7.A turntable reaches an angular speed of 33.3 rpm, in 2.1 s, starting from rest.
(a) Assuming the angular acceleration is constant, what is its magnitude?
b How many revolutions does the turntable make during this time interval?
V = 33.3rev/1min = 33.3rev/60s - 0.555rev/s.
a. a = (Vf - Vi) / t,
a = (0.555 - 0) / 2.1 = 0.264rev/s^2.
b. rev = 0.555rev/s * 2.1s = 1.17.
post
b) is incorrect above. You need to divide by 2 because the answer is the average velocity multiplied by time.
The correct answer is 0.585 revolutions.
To find the magnitude of the angular acceleration, you can use the formula:
angular acceleration (α) = (final angular speed (ωf) - initial angular speed (ωi)) / time (t)
First, convert the final angular speed from rpm (revolutions per minute) to rad/s (radians per second):
ωf = 33.3 rpm * (2π rad/1 rev) * (1 min / 60 s)
≈ 3.49 rad/s
The initial angular speed is zero since the turntable starts from rest:
ωi = 0
Next, substitute the values into the formula:
α = (3.49 rad/s - 0) / 2.1 s
≈ 1.66 rad/s²
So, the magnitude of the angular acceleration is approximately 1.66 rad/s².
To find the number of revolutions the turntable makes during this time interval, you can use the formula:
number of revolutions = final angular speed / (2π)
Substitute the values into the formula:
number of revolutions = 3.49 rad/s / (2π)
Simplify:
number of revolutions ≈ 0.556 revolutions
Thus, the turntable makes approximately 0.556 revolutions during the 2.1-second time interval.