A rotating water pump works by taking water in at one side of a rotating wheel, and expelling it from the other side. If a pump with a radius of 0.120 m starts from rest and accelerates at 30.5 rad/s2, how fast will the water be traveling when it leaves the pump after it has been accelerating for 9.00 seconds?
To find the speed at which the water will be traveling when it leaves the pump, we need to use the principles of rotational motion. We can use the equation that relates angular acceleration, time, initial angular velocity, and final angular velocity:
ωf = ωi + αt,
where ωf is the final angular velocity, ωi is the initial angular velocity, α is the angular acceleration, and t is the time.
Given:
Radius of the pump (r) = 0.120 m.
Angular acceleration (α) = 30.5 rad/s^2.
Time (t) = 9.00 seconds.
First, let's calculate the initial angular velocity (ωi) of the pump. Since the pump starts from rest, ωi = 0.
Now, using the above equation, we can find the final angular velocity (ωf):
ωf = ωi + αt
ωf = 0 + (30.5 rad/s^2)(9.00 s)
ωf = 274.5 rad/s
The final angular velocity (ωf) represents the speed at which the pump is rotating when the water leaves it. Since the water is expelled from the other side of the rotating wheel, the speed of the water will be proportional to the linear speed of the pump's circumference.
The linear speed of the pump's circumference is given by the formula:
v = rω,
where v is the linear speed, r is the radius of the pump, and ω is the angular velocity.
Substituting the values:
v = (0.120 m)(274.5 rad/s)
v ≈ 32.94 m/s
Therefore, the water will be traveling at approximately 32.94 m/s when it leaves the pump after accelerating for 9.00 seconds.