# I can't do space curvilinear motion

1. Aug 15, 2011

### EddieHimself

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

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

2. Relevant equations

aR = $\ddot{R}$ - 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}$2sin$\phi$cos$\phi$

3. The attempt at a solution

Last edited by a moderator: May 5, 2017
2. Aug 16, 2011

### 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?

3. Aug 16, 2011

### EddieHimself

i've already done that bit. I worked out the normal acceleration for the loop, $\dot{\theta}$ as 0.1437 rad/s, ar = 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.

4. Aug 16, 2011

### 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?

5. Aug 17, 2011

### EddieHimself

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?

Last edited: Aug 17, 2011
6. Aug 17, 2011

### 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.