Suppose a revolving door at the bamk is spinning around once every 5 seconds. As you walk through, you push on the door for 2 seconds, which increases the rotational speed so that it rotates once a second. What was the rotational acceleration due to your push?
T = 5 so f = 1/5 and 2 pi f = (2/5)pi rad/s
new T = 1 so f = 1 and 2 pi f = 2 pi rad/s
change in angular velocity =2 pi(4/5)
change in angular velocity per second = angular acceleration = 2 pi (4/5)/2
= 4 pi/5 rad/s^2
To find the rotational acceleration due to your push, we need to determine the change in rotational speed and the time over which this change occurs.
First, let's determine the initial rotational speed (ω_initial) of the revolving door. The problem states that it is spinning around once every 5 seconds, which means the initial angular velocity (ω_initial) is given by:
ω_initial = 2π radians / 5 seconds
Now, your push on the door increases its rotational speed. Let's determine the final rotational speed (ω_final) of the revolving door. The problem states that after your push, it rotates once per second, so the final angular velocity (ω_final) is:
ω_final = 2π radians / 1 second
The change in rotational speed (Δω) is the difference between the final and initial angular velocities:
Δω = ω_final - ω_initial
Now, let's determine the time over which this change in rotational speed occurs. According to the problem, you push on the door for 2 seconds.
Now, we can calculate the rotational acceleration (α) using the formula:
α = Δω / Δt
Substituting the values we found:
α = (2π/1 - 2π/5) radians / 2 seconds
Simplifying:
α = 8π / 10 radians / 2 seconds
Reducing the fraction:
α = 4π / 10 radians / second^2
Therefore, the rotational acceleration due to your push on the revolving door is 4π / 10 radians per second^2.