Solving an Old Phonograph Problem: Finding Angular Velocity, Speed & Period

AI Thread Summary
An old phonograph record spins at 77.6 rpm, leading to an angular velocity of approximately 8.124 radians per second. The speed of the record below the needle, positioned 7.4 cm from the spindle, is calculated to be about 0.6012 m/sec. The period for one complete rotation of the record is approximately 0.7734 seconds. The calculations involve converting revolutions per minute to revolutions per second and applying the formulas for angular velocity, linear speed, and period. Understanding these concepts is essential for solving similar problems involving rotational motion.
Jayhawk1
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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.
 
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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?
 
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Jayhawk1 said:
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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