A flywheel with a diameter of 2.45 m is rotating at an angular speed of 74.1 rev/min. (a) What is the angular speed of the flywheel in radians per second? (b) What is the linear speed of a point on the rim of the flywheel? (c) What constant angular acceleration (in revolutions per minute-squared) will increase the wheel's angular speed to 1260 rev/min in 102 s? (d) How many revolutions does the wheel make during that 102 s?

Answer 1

a. C = pi*D = 3.14 * 2.45 = 7.7m.
V=74.1rev/min * 6.28rad/rev *(1/60min/s
= 7.76 rad/s.

b. V = 7.76rad/s * 7.7m/6.28rad=9.51m/s

c. t = 102s / 60s/min = 1.7 min.
V = Vo + at = 1260rev/min.
74.1 + 1.7a = 1260,
1.7a = 1260 - 74.1 = 1185.9,
a = 698rev/min^2.

d. rev = 1260rev/min * 1.7min = 2142