The shaft of a pump starts from rest and has an angular acceleration of 3.40 rad/s2 for 11.0 s.
(a) At the end of this interval, what is the shaft's angular speed?
rad/s
(b) What is the angle through which the shaft has turned
Disregard. I figured it out
To find the answers to your questions, we can use the kinematic equation for rotational motion:
ω = ω0 + αt
where:
ω = final angular speed
ω0 = initial angular speed (which is 0 in this case since the shaft starts from rest)
α = angular acceleration
t = time
(a) To find the angular speed at the end of the 11.0 second interval, we substitute the given values into the equation:
ω = ω0 + αt
ω = 0 + 3.40 rad/s^2 × 11.0 s
ω = 37.4 rad/s
Therefore, at the end of the interval, the shaft's angular speed is 37.4 rad/s.
(b) To find the angle through which the shaft has turned, we can use another kinematic equation:
θ = ω0t + (1/2)αt^2
where:
θ = angle
ω0 = initial angular speed
α = angular acceleration
t = time
Since the shaft starts from rest, ω0 is 0. We substitute the given values and solve for θ:
θ = ω0t + (1/2)αt^2
θ = (1/2)αt^2
θ = (1/2) × 3.40 rad/s^2 × (11.0 s)^2
θ = 200.35 rad
Therefore, the angle through which the shaft has turned is 200.35 radians.