Question

A flywheel with a diameter of 1.39 m is rotating at an angular speed of 194 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 1280 rev/min in 128 s? (d) How many revolutions does the wheel make during that 128 s?

Answer 1

1 revolution = 2 radians

1 minute = 60 seconds

Diameter of the flywheel = D = 1.39 m

Radius of the flywheel = R = D/2 = 1.39/2 = 0.695 m

Initial angular speed of the flywheel = ₁ = 194 rev/min

Converting from rev/min to rad/s,

₁ = 20.32 rad/s

Linear speed of a point on the rim of the flywheel = V

V = ₁R

We will use the angular speed of the flywheel in rad/s so that we get the speed in m/s.

V = (20.32)(0.695)

V = 14.12 m/s

Final angular speed of the flywheel = ₂ = 1280 rev/min

Time period of acceleration = T = 128 sec = 128/60 min = 2.133 min

Angular acceleration of the flywheel =

₂ = ₁ + T

1280 = 194 + (2.133)

= 509.1 rev/min²

Number of revolutions the wheel makes in the 128 sec = n

₂² = ₁² + 2n

(1280)² = (194)² + 2(509.1)n

n = 1572 rev

a) Angular speed of the flywheel in rad/s = 20.32 rad/s

b) Linear speed of a point on the rim = 14.12 m/s

c) Angular acceleration of the wheel = 509.1 rev/min²

d) Number of revolutions the wheel makes during the 128 sec = 1572 rev