Applying a constant force on pendulum and integrating for velocity

Question

We apply a constant horizontal force $ F $ to a weight that's connected to a solid rod. I'd like to find out the velocity of the weight when it's at its horizontal position. See image below.

My attempt at solving it:

$$ {dv\over dt} = {F(\theta) \over m} $$ $$ {dv\over ds}{ds\over dt} = {F(\theta) \over m} $$ $$ {dv}{ds\over dt} = {F(\theta) \over m} ds $$ $$ {dv}\cdot v = {r \over m} F(\theta) d\theta $$ $$ {dv}\cdot v = {rF \over m} \cos(\theta) d\theta $$

Solving the integral I get

$$ {v^2 \over 2} = {rF \over m} \sin(\pi/2) = {rF \over m} $$

I'm pretty sure I messed up somewhere because this give's the wrong answer.

Any help is appreciated. Thanks!

Edit: I just realized I didn't take into account $ g $ that's working against us!

Answer 1

Work done by $F$ is $rF$

Work done by gravity is $-mgr$

Kinetic energy of the mass is $1/2 m v^2$

$$rF-mgr=\frac12 m v^2$$

Answer 2

You forgot the $mgL$ term in your setup. It should be $$\frac{dv}{dt}=\frac{F\cos\theta}{m}-g\sin\theta$$

Then do the same thing you will get the same answer as the one mentioned by grdgfgr.