Question

A gyroscope flywheel of radius 2.05 cm is accelerated from rest at 14.1 rad/s2 until its angular speed is 2390 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

r = 0.0205 m

a = 14.1 rad/s^2

angular speed , w = 2390 rev/min

w = 2390 * 2pi/60 rad/s

w = 250.2 rad/s

a) tangential acceleration = a * r

tangential acceleration = 14.1 * 0.0205

tangential acceleration = 0.29 m/s^2

b) radial acceleration = w^2 * r

radial acceleration = 250.2^2 * 0.0205

radial acceleration = 1283 m/s^2

c)

for the angle rotated

angle = w^2/(2 ( a))

angle = 250.2^2/(2 * 14.1)

angle = 2219 radian

distance = 2219 * 0.0205

distance = 45.5 m

the distance moved is 45.5 m