Rotational Dynamics: Turbine Spins 3600 RPM to Coast

[novicephysics]

An electric-generator turbine spins at 3600 rpm. Friction is so small that it takes the turbine 10.0 min to coast to a stop. How many revolutions does it make wile stopping?

 

[blochwave]

For these problems you can use the rotational versions of all your linear equations you've been using

So instead of d, you have theta, instead of v, angular velocity, and so on.

So the equation d=1/2at^2-Vi*t can be used but with the angle traveled in radians for d, angular acceleration for a, and initial angular velocity for Vi. That's not the particular equation you should use here though. This is similar to a problem where you know initial velocity and time, and want to find a. Then you can find d(which in this case is the total angle traveled in radians, so for example if it were like 6pi, the answer would be 3 revolutions)

 

[Moetasim]

Based on the given information, we can calculate the number of revolutions the turbine makes while coasting to a stop using the equation ω = ω0 + αt, where ω is the final angular velocity, ω0 is the initial angular velocity, α is the angular acceleration, and t is the time.

Since the turbine spins at 3600 rpm initially and comes to a stop after 10.0 min, we can convert the initial angular velocity to rad/s by multiplying it by 2π/60, which gives us ω0 = 120π rad/s. We also know that the final angular velocity is 0 rad/s. Therefore, the angular acceleration can be calculated as α = ω/t = (0 - 120π)/600 s = -0.2π rad/s^2.

Now, using the equation ω = ω0 + αt, we can solve for the time it takes for the turbine to come to a stop, which is t = (ω - ω0)/α = (0 - 120π)/(-0.2π) = 600 s.

Since we are looking for the number of revolutions the turbine makes while stopping, we can use the formula N = ω0t + 1/2αt^2, where N is the number of revolutions. Substituting the known values, we get N = 120π(600) + 1/2(-0.2π)(600)^2 = 36000π - 360000π = -324000π.

Therefore, while coasting to a stop, the turbine makes approximately 324000π revolutions. This value may seem negative, but it simply indicates that the turbine rotates in the opposite direction while slowing down.