Angular velocity, speed, and period of a phonograph record at 77.6 rpm

[Jayhawk1]

I was curious as to how to approach this problem... An old phonograph record is spinning at 77.6 rpm. The phonograph needle is stuck 7.4 cm from the spindle. a) What is the angular velocity of the record? b) What is the speed of the record below the needle? c) What is the period corresponding to one complete rotation of the record? I think the easiest way to approach it would be by finding the frequency but I am not quite sure how to get it. Any help would be greatly appreciated.

 

[whozum]

a) 77.6 revolutions per minute. 77.6/60 revolutions per second. Sounds like an angular velocity to me.

b) You know the angular velocity. $$v_{linear} = r\omega$$

c) 77.6 revolutions in one minute. How long does it take for one revolution?

 

[xanthym]

I was curious as to how to approach this problem... An old phonograph record is spinning at 77.6 rpm. The phonograph needle is stuck 7.4 cm from the spindle. a) What is the angular velocity of the record? b) What is the speed of the record below the needle? c) What is the period corresponding to one complete rotation of the record? I think the easiest way to approach it would be by finding the frequency but I am not quite sure how to get it. Any help would be greatly appreciated.

From the problem statement:
{Frequency of Rotation} = f = (77.6 rpm) =
= {(77.6)/60 rev/sec)
= (1.293 rev/sec)

Question a):
{Angular Velocity} = ω = 2*π*f =
= 2*π*(1.293 rev/sec) =
= (8.124 radians/sec)

Question b):
{Speed of Record below Needle} = v = ω*r =
= (8.124 radians/sec)*(7.4 cm) =
= (60.12 cm/sec) = (0.6012 m/sec)

Question c):
{Period} = T = 1/f =
= 1/(1.293 rev/sec) =
= (0.7734 sec)


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