An old-fashioned single-play vinyl record rotates on a turntable at 40.0 rpm. What are (a) the angular velocity in rad/s and (b) the period of the motion?
1rev=2pi rad
1min=60sec
40rev/min=80pirad/60
=4.189ras/s
To find the angular velocity and period of the motion of an old-fashioned single-play vinyl record rotating on a turntable at a given RPM, we can use the following formulas:
(a) Angular velocity (ω) = 2π × RPM / 60
(b) Period (T) = 1 / (RPM / 60)
(a) To find the angular velocity in rad/s, we can use the formula:
Angular velocity (ω) = 2π × RPM / 60
Given that the record rotates at 40.0 RPM, we can substitute this value into the formula to find the result:
ω = 2π × 40.0 / 60
= (2 × 3.14159 × 40.0) / 60
= 8.37758 rad/s (rounded to five decimal places)
Therefore, the angular velocity of the vinyl record is approximately 8.37758 rad/s.
(b) To find the period of the motion, we can use the formula:
Period (T) = 1 / (RPM / 60)
Substituting the given value of RPM into the formula:
T = 1 / (40.0 / 60)
= 1 / (2/3)
= 3/2
= 1.5 seconds
Therefore, the period of the motion is 1.5 seconds.