OpenStudy (anonymous):

i am trying to solve for angular acceleration by deriving this position equation: x = r/sin(theta). I am having trouble taking the 2nd derivative so if anyone could lend a hand thanks!

6 years ago
OpenStudy (anonymous):

\[v=dx/dt =[(r*\cos \theta) d \theta/dt]/\sin^2\theta \]

6 years ago
OpenStudy (anonymous):

Is r a function of theta?

6 years ago
OpenStudy (kropot72):

Assuming your position equation is as follows: \[x=(r)\div \sin \theta\]where is t?

6 years ago
OpenStudy (anonymous):

r is constant, as well you will do chain rule to make it a function of time. similar how i solved for velocity, v.

6 years ago
OpenStudy (dumbcow):

you are missing a neg sign....d/dx 1/x = -1/x^2 also i believe this holds true \[a=\frac{dx^{2}}{d^{2}t} = \frac{dx}{d \theta}*\frac{d \theta^{2}}{d^{2}t}\] \[\frac{dx^{2}}{d^{2}t} = -\frac{r \cos \theta}{\sin^{2} \theta}*\frac{d \theta^{2}}{d^{2}t}\]

6 years ago