I am reading the two concepts mentioned in the title. According to the definition of torque and moment of inertia, it would appear that if I pushed on a door, with the axis of rotation centered about its hinges, at the door-knob, it would be difficult, relative to me applying a force nearer to the hinges. However, I just experimented with my front door, and it would appear to be the converse, what am I misunderstanding?
Definition of torque: $\vec \tau=\vec r\times\vec{F}$, where r is the length of the lever arm (the distance from the axis of rotation to the point of application of the force).
EDIT:
Also, what does it mean for torque to be perpendicular to the rotation?
Best Answer
You are right in observing the converse to what you were expecting.
By definition $\tau=r \times F$. But also $\tau=I\alpha$, with $\alpha$ the angular acceleration (an analogy with Newton's second law $F=ma$).
Now suppose we want to achieve a given angular acceleration $\alpha$. The two equalities above can be combined to give $r\times F=I\alpha$. By assumption both $I$ and $\alpha$ are constants. For the case of pushing near the hinge, we see that, to get the same product on the right side, we must exert a large force $F$, because $r$ is small and we want to achieve a certain angular acceleration. For the case at the doorknob, the force $F$ needed is much smaller as $r$ is bigger and $F$ needn't be so big to give the same angular acceleration.
You encounter this phenomenon if you go on a seesaw as well. Sit near the edge and you will make the seesaw accelerate faster.
Torque is perpendicular to the rotation by definition. I don't think there is much physical meaning to the fact that torque is perpendicular the rotation, but it fits in with the angular velocity also being defined perpendicular to the rotation.