Question

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!

Answer 1

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

Answer 2

Is r a function of theta?

Answer 3

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

Answer 4

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

Answer 5

you are missing a neg sign....d/dx 1/x = -1/x^2

also i believe this holds true
\[a=\frac{dx^{2}}{d^{2}t} = \frac{dx}{d \theta}*\frac{d \theta^{2}}{d^{2}t}\]
\[\frac{dx^{2}}{d^{2}t} = -\frac{r \cos \theta}{\sin^{2} \theta}*\frac{d \theta^{2}}{d^{2}t}\]