A popular carnival ride consists of seats attached to a central disk through cables. The passengers travel in uniform circular motion. As shown in the figure, the radius of the central disk is R0 = 3.00 m, and the length of the cable is L = 1.43 m. The mass of one of the passengers (including the chair he is sitting on) is 56.1 kg.

Question

a) If the angle θ that the cable makes with respect to the vertical is 30°, what is the speed, v, of this passenger?

b) What is the magnitude of the force exerted by the cable on the chair?

I got b), F=635.5 N, but I can't figure out a. I tried setting 635.5=mv^2/r, but that didn't work. I also tried mg=mv^2/r, v=sqrt(gr) but that also didn't work. Any help or direction would be greatly appreciated!

Here's the figure: i41.tinypic . com/2ho9v69.png

Answer 1

I cannot find the web site of your figure, but thanks for showing you work.

Let the cable force be T.

T cos30 = M g
T = 634.8 N

T sin30 = M V^2/R = T/2
R is the orbit radius, which is probably
R = Ro + Lsin30 = 3.72 m
(Without the figure, that is an educated guess)

Solve for V

Answer 2

Thank you so much, you guessed right without the figure. I really appreciate the help!