a dentists drill starts from rest. after 5.39 seconds of constant angular accleration, it is rotating at a rate of 2.85E4 rpm.
through what angle did it rotate in the process?
(i just don't know what formula to use)
Theta= wi*time + 1/2 angaccel*time^2
To determine the angle through which the dentist's drill rotated, you can use the following formula:
θ = ωi * t + 0.5 * α * t^2
where:
θ represents the angle rotated,
ωi is the initial angular velocity,
t is the time,
α is the angular acceleration.
Since the drill starts from rest, the initial angular velocity (ωi) is 0. The final angular velocity (ωf) can be obtained by converting 2.85E4 rpm to rad/s.
1 rpm = (2π/60) rad/s
2.85E4 rpm = 2.85E4 * (2π/60) rad/s
Now, let's calculate ωf:
ωf = 2.85E4 * (2π/60) rad/s
You also have the value for the time (t) = 5.39 seconds.
Now, you need to find the angular acceleration (α). To find that, we can use the equation:
ωf = ωi + α * t
Since ωi = 0, the equation simplifies to:
ωf = α * t
Now, solve for α:
α = ωf / t
To find the angle rotated (θ), substitute the values into the first equation:
θ = (ωi * t) + (0.5 * α * t^2)
Calculate ωf, α, and substitute the values:
ωf = 2.85E4 * (2π/60) rad/s
α = ωf / t
θ = (0 * 5.39) + (0.5 * (2.85E4 * (2π/60)) * (5.39)^2
By evaluating this expression, you will obtain the angle through which the dentist's drill rotated.