Question

A person who walks through the revolving door exerts a horizontal force of magnitude 65-N on one of the four door panels and keeps the 11° angle constant relative to a line which is normal to the panel. If each panel is modeled by a 55-kg uniform rectangular plate which is 1.1 m in length as viewed from above, determine the magnitude of the final angular velocity ω of the door if the person exerts the force for 2.7 seconds. The door is initially at rest and friction may be neglected.

Answer 1

the final angular velocity = 1.942 rad/s²

Step-by-step explanation:

Let assume that the diameter is 1.0 m since we are not given the diagram.

So the force applied on one door is 65 N at 11° and at a distance of 1.0 m

Mass of each panel plate is 55 kg

Moment of inertia of the door about the center point O is:

`I_o = 4((1)/(3) ml^2)`

`I_o = 4((1)/(3) *55*1.1^2)`

`I_o = 88.73 \ kg.m^2`

Using the conservation of angular momentum ; we have:

`\int\limits^(t_2)_(t_1) \ \ M_o dt = (H_o)_2 -(H_o)_1`

Then the moment of force about center point O is:

`M_o` = (65 cos 11° )(1.0)

`M_o` = 63.81 N-m

as moment is constant , it does not vary with time ; as much:

`M_o(t_2-t_1) = I_o(\omega_2 - \omega_1)`

`63.81(2.7-0) = 88.73( \omega -0)`

172.287 = 88.73 ω

ω = 172.287/88.73

ω = 1.942 rad/s²

Hence, the final angular velocity = 1.942 rad/s²