A record turntable rotating at 33 1/3 rev/min slows down and stops in 36s after the motor is turned off.
(a) Find its (constant) angular acceleration in revolutions per minute-squared.
(b) How many revolutions does it make in this time?
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I tried using a = w/t where
a = 0-33.3 / .6, but don't quite know where to go from there.
angAcc=changeinW/time
= (-33.3)/(36/60)
revolutions
D=avgvelocity*time=33.3/2 * (36/60)
To find the angular acceleration of the turntable, you can use the following equation:
ωf = ωi + αt
Where:
ωf is the final angular velocity
ωi is the initial angular velocity
α is the angular acceleration
t is the time
In this case, the final angular velocity (ωf) is 0 because the turntable stops rotating. The initial angular velocity (ωi) is given as 33 1/3 rev/min. The time (t) is given as 36 seconds.
So the equation becomes:
0 = 33 1/3 + α * 36
To isolate α, rearrange the equation:
α * 36 = -33 1/3
Now simplify the right side:
α * 36 = -100/3
Finally, divide both sides by 36:
α = (-100/3) / 36
Simplify further:
α = -100/108
Now, to convert the angular acceleration to revolutions per minute-squared, divide α by 60 (since there are 60 seconds in a minute):
α = (-100/108) / 60
Simplify further:
α = -5/54 rev/min²
So the angular acceleration of the turntable is -5/54 rev/min².
To find the number of revolutions the turntable makes in 36 seconds, you can use the formula:
θf = θi + ωit + 0.5αt²
Where:
θf is the final angle (in revolutions)
θi is the initial angle (which is 0 in this case)
ωi is the initial angular velocity (33 1/3 rev/min)
α is the angular acceleration (-5/54 rev/min²)
t is the time (36 seconds)
Since the turntable starts from rest (θi = 0) and the final angle is the unknown, the equation simplifies to:
θf = ωit + 0.5αt²
Substituting the given values:
θf = (33 1/3) * 36 + 0.5 * (-5/54) * (36²)
Calculating the right side:
θf = 120 + 0.5 * (-5/54) * 1296
θf = 120 - 120
Therefore, the turntable makes 0 revolutions in the given time of 36 seconds.