An EF 151 GTA sits on the EF stool and rotates CCW with an initial speed of 60 rpm. After 28 revolutions he stops. Assuming constant acceleration, how long was he rotating?
To find the time it took for the EF 151 GTA to stop rotating, we can use the equation of motion that relates angular displacement, initial angular velocity, final angular velocity, and time. The equation is:
θ = ω_i * t + (1/2) * α * t^2
Where:
θ is the angular displacement (in radians),
ω_i is the initial angular velocity (in radians per second),
t is the time (in seconds), and
α is the angular acceleration (in radians per second squared).
In this case, we know that the EF 151 GTA rotated 28 revolutions, which is equivalent to an angular displacement of 28 * 2π radians.
Since the EF 151 GTA started with an initial speed of 60 rpm (revolutions per minute), we need to convert it to radians per second. There are 2π radians in one revolution and 60 minutes in one hour. So, the initial angular velocity (ω_i) is:
ω_i = (60 rpm) * (2π radians / 1 revolution) * (1 revolution / 60 seconds)
= 2π radians per second
We also know that the final angular velocity (ω_f) is zero because the EF 151 GTA stopped rotating.
Now, let's calculate the angular acceleration (α). Since the rotation is assumed to have a constant acceleration, we can use the kinematic equation for uniformly accelerated angular motion:
ω_f^2 = ω_i^2 + 2αθ
Since ω_f is zero and θ is 28 * 2π radians, we can solve for α:
0 = (2π radians per second)^2 + 2α * (28 * 2π radians)
α = - (2 * (2π radians per second)^2) / (2 * 28 * π radians)
α = - (2π radians per second squared) / 28
By substituting the values into the equation of motion, we can solve for the time (t):
(28 * 2π radians) = (2π radians per second) * t + (1/2) * (- (2π radians per second squared) / 28) * t^2
Simplifying this equation will give us the value of t, which represents the time it took for the EF 151 GTA to stop rotating.