Question

A gyroscope flywheel of radius 3.21 cm is accelerated from rest at 13.2 rad/s2 until its angular speed is 2450 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

Part a)

`a_t = 0.423 m/s^2`

Part b)

`a_c = 2113 m/s^2`

Part c)

`d = 80 m`

Step-by-step explanation:

Part a)

as we know that angular acceleration of the wheel is given as

`\alpha = 13.2 rad/s^2`

now the radius of the wheel is given as

R = 3.21 cm

so the tangential acceleration is given as

`a_t = R\alpha`

`a_t = (0.0321)(13.2)`

`a_t = 0.423 m/s^2`

Part b)

frequency of the wheel at maximum speed is given as

`f = 2450 rev/min`

`f = (2450)/(60) = 40.8 rev/s`

now we know that

`\omega = 2\pi f = 2\pi(40.8) = 256.56 rad/s`

now radial acceleration is given as

`a_c = \omega^2 r`

`a_c = (256.56)^2(0.0321) = 2113 m/s^2`

Part c)

total angular displacement of the point on rim is given as

`\Delta \theta = \omega_0 t + (1)/(2)\alpha t^2`

here we know that

`\omega = \omega_0 + \alpha t`

`256.56 = 0 + 13.2 t`

`t = 19.4 s`

now angular displacement will be

`\Delta \theta = 0 + (1)/(2)(13.2)(19.4)^2`

`\Delta \theta = 2493.3 rad`

now the distance moved by the point on the rim is given as

`d = R\theta`

`d = (0.0321)(2493.3)`

`d = 80 m`