Answer to Question #115563 in Mechanics | Relativity for Jocelyn

Question

A machine part rotates at an angular speed of ωi. Its speed is then increased to ωf at an angular acceleration of α. If both the initial and final angular speeds are now doubled and the angular acceleration remains the same, by what factor is the angular displacement changed?

Answer 1

The problem is similar to determining the distance for the movement with constant acceleration.

Let us write the dependence of angular velocity on time:

$\omega(t) = \omega_i + \alpha t.$ Therefore, displacement is $\Delta \varphi = \omega_i t + \dfrac{\alpha t^2}{2}$ .

Let $t_0$ be the time of movement, therefore $\omega_f = \omega_i + \alpha t_0.$ So $t_0 = \dfrac{\omega_f-\omega_i}{\alpha}.$ Therefore,

$\Delta \varphi = \omega_i t_0 + \dfrac{\alpha t_0^2}{2} = \dfrac{\omega_f^2-\omega_i^2}{2\alpha}.$

If we double the velocities and the acceleration remains the same, then

$\Delta_2 \varphi = \dfrac{4\omega_f^2-4\omega_i^2}{2\alpha} = 4\dfrac{\omega_f^2-\omega_i^2}{2\alpha}.$ So the factor is 4.