Equations for rotational motion with constant acceleration

[DH214]

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 35.0 rad/s2, how fast will the water be traveling when it leaves the pump after it has been accelerating for 9.00 seconds?

 

[djeitnstine]

$$\alpha$$ is measured in radians per second sqared $$(rad s^{-2} )$$ my guess is that you are either trying to find the angular velocity or the linear velocity.

angular acceleration = $$\alpha = \frac{\Delta \omega}{t}$$
angular velocity = $$\omega = \frac{\Delta \theta}{t}$$

the relationship between linear velocity/acceleration and angular velocity/acceleration is

linear velocity v

radius r

$$v= r \omega$$ m/s

$$a= r \alpha$$ m/$$s^{2}$$

 

[thomate1]

The equations for rotational motion with constant acceleration can be used to solve this problem. The first equation is ω = ω0 + αt, where ω is the final angular velocity, ω0 is the initial angular velocity (in this case, 0 since the pump starts from rest), α is the angular acceleration, and t is the time. Plugging in the given values, we get ω = 35.0 rad/s2 * 9.00 s = 315 rad/s.

The second equation is θ = θ0 + ω0t + 1/2 αt^2, where θ is the angular displacement, θ0 is the initial angular displacement (in this case, 0 since the pump starts from rest), ω0 is the initial angular velocity (again, 0), α is the angular acceleration, and t is the time. We can rearrange this equation to solve for ω, which gives us ω = √(2αθ - αt^2).

Since we know the radius of the pump (0.120 m) and the angular displacement (2π radians, since the water travels around the entire wheel), we can plug those values into the equation and solve for ω. This gives us ω = √(2 * 35.0 rad/s2 * 2π - 35.0 rad/s2 * 9.00 s^2) = 22.9 rad/s.

Finally, we can convert this angular velocity to linear velocity by using the equation v = ωr, where v is the linear velocity, ω is the angular velocity, and r is the radius. Plugging in the values, we get v = 22.9 rad/s * 0.120 m = 2.75 m/s.

Therefore, after 9.00 seconds of acceleration at 35.0 rad/s2, the water will be traveling at a speed of 2.75 m/s when it leaves the pump.