Can someone help me to give me the full solution to this problem?
Advanced thank you for the one s who are willing to help me.
A manufacturer of CD-ROM drives claims that the player can spin the disc as frequently as 1200 revolutions per minute.
a. If spinning at this rate, what is the speed of the outer row of data on the disc; this row is located 5.6 cm from the center of the disc?
b. What is the acceleration of the outer row of data?
billrussell42
Favorite Answer
1200 rev/min x 1min/60s = 20 rev/s
circumference = 2πR = 2π5.6 = 35.19 cm, one rev
20 rev/s x 35.19 cm/rev = 703.7 cm/s = 7.037 m/s
acceleration (outward) at rim is a = rω² = V²/r
V is the tangental velocity in m/s
r is radius of circle in meters
ω is angular velocity in radians/sec
a is in m/s²
ω = 20 rev/s x 2π rad/rev = 125.7 rad/s
a = rω² = 0.056•125.7² = 884 m/s²
electron1
As the disk rotates one time an object on the outer row of data will move a distance that is equal to the circumference of a circle.
C = 2 * π * 0.056 = π * 0.112
The circumference of the circle is approximately 0.35 meters. The object moves 1200 times this distance in 60 seconds.
v = 1200 * π * 0.112 * π ÷ 60 = 2.24 * π
The velocity is approximately 7.04 m/s. Since the outer row of data is moving around a circle at a constant speed, it has centripetal acceleration.
a = v^2 ÷ r = 5.0176 * π^2 ÷ 0.056 = 89.6 * π^2
The centripetal acceleration is approximately 884 m/s^2. I hope this is helpful for you.
az_lender
(a) 1200 rev/min = 20 rev/sec.
(20 rev/s)*(2*pi*0.056 m/rev)
= maybe 7 m/s, use calculator.
(b) a = v^2/r, where r = 0.056m and v comes from part (a) above.
oldschool
V = ω*r = 1200rev/60s * 2πrad/rev *0.056 = 1.12m/s <<<<< a.
a = V²/r = 1.12²/0.056 = 22.4m/s² <<<<< b.