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 the EF 151 GTA was rotating, we can use the equation for angular displacement:
θ = ω₁t + (1/2)αt²
Where:
θ is the angular displacement (in radians)
ω₁ is the initial angular velocity (in radians per second)
t is the time (in seconds)
α is the angular acceleration (in radians per second squared)
First, we need to convert the initial speed from rpm to radians per second. Since 1 revolution is equal to 2π radians, we can use the conversion factor:
Initial angular velocity (ω₁) = (Initial speed in rpm) * (2π radians / 1 minute) * (1 minute / 60 seconds)
ω₁ = 60 rpm * (2π radians / 1 minute) * (1 minute / 60 seconds) = 2π radians per second
The problem states that the EF 151 GTA stops after 28 revolutions. Since 1 revolution is equal to 2π radians, the angular displacement (θ) can be calculated as:
θ = (Number of revolutions) * (2π radians / 1 revolution) = 28 revolutions * (2π radians / 1 revolution) = 56π radians
Since the EF 151 GTA stops, its final angular velocity (ω₂) is 0.
Now we can rearrange the equation to solve for time (t):
θ = ω₁t + (1/2)αt²
56π radians = (2π radians per second) * t + (1/2)αt²
56π radians = 2πt + (1/2)αt²
Since α is constant and can be assumed to be the same for the entire rotation, we can factor it out:
56π radians = (2π + (1/2)αt) * t
Now we can solve for time using quadratic equation:
0 = (1/2)αt² + 2πt - 56π
Using the quadratic formula (ax² + bx + c = 0), where:
a = (1/2)α
b = 2π
c = -56π
t = (-b ± √(b² - 4ac)) / (2a)
Substituting the values:
t = (-(2π) ± √((2π)² - 4 * (1/2)α * (-56π))) / (2 * (1/2)α)
t = (-2π ± √(4π² + 2α * 56π)) / α
t = (-2π ± √(4π² + 112απ)) / α
Since we don't have the value for α (angular acceleration), we cannot calculate the exact time it took for the EF 151 GTA to stop rotating. Additional information is required to find the value of α and determine the time.