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## Main Question or Discussion Point

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?

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?