

Homework Equations

a_c=r*w^2
a_t=r*alpha

The Attempt at a Solution

I tried to find the acceleration at the top using those formulas and couldn't get it. I thought I should try total acceleration using a= sqrt(a_c^2+a_t^2) and got 0.260 m/s, but that was no good either. Honestly what is really confusing me is what exactly I'm finding; I've calculated several types of acceleration during this homework so I'm not sure what just the "passenger's acceleration" is. Because of part two I'm guessing I need to incorporate vectors somehow, but I can't figure out what approach to take.

Thanks for any help!

fleabass123 said:

I thought I should try total acceleration using a= sqrt(a_c^2+a_t^2)

That's correct.

and got 0.260 m/s, but that was no good either.

Show your calculations for a_c and a_t.

I did:

a= sqrt(a_c^2+a_t^2)

a_c^2=r*w^2=9*.169815819^2=.259536711

(I calculated w by doing 2pi/37seconds)

a_t^2=r*theta=9*(-.22)^2=.4356

(I used the given deceleration as theta, but I'm not sure if that works)

so a= sqrt((.259536711^2)+(.4356^2))= .507

As you can see I got a different answer on this new attempt. I suppose I'll try that, but I'm still not confident my decision to use -.22 as my acceleration works.

fleabass123 said:

a_t^2=r*theta=9*(-.22)^2=.4356

I assume you meant this to be a_t, not a_t^2. Why did you square the angular acceleration? (Angular acceleration is usually represented by alpha, not theta.)

fleabass123 said:

Is that squared term the root of my problems?

Yes, it looks like the main problem in that last attempt.

fleabass123 said:

I tried arctan(a_c/a_t), but that was no good.

That would work just fine, but you need to know how they want you to describe the direction. With respect to the vertical or the horizontal? (The angle that you calculated is with respect to what?)

fleabass123 said:

"_____degrees below the direction of motion" is what the field looks like. I thought my trig function would work in that situation.

Yes, it should work. Show the details. What did you put for a_c, a_t, and theta?

fleabass123 said:

I did "arctan theta=.259536711/1.98" and I got 7.47 for theta.

Looks good, but what's the direction of motion? Remember that the acceleration is negative.

perrytones said:

180-arctan(a_c/a_t) might give you your answers.