Question

A large, heavy door is swinging closed with an angular speed of 0.3 rad/s. When the door is at a 90° angle from being shut, you attempt to stop it by pushing so it experiences a constant angular acceleration of 0.06 rad/s2 . Are you able to stop the door before it closes? If so, how much time did it take to stop the door and where does the door end up? If not, how much time did it take for the door to close?

Answer 1

Time taken to stop the door is 5 s

Solution:

As per the question:

Angular speed,
`\omega = 0.3 rad/s`

`theta = 90^(\circ)`

Angular acceleration,
`\alpha = 0.06\ rad/s^(2)`

Now,

Using Kinematic eqn for rotational motion:

`\omega'^(2) = \omega^(2) + 2\alpha \theta`

`0 = 0.3^(2) + 2* - 0.06* \theta`

`\theta = 0.75\ rad`

Since, the value of
`\theta` <
`(\pi)/(2)`, thus we will able to stop the door before it closes.

Time taken to stop the door, t :

`\omega'  = \omega + \alpha t`

`0 = 0.3 - 0.06 t`

t = 5 s