Question

A gyroscope flywheel of radius 1.57 cm is accelerated from rest at 17.0 rad/s2 until its angular speed is 2920 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

Step-by-step explanation:

Tangential acceleration = angular acceleration x radius

= 17 x 1.57 x 10⁻²

= .267 m /s².

b ) angular velocity = 2πn , n is no of revolution per second

= 2 x 3.14 x 2920 / 60

= 305.62 rad / s

tangential velocity v = angular velocity x radius

v = 305.62 x .0157

v = 4.8 m /s

radial acceleration = v² / r , r is radius

= 4.8 x 4.8 / .0157

= 1467.5 m /s²

c )

tangential final velocity v = 4.8 m /s

tangential initial velocity u = 0

tangential acceleration a = .267 m /s

distance travelled = s

v² = u² + 2as

4.8² = 2 x .267 x s

s = 4.8² / 2 x .267

= 43.15 m .

=