Torque And Moment Of Inertia

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: $\tau=r\times\vec{F}$, 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?

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Regardless on where you press on the door, the moment of inertia is the same. So what you are observing is the effect of higher torques the further away you push from the hinges. –  ja72 Nov 26 '12 at 14:51

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.

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Actually $\vec\tau = I \vec \alpha + \vec \omega \times I \vec \omega$, but for a planar case the last part drops out. –  ja72 Nov 26 '12 at 14:47
Thank you for this comment. I did not know this myself, as I have only recently started studying this. You have a nice answer so i upvoted it. –  user16228 Nov 26 '12 at 17:19

What you are experiencing is the effective mass of the swinging door at the point of contact. The calculate this value use the linear and angular equations of motion (starting at rest). Sum of forces equals mass $m$ times acceleration at c.g. $a_{cg}$, and sum of moments about c.g. equals inertia $I_{cg}$ times angular acceleration $\alpha$

$$F + R = m\, a_{cg} = m \left( \frac{L}{2} \alpha \right) \\ \left(x-\frac{L}{2}\right) F - \frac{L}{2} R = I_{cg} \alpha$$

where $F$ is the applied force (you) at a distance $x$ from hinges and $R$ is the hinge reaction force. The linear acceleration at the point of contact is $a = x\,\alpha$

The above is solved as

$$F = \left( \frac{I_{cg}}{x^2} + \frac{m L^2}{4 x^2} \right) a \\ R = \left( \frac{m L}{2 x} - \left( \frac{I_{cg}}{x^2} + \frac{m L^2}{4 x^2} \right) \right) a$$

So the effective mass is $F = m_{eff}\,a$ $$m_eff = \left( \frac{I_{cg}}{x^2} + \frac{m L^2}{4 x^2} \right)$$ and the rection $R = \left(\frac{m L}{2 x}-m_{eff}\right) a$

As a result the closer to the hinges you push (smaller $x$) the higher the effective mass. Also note that when $x=\frac{L}{2}+\frac{2 I_{cg}}{m L}$ the reaction force is zero and $x$ is the center of percussion of the door.

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What is effective mass, and what is c.g.? –  Mack Nov 30 '12 at 15:41
c.g. is center of gravity. effective mass is the mass you feel when you push the door. If you closed your eyes and pushed this is how much inertial mass you would feel reacting on your hand. –  ja72 Nov 30 '12 at 17:37