# Amusement park ride physics problem

1. Oct 15, 2007

### grouchy

1. The problem statement, all variables and given/known data

a amusement park ride consists of a rotating circular platform 11.8 m in diameter from which 10 kg seats are suspended at the end of 1.18 m massless chains. When the system rotates, the chains make an angle of 38.1 degrees with the vertical. The acceleration of gravity is 9.8 m/s^2.

What is the speed of each seat? Answer in units of m/s.

http://geocities.com/grouchy187/figure.bmp

2. Relevant equations

F = ma ; F = mv^2/r ???

3. The attempt at a solution

i tried to make a free-body diagram to find F and then I plugged it into the above equation and got 7.57582 m/s but that was wrong...

2. Oct 16, 2007

### learningphysics

3. Oct 16, 2007

### grouchy

I erased my old work but I did the following this time. Only thing is I want to double check with someone else to see if it makes sense before I try submitting it (only get a certain amount of chances and each one is less credit)

I did this but again I ain't sure if its done right.... Horizontal force equals mg tan of angle. This is balanced with centripetal force so mv^2/r = mg tan (angle).

So v = sqrt of (g*r*tan (angle))
v = 6.733251157 m/s (don't need sig. figures)

So does this look right to anyone else? Thx.

4. Oct 16, 2007

### Staff: Mentor

OK. But horizontal force isn't balanced with centripetal force, it is the centripetal force.

What did you use for r?

5. Oct 16, 2007

### grouchy

I used 5.9 for r since the diameter was given as 11.8

6. Oct 16, 2007

### Staff: Mentor

That's the radius of the circular platform. You need the radius of the circular path swept out by the seats. How far is each seat from the axis?

7. Oct 16, 2007

### grouchy

OH! ok, so it would be 1.18 sin (theta) + 11.8/2 = 6.6281. And I got 7.1365 m/s as my speed which was correct. THX!

If you could help with the second part that would be great. it asks...

" If a child of mass 52.1kg sits in a seat, what is the tension in the chain (for the same angle)? Answer in units of N."

I tried ...

Fx = T - mg cos (theta)
T = mg cos (theta)
T = 478.9129157

but that was wrong... also tried

Fx = T - mg sin (theta)
T = mg sin (theta)
T = 375.5156929 but this was wrong too..

Last edited: Oct 16, 2007
8. Oct 16, 2007

### Staff: Mentor

Rethink this. The tension acts at an angle and the weight acts down.

You might want to consider the vertical forces.

9. Oct 16, 2007

### grouchy

humm...if I do it with the vertical forces

Fy = T cos (theta) - mg = 0
T cos (theta) = mg
T = mg /cos (theta)
T = 62.1(9.8)/ cos (38.1)
T = 773.3548298 N

kinda reluctant to try this since each attempt is less points but would that be it?

10. Oct 16, 2007

### Staff: Mentor

If you're not sure, figure it out using horizontal forces and compare.

11. Oct 16, 2007

### grouchy

Got it! Thx for all the help! Might be back later with another one lol