A turntable is a uniform disc of mass m and radius R. The turntable is initially spinning clockwise when looked down on from above at a constant frequency f0. The motor is turned off at t=0 and the turntable slows to a stop in time t with constant angular deceleration.
(a) What is the magnitude of the initial angular velocity ω0 of the turntable? Express your answer in terms of f0 (enter pi for π, and f_0 for f0).
ω0=
(b) What is the magnitude of the angular acceleration α of the turntable? Express your answer in terms of f0 and t (enter pi for π, f_0 for f0, and t for t).
α=
(c) What is the magnitude of the total angle Δθ in radians that the turntable spins while slowing down? Express your answer in terms of f0 and t (enter pi for π, f_0 for f0, and t for t).
Δθ=
To answer these questions, we'll use the formulae of rotational kinematics:
1. ω = ω0 + αt
2. Δθ = ω0t + (1/2)αt^2
3. ω^2 = ω0^2 + 2αΔθ
(a) To find the magnitude of the initial angular velocity ω0, we'll use the fact that the turntable is initially spinning clockwise at a constant frequency f0. The frequency (f) is the number of complete rotations per unit of time, so ω0 = 2πf0.
ω0 = 2πf0
(b) To find the magnitude of the angular acceleration α, we'll consider that the turntable slows down to a stop in time t with constant angular deceleration. Angular deceleration is the negative value of angular acceleration, so α = -ω0/t.
α = -ω0/t = -(2πf0)/t
(c) To find the magnitude of the total angle Δθ that the turntable spins while slowing down, we'll use the third equation and rearrange it to solve for Δθ:
Δθ = (ω^2 - ω0^2) / (2α)
Using the values of ω0 and α we found earlier:
Δθ = (0^2 - ω0^2) / (2α) = -ω0^2 / (2α) = -(ω0^2t) / (2(2πf0)/t) = -ω0^2t^2 / (4πf0)
Δθ = -ω0^2t^2 / (4πf0)
So the answers are:
(a) ω0 = 2πf0
(b) α = -(2πf0)/t
(c) Δθ = -ω0^2t^2 / (4πf0)