I can't do space curvilinear motion

[EddieHimself]

Homework Statement

[PLAIN]http://a4.sphotos.ak.fbcdn.net/hphotos-ak-ash4/s720x720/300201_1919471514513_1473790586_31632884_1692676_n.jpg

Homework Equations

a_R = \ddot{R} - R\dot{\phi}² - R\dot{\theta}²cos²(\phi)

a_\theta = \frac{cos(\phi)}{R}\frac{d}{dt}(R²\dot{\theta}) - 2R\dot{\theta}\dot{\phi}sin\phi

a_\phi = \frac{1}{r}\frac{d}{dt}(R²\dot{\phi}) + R\dot{\theta}²sin\phicos\phi

The Attempt at a Solution

 

[Spinnor]

You can calculate y,tt (the acceleration in the y direction) = v^2/r = 166*166/1200 = 23m/s^2 the acceleration in the y direction. With this you can find a_r and a_phi by some trig. You know at the instant of calculation that phi,t = 0, R,t = 0, and using some trig you can calculate theta,t.

Does this help?

 

[EddieHimself]

i've already done that bit. I worked out the normal acceleration for the loop, \dot{\theta} as 0.1437 rad/s, a_r = 8.601 m/s² and a_\phi as 21.49 m/s² but the problem is when i try to input all this into working out the values of \ddot{R} and \ddot{\phi} that i seem to come out with something completely different.

 

[Spinnor]

The units you have for a_phi and the units for a_phi in the answer are different. I think you are off by a factor of R?

 

[EddieHimself]

The units you have for a_phi and the units for a_phi in the answer are different. I think you are off by a factor of R?

It's asking me to work out \ddot{\phi} (rad/s²), which is a different quantity to a_\phi, (m/s²). If i just divide 21.49/1077 that equals 0.0199 which is not the right answer. My problem is that i don't know whether the value i have for a_\phi is wrong, or whether I've done something wrong in the equation or what?

 

[Spinnor]

Maybe I'm doing something wrong but I suspect the answer in the book. I got an acceleration in the y direction of about 23m/s^2 and R,tt should be a fraction of that number and not the 34.4m/s^2 in the book. I'm stumped, if you get an answer please let us know, if I'm doing something wring I would like to know.