Opening a Door

Classical Mechanics Level 3

In the $xyz$ -coordinate system, a $20 \text{ kg}$ door is hinged so that it can rotate freely about the $z$ -axis (its length is along the $z$ -axis). The door is $1 \text{ m}$ wide, and its width is initially aligned with the $x$ -axis. The door is initially at rest.

A constant force $(F_x,F_y) = \big(-10 \text{ N}, +10 \sqrt{3} \text{ N}\big)$ is applied to the door at its far edge (farthest from the $z$ -axis).

What is the door's angular speed $($ in $\text{rad/s})$ when its width is first aligned with the $y$ -axis?

Details and Assumptions:

Neglect gravity and air resistance.
The door's mass is uniformly distributed over its width.
Give your answer to 3 decimal places.

The answer is 2.863.

1 solution

Sardor Yakupov
Jul 21, 2017

$\overrightarrow { M } =\overrightarrow { r } \times \overrightarrow { F } ;\\ \overrightarrow { M } ({ r }_{ y }{ F }_{ z }-{ F }_{ y }{ r }_{ z };{ r }_{ z }{ F }_{ x }-{ F }_{ z }{ r }_{ x };{ r }_{ x }{ F }_{ y }-{ F }_{ x }{ r }_{ y });\\ \overrightarrow { M } (0;0;10\sqrt { 3 } );\\ I\beta =\overrightarrow { M } ;\\ \beta =\frac { \overrightarrow { M } }{ I } ;$

Here we apply formula of accelerated motion, angle is pi/2. Also, $I=\frac { 1 }{ 3 } m{ l }^{ 2 }$

$\varphi =\frac { { \omega }^{ 2 } }{ 2\beta } \\ \omega =\sqrt { \pi \beta } =\sqrt { \frac { \pi \overrightarrow { \left| M \right| } }{ I } } =\sqrt { \frac { 10\sqrt { 3 } \pi }{ \frac { 1 }{ 3 } 20 } } \approx 2.86rad/s\\ \\$

Opening a Door

The answer is 2.863.

1 solution

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