4
$\begingroup$

As I get closer and closer to the hinges of a door, it becomes harder and harder to open. However, the distance to the hinges is getting smaller and smaller, where the rotation is occurring. All the force I am applying to the door is parallel. In my experiment, I can push in one place, and then another place closer to the hinges. So the forces are parallel. So, by the parallel axis theorem: $I'=I+md^2$ and $d^2$ is getting smaller as I get closer to the hinges. I' is going down, so moment of inertia goes down and the door should be easier to open as I get closer to the hinges. However, the opposite is happening. Why does it appear that opening the door is breaking the parallel axis theorem?

$\endgroup$
2
  • 4
    $\begingroup$ Changing where you push does nothing to change the axis around which the door is spinning. It does, however, change the torque. $\endgroup$ Commented Feb 21, 2023 at 5:16
  • 5
    $\begingroup$ "$d^2$ is getting smaller as I get closer to the hinges" - is it? Look up what $d$ represents. It's not the distance between the hinges and the place where you push. $\endgroup$ Commented Feb 21, 2023 at 17:28

4 Answers 4

23
$\begingroup$

How does the parallel axis theorem explain the opening of a door? - It doesn't.

The moment of inertia of a door does not depend on where you push it.
The torque that you apply to a door depends on the magnitude of the force $F$ and the distance $d$ between the force and the hinges.
Suppose that you need to apply a certain torque $\tau_{\rm applied}$ to open the door in a reasonable time whilst at the same time overcoming frictional torques at the hinges.
$\tau_{\rm applied} = F\,\times d$, so as the distance between force and hinge $d$ is reduced, the force that you apply needs to increase and if the distance is reduced too much you will not be able exert enough force on the door to open it.

$\endgroup$
4
  • 10
    $\begingroup$ The door's moment of inertia does depend on where the hinges are. If the door swings about an axis other than the edge, the moment of inertia will be reduced and the door will be easier to open (sometimes outside doors in office buildings are like this). However, the force required will still depend on what part of the door you push on. $\endgroup$
    – Rich006
    Commented Feb 21, 2023 at 14:17
  • 7
    $\begingroup$ The context of the question is a fixed door, whose hinges are in a fixed position. @Rich006 $\endgroup$
    – Lee Mosher
    Commented Feb 21, 2023 at 16:10
  • 5
    $\begingroup$ @LeeMosher, yes but when stating that the theorem doesn't apply here, it's useful to give an example where it does apply. $\endgroup$
    – Rich006
    Commented Feb 21, 2023 at 19:33
  • 1
    $\begingroup$ @LeeMosher - you're right that this is the contextual setup of the question, but the actual question is "Why does it appear that opening the door is breaking the parallel axis theorem?", and Rich006 is just addressing the root cause of this misconception, as this answer glosses over it. $\endgroup$ Commented Feb 22, 2023 at 17:11
5
$\begingroup$

The $d$ in the parallel axis theorem is the distance between the center of mass of the object and the axis around which this object is to rotate. The torque, on the other hand, is defined by the moment arm, which is a (projected) distance between the axis and the point of application of the force.

The position of the center of mass doesn't depend on where the force is applied, so parallel axis theorem is not applicable here.

$\endgroup$
4
$\begingroup$

It's because you are ignoring the inertia of the door, which resists its swinging open. also, as you apply force closer and closer to the hinge point, the torque you are applying to rotate the door becomes less and less and the amount of force needed to develop enough torque goes up and up.

$\endgroup$
1
$\begingroup$

"As I get closer and closer to the hinges of a door, it becomes harder and harder to open."

True, but it does not become harder to open due to inertia, but rather the hinge reaction. Well, hinge reaction & distance to the center of mass.

Mass moment of inertia is a property of the body, and it has a different value depending on where on the body it is summed. The lowest value (and most commonly referred to) is the MMOI at the center of mass.

The parallel axis theorem gives us the value of MMOI at any other point displaced from the COM. In 2D this is simply $$I = I_{\rm COM} + m ( x^2 + y^2)$$

What you are asking about, is not the mass moment of inertia though. You are describing the effective mass of the door at the point of contact.

Imagine two scenarios, first a free floating door in space at rest, pushed by a unit force at some arbitrary location A, and the second scenario where the door is hinged in one end.

fig1

If you calculate the translational acceleration of point A due to a unit force $F$, you would have calculated the effective mass of the system at A. Sometimes this is called the reduced mass. It is estimated from $F = m_{\rm reduced} a_A$ where $a_A$ is the acceleration of A and $F$ is the applied force there.

  1. Free Floating Door has no reaction forces so the equations of motion (simplified) are

    $$\left.\begin{aligned}F & =m\left(a_{A}+d\,\alpha\right)\\ d\,F & =I\alpha \end{aligned} \right\} \;\begin{aligned}a_{A} & =F\left(\tfrac{1}{m}+\tfrac{d^{2}}{I}\right)\\ \alpha & =\tfrac{d\,F}{I} \end{aligned}$$

    From the above the effective mass of free floating door, pushed at a distance $d$ from the center of mass is

    $$\boxed{ m_{\rm free} = \frac{m\,I}{I+m d^2} }$$

    The above means that the maximum resistance we feel on a free-floating door is when we push at the center of mass and $d=0$. At that value $m_{\rm free} = m$. Any other value for $d$ increases the denominator and makes the resistance we feel less.

  2. Pivoted Door Similar to the scenario above, with with an additional reaction force $R$, and a kinematic relationship between $a_A$ and $\alpha$ the rotational acceleration

    $$ \left.\begin{aligned}a_{A} & =\left(c+d\right)\alpha\\ F+R & =m\left(a_{A}-d\,\alpha\right)\\ d\,F-c\,R & =I\alpha \end{aligned} \right\} \;\begin{aligned}a_{A} & =F\left(\tfrac{\left(c+d\right)^{2}}{I+mc^{2}}\right)\\ \alpha & =\tfrac{\left(c+d\right)\,F}{I+mc^{2}} \end{aligned}$$

    From the above the effective mass of pivoted door, pushed at a distance $c+d$ from the pivot is

    $$\boxed{ m_{\rm pivot} = \frac{I + m c^2}{\left(c+d\right)^{2}} } $$

    The above means that when we push at the pivot and $d=-c$ the effective mass we feel is infinite (as you mentioned in the question). From that point on any location $d>-c$ would reduce this value. The specific distance where $m_{\rm pivot} = m$ is to be found slightly to the right of the center of mass $d = \sqrt{c^2 + \tfrac{I}{m}} - c > 0$. Specifically for a thin slender object like a door you have $c=L/2$ and $I=\tfrac{m}{12} L^2$ which means the point where you feel the mass of the door is $d = \left( \tfrac{1}{\sqrt{3}} - \tfrac{1}{2} \right) L \approx 0.0777\,L$ or about 8% of the total length past the center of mass (middle).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.