1
$\begingroup$

enter image description here

According to the solution to the related example problem in the book, points B and D have no linear velocity when the door is fully closed, that is, when point E strikes the floor, because they are at the lower limit of their respective motion ranges.

From this reasoning, when an object dropped from rest from a height h strikes the floor it too is at the lower limit of its motion range but its velocity is not zero but sqrt(2gh) which is not consistent with how this reasoning is applied to the panels' centers of mass. What is the reason for this discrepancy?

$\endgroup$
0
$\begingroup$

The garage door is an example of constrained motion. End A is constrained to move in a horizontal line. Points B, D are constrained to move in a vertical line and to stay in a rigid position relative to A, C and C, E respectively.

The object dropped from a height is not constrained while it is freely falling.

A fairer comparison would be to examine the falling object when it touches the rigid floor. Now it is subject to a constraint, which prevents it from moving any lower. Usually the object is deformable like a ball. Its centre of mass moves lower but slows down from a velocity of $\sqrt{2gh}$ downward to zero over a short distance, usually much less than the radius of the ball, then reverses direction. Like the garage door, at the lower limit of its motion its vertical velocity is zero.

For the ball its gravitational PE of $mgh$ has been transformed first into the same amount of KE, then into the same amount of elastic PE as it deforms. Assuming there are no losses of energy due to friction, air resistance, hysteresis, etc then it will bounce back up to its original height.

For the garage door its gravitational PE when in the fully open position has also been transformed into the same amount of KE when the door reaches the lowest limit of its movement. The COM of the door is no longer moving vertically, but the two panels AC and CE are rotating about B and D respectively - so the door now has rotational KE. If there are no friction losses the door will - like the ball - swing back up to its initial height, its rotational KE being transformed back into linear KE and finally to gravitational PE.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for your answer $\endgroup$ – user436788 Apr 5 at 21:35
  • $\begingroup$ From what I understand the main difference is whether the free fall or constrained motion is considered. For example, if the object dropped from height h were not deformable its final free fall velocity would be nonzero, but at the same instant it hits the ground its center of mass would be fixed so its constrained velocity would be zero. $\endgroup$ – user436788 Apr 5 at 21:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.