• Support PF! Buy your school textbooks, materials and every day products Here!

Ferris wheel Rotational Kinematics Question

  • #1

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!
 

Answers and Replies

  • #2
Doc Al
Mentor
44,905
1,169
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.
 
  • #3
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.
 
  • #4
Doc Al
Mentor
44,905
1,169
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.)
 
  • #5
Yes, I meant alpha. And I squared it because I was accidentally looking at the formula for centripetal acceleration when I was writing the tangential acceleration formula. Is that squared term the root of my problems?
 
  • #6
Doc Al
Mentor
44,905
1,169
Is that squared term the root of my problems?
Yes, it looks like the main problem in that last attempt.
 
  • #7
That was right! Thanks so much. If you wouldn't mind I still can't get part two. I know that a_c and a_t are perpendicular to each other, so I was thinking I should use trigonometry to find the direction of acceleration. I tried arctan(a_c/a_t), but that was no good.
 
  • #8
Doc Al
Mentor
44,905
1,169
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?)
 
  • #9
"_____degrees below the direction of motion" is what the field looks like. I thought my trig function would work in that situation.
 
  • #10
Doc Al
Mentor
44,905
1,169
"_____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?
 
  • #11
I did "arctan theta=.259536711/1.98" and I got 7.47 for theta.
 
  • #12
180-arctan(a_c/a_t) might give you your answers.
 
  • #13
Doc Al
Mentor
44,905
1,169
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.

180-arctan(a_c/a_t) might give you your answers.
:wink:
 
  • #14
That was right! Thank you both very much. I really appreciate it! :D
 

Related Threads on Ferris wheel Rotational Kinematics Question

  • Last Post
3
Replies
67
Views
11K
  • Last Post
Replies
6
Views
6K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
4
Views
18K
Replies
1
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
13
Views
7K
  • Last Post
Replies
8
Views
3K
Top