Question

A flywheel with a very low friction bearing takes 1.6 h to stopafter the motor power

Answer 1

Answer:

(I). The initial rotation rate is 4.29 rad/s.

(II). The revolutions is 3932.

Explanation:

Given that,

Time = 1.6 h

Angular velocity = 41 rpm

(I). We need to calculate the initial rotation rate in rad/s

$\omega=41\ rev/m$

$\omega=\dfrac{41\times 2\pi}{60}\ rad/s$

$\omega=4.29\ rad/s$

(II). We need to calculate the revolutions

Using formula of revolutions

$\theta=\omega t$

$\theta=4.29\times1.6\times3600$

$\theta=24710\ rad$

$\theta=\dfrac{24710}{2\pi}$

$\theta=3932\ revolution$

Hence, (I). The initial rotation rate is 4.293 rad/s.

(II). The revolutions is 3932.