Vertical pendulum quick question

1. Jan 10, 2015

joshmccraney

hi pf!

i am looking at a simply problem dealing with an upside down vertical pendulum of length $L$ having mass $m$ at the top. i believe my professor wrote that a torque balance yields $m \ddot{\theta} = mg\sin \theta + f(t)$ where $f$ is a torque (i think) and $\theta$ is the angle the pendulum makes with the vertical axis.

my question is how the left hand side works? isn't newton's second law extrapolated for angular rotation as moment of inertia times angular acceleration equals sum of torques? if so, wouldn't we have $m L^2 \ddot{\theta} = mgL\sin \theta + f(t)$ as the torque balance?

2. Jan 10, 2015

BruceW

hey, yeah, I think you're right. In any case, without the torque it should be
$$\ddot{\theta}=\frac{g}{L} \sin{\theta}$$
So maybe your professor forgot to write the $L$ in there.

edit: p.s. I'm guessing the angle is being measured from the highest point, hence no negative sign on the right hand side

3. Jan 11, 2015