Force/velocity (second opinion)?

Question

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.

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

[I did this but I aint 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.]

Answer 1

The mass of the seats is immaterial. The effective radius of rotation is 11.8/2 + 1.18 sin(38.1) = 6.6281 meters. Let the rotation be omega. Then the centrifugal force is 6.6281 omega^2 per kilo. The vertical force is 9.8 per kilo. These must be in the ratio of tan(38.1) = 0.7841. Putting these together, we have 6.6281 omega^2 = 0.7841 x 9.8, giving 1.1593 for omega^2 or 1.0767 for omega. The speed will be this times the effective radius, or 7.1365 meters per second.

Answer 2

Centripetal force is
Fc=ma
where
a=V^2/R where
R= D/2 + L sin(38.1)

also
Fc= mg tan(38.1)

finally
V= sqrt( aR)
V=sqrt( Fc/m R)
V= sqrt( mg tan(38.1) / m R)
V=sqrt (g R tan(38.1))
where R= D/2 + L sin(38.1)
So your stuff looks good to me.

Answer 3

Component of tension T cos 38.1 balances (vertical force) weight mg and component Tsin 38.1 provides (horizontal force) centripetal force mv^2/r

T cos 38.1 = mg ---------------------(1)

Tsin 38.1 = mv^2/r------------------(2)

dividing (2) by (1)

v^2/r g = tan38.1

v = sq st (rg tan 38.1)

v = 6.7332511 m/s