Question

d L Why are door stops there? [6 pt). A well designed door stopper anchored to a wall is placed in such a way as to minimize damage to the door hinges. Instead of placing the door stop at the maximum door length (to apply the maximum torque), it is better to place it at a closer distance d <L from the hinges, so that the hinges do not need to provide a horizontal force to stop the door. For this problem, assume that the door acts like a linear stick with mass M= 33 kg and length L = 0.90 m, that it is pivoted about one end (at the hinges), and that it is rotating with angular speed w = 1.5 radian/second as it is being closed. The door is stopped only by the normal force from the door stop (the hinge provides no horizontal force) in a perfectly inelastic collision that takes At = 0.50 s. (a) [2 pt] Determine the tangential velocity and momentum of the center of mass of the door just before it strikes the stopper. In this case, the center of mass is at the center of the door. (b) [2 pt] Use the translational momentum-impulse equation to determine the magnitude of the normal force exerted by the door stop to stop the door's tangential motion. (©) [2 pt] Use the angular momentum-impulse equation to determine the optimal distance d for the door stop.

Answer 1

Solution:

Part(a): The tangential velocity of the centre of mass of the door just before to strike the stopper is given by

$v=r\omega \\\\$ (1)

Where,

r = distance of the centre of mass from the hing

w = angular velocity of the dore

Given that:

L = 0.90m w = 1.5rad/s

So,

$r=\frac{L}{2}=\frac{0.90}{2}=0.45m\\\\$

Putting these valuese in eqⁿ (1), we get

$v=0.45\times 1.5\\\\ \Rightarrow v=0.675m/s$ (2)

And the tangential momentum of the centre of mass just before to strike the stopper is given by

(3)

P= my (: m= 33kg and using equation (2) p= 33 x 0.675 p=22.275kgm/s

Part(b): The magnitude of normal force exerted by the door stopper on the door to stop its tangential motion is given by

$F=\left | \frac{p_{f}-p_{i}}{t_{f}-t_{i}} \right |\\\\ \Rightarrow F=\left | \frac{m\left (v_{f}-v_{i} \right )}{\Delta t} \right |\\\\ \left ( \because \Delta t=0.5s, \ v_{f}=0.0m/s, \ v_{i}=0.675m/s, \ and \ m=33kg \right )\\\\ \Rightarrow F=\left | \frac{33\left ( 0.0-0.675 \right )}{0.5} \right |\\\\ \Rightarrow F=44.55N$ (4)

Part(c): Let the door stopper is at a distance d from the door hinge, then angular impulse exerted by the door stopper on the door to stop tangential motion is given by

$I_{ang}=L_{f}-L_{i}\\\\ \Rightarrow I_{ang}=mvd-mvL\\\\$ (5)

For optimum value of distance d:

$I_{ang}=0\\\\ Using \ equation \ (5), we \ get\\\\ mvd_{opt}-mvL=0\\\\ \Rightarrow d_{opt}=L\\\\ \left ( \because L=0.90m \right )\\\\ \Rightarrow d_{opt}=0.90m\\\\$ (6)