Angular velocity questions. PLEASE HELP.?

Question

1. A helicopter blade is designed to rotate at such an angular velocity that the tip of the blade is less than the speed of sound in air, which is 340 ms-1. The blades rotate at 260 r.p.m Calculate the maximum possible length of the blades.

2. A Cd player varies the rate of rotation of the dics in order to keep the track from which the music is being reproduced moving at a constant linear speed of 1.30ms-1. Clacutae the rates of rotation of a 12.0 cm disc when the music is being read from
(a) the outer edge of the disc
(b) a point 2.55cm from the centre of the disc.
Give answers in rads-1 and in rev min-1

Answer 1

1. 260 rpm = 260/60 rps = 13/3 = 4.33333 rps

C = circumference of the circle the blade tip makes as it rotates
R = length of blade (radius of circle)
C = 2piR

340 = (13/3)C = (13/3)2piR
R = 340*3/(26pi) = 12.488 meters

2.
w = angular speed in rads/s
R = radius of the disc (6 cm = 0.06 m)
v = linear speed (constant at 1/2 m/s)

v = wR = 1.3 and this gives w = 1.3/R

a. R = 6 cm = .06 meters (edge to the the center)
w = 1.3/.06 = 21.6667 rads/s or 3.448 rps = 206.901 rpm

b. R = 2.55 cm = .0255 m
w = 1.3/.0255 = 50.98 rads/s or 8.114 rps = 486.827 rpm

Answer 2

1.
v = w * r
v - actual speed of the tip of the blade
w -angular speed (rad/s)
r - radius of the circle (length of the blade)
w = 260 rev/min = 260*2*pi / 60 (rad/s)=27.227 rad/s
r = v / w = 340 / 27.227 = 12.49 m

2. 12 cm disc (assuming this is diameter)
d = 12 cm
v = 1.3 m/s

r1 = d /2 = 6 cm = 0.06m
r2 = 2.55 cm = 0.0255 m

w1 = v / r1 = 1.3 / 0.06 = 21.667 rad/s
w1 = 21.667 * 60 / (2*pi) = 206.9 rev/min

w2 = v / r2 = 1.3 / 0.0255 = 50.98 rad/s
w2 = 486.8 rev/min