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=

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(a) To find the initial angular velocity ω0, we need to relate it to the frequency f0. The relationship between angular velocity (ω) and frequency (f) is given by:

ω = 2πf

Given that the turntable is spinning at a constant frequency f0, the initial angular velocity can be expressed as:

ω0 = 2πf0

Therefore, the magnitude of the initial angular velocity ω0 is 2π times the frequency f0.

Answer: ω0 = 2πf0

(b) We are told that the turntable slows down with a constant angular deceleration. The magnitude of the angular acceleration α can be obtained using the following equation:

α = (ω - ω0) / t

Since the turntable slows down and eventually comes to a stop, the final angular velocity ω is zero. Substituting this value into the equation, we get:

α = (0 - ω0) / t = -ω0 / t

Thus, the magnitude of the angular acceleration α is equal to the negative initial angular velocity ω0 divided by the time t.

Answer: α = -ω0 / t

(c) To find the magnitude of the total angle Δθ, we can use the equation that relates angular velocity, angular acceleration, and angle:

Δθ = ω0t + (1/2)αt^2

Substituting the expressions for ω0 and α obtained in parts (a) and (b), we have:

Δθ = (2πf0)t + (1/2)(-ω0 / t)t^2
= 2πf0t - (1/2)ω0t

Now, we can simplify the expression:

Δθ = (2πf0 - (1/2)ω0)t

Therefore, the magnitude of the total angle Δθ is given by the difference between 2π times the initial frequency f0 and (1/2) times the initial angular velocity ω0, all multiplied by the time t.

Answer: Δθ = (2πf0 - (1/2)ω0)t

To solve this problem, we need to use the equations of rotational kinematics. Here are the step-by-step explanations to find the answers to each part:

(a) To find the magnitude of the initial angular velocity ω0, we can use the relationship between frequency (f) and angular velocity (ω):
ω = 2πf

Given that the initial spinning frequency is f0, we can find ω0 by substituting it into the equation:
ω0 = 2πf0

Thus, the magnitude of the initial angular velocity ω0 is 2πf0.

(b) To find the magnitude of the angular acceleration α, we can use the equation:
ω = ω0 + αt

At the point the turntable stops, the final angular velocity (ω) is 0, and the initial angular velocity (ω0) is 2πf0. The time it takes to slow down is t.

Substituting the values into the equation, we have:
0 = 2πf0 + αt

Solving for α, we get:
α = -2πf0/t

Therefore, the magnitude of the angular acceleration α is -2πf0/t.

(c) To find the magnitude of the total angle Δθ, we can use the equation:
Δθ = ω0t + 0.5αt^2

Substituting the values we have:
Δθ = (2πf0)t + 0.5(-2πf0/t)t^2

Simplifying the equation, we get:
Δθ = 2πf0t - πf0t

Therefore, the magnitude of the total angle Δθ is πf0t.

To summarize:

(a) ω0 = 2πf0
(b) α = -2πf0/t
(c) Δθ = πf0t