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.

Answer 1

Answer:

Part a)

$a_t = 0.423 m/s^2$

Part b)

$a_c = 2113 m/s^2$

Part c)

$d = 80 m$

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

$f = 2450 rev/min$

$f = \frac{2450}{60} = 40.8 rev/s$

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

$\Delta \theta = \omega_0 t + \frac{1}{2}\alpha t^2$

here we know that

$\omega = \omega_0 + \alpha t$

$256.56 = 0 + 13.2 t$

$t = 19.4 s$

now angular displacement will be

$\Delta \theta = 0 + \frac{1}{2}(13.2)(19.4)^2$

$\Delta \theta = 2493.3 rad$

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

$d = R\theta$

$d = (0.0321)(2493.3)$

$d = 80 m$