How Does Torque Balance Apply to an Upside Down Vertical Pendulum?

[member 428835]

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?

 

[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

 

[member 428835]

thanks for your input! makes me feel better about it.