How to find the acceleration with polar coordinates?

[TonyTheTech]

Homework Statement

The quality of the image is bad so here's the statement:

For an interval of motion the drum of radius b turns clockwise at a constant rate ω in radians per second and causes the carriage P to move to the right as the unwound length of the connecting cable is shortened. Use polar coordinates r and θ and serive expressions for the velocity v and acceleration a of P in the horizontal guide in terms of the angle θ. Check your solution with time of the relation x²+h²=r²

Homework Equations

I first found that the velocity of the carriage is v=bω/sin(θ)

The Attempt at a Solution

I attempted to directly derivate the equation which give me -bωcos(θ)/sin²(θ)

However, in the book answers, the answer is supposer to be b²ω²/h *cot³(θ).

I think I have to do something with the vectors, like derivate v=v_re_r + v_θe_θ but I don't understand much.

 

[fgb]

It seems to me that you just forgot a term in the chain rule to differentiate your expressior for the speed. Realize that as r is changing, so is $\theta$, so $\theta$ is also a function of time.

You have $v = -\dfrac{b\omega}{\sin{\theta}}$

$a=\dfrac{dv}{dt} = \dfrac{b\omega\cos{\theta}}{\sin{\theta}^2}\dfrac{d\theta}{dt}$

There is probably a simpler way to do this, but to obtain $\frac{d\theta}{dt}$ you can realize that $r*\cos{\theta} = h$, where h is a constant. If you differentiate both sides, you get

$\dfrac{dr}{dt}*cos{\theta} + r(-\sin{\theta})\dfrac{d\theta}{dt} = 0$

You can solve for $\dfrac{d\theta}{dt}$ and plug it into the expression for a, it gives you the right answer :) Realize that $\frac{dr}{dt} = -b\omega$, since it is simply the rate at which the rope is being pulled.

Did I make any sense? :P