Question

Chapter 10, Problem 030 GO A gyroscope flywheel of radius 2.35 cm is accelerated from rest at 12.7 rad/s2 until its angular speed is 2790 rev/min. (a) What is the tangential acceleration of a point on the rim of the flywheel during this spin-up process? (b) What is the radial acceleration of this point when the flywheel is spinning at full speed? (c) Through what distance does a point on the rim move during the spin-up?

Answer 1

Given,

Radius r = 2.35 cm = 0.0235 m

Angular acceleration = 12.7 rad/s²

Initial angular speed w₁= 0

Final angular speed w₂ = 2790 rev/min

w₂ = 292.29 rad/s

a) Tangential acceleration a_t is given by

a_t = r

a_t = 12.7 0.0235

a_t = 0.298 m/s²

a_t= 0.30 m/s²

b) Radial acceleration a_ris given by

a_r = w₂² r

a_r = (292.29)² 0.0235

a_r = 2007.69 m/s²

c)

Let , be the angle through which the flywheel rotates during spin-up process

w₂² = w₁² + 2

(292.29)² = 0 + 212.7

= 3363.52 rad

Distance s through which the point moves is given by

s = r

s = 3363.52 0.0235

s = 79.04 m