Answer to Question #122401 in Physics for jacob

Question

A flywheel turns through 32 rev as it slows from an angular speed of 6.6 rad/s to a stop. (a) Assuming a constant angular acceleration, find the time for it to come to rest. (b) What is its angular acceleration? (c) How much time is required for it to complete the first 16 of the 32 revolutions?

Answer 1

The kinematic equation of the angular motion is the following:

$$\Delta\varphi = \omega_0 t + \dfrac{\varepsilon t^2}{2}$$

where $\Delta\varphi = 32\cdot 2\pi = 64\pi\space rad$ is a changing in the angle of revolution (as far as one revolution implies changing the angle by $2\pi$ rad), $\omega_0 = 6.6rad/s$ ​is the initial angular speed, $\varepsilon$ is the constant angular acceleration and $t$ is the interval, during which the motion is considered.

The changing in angular velocity and the constant angular acceleration, by definition, are connected in the following manner:

$$\varepsilon = \dfrac{\Delta\omega}{t} = \dfrac{\omega_0}{t}$$

where $\Delta\omega = \omega_0-0 = \omega_0 = 6.6 rad/s$, as far as the flywheel slows down from the 6.6 rad/s to the rest.

a), b)

Let's solve two previous equations simultaneously with respect to $t$ and $\varepsilon$. Expressing $t$ from the second equation and substituting it into the first one, obtain:

$$\Delta\varphi = \omega_0 \dfrac{\omega_0}{\varepsilon} + \dfrac{\varepsilon}{2}\dfrac{\omega_0^2}{\varepsilon^2}$$ $$\Delta\varphi =\dfrac{3}{2}\dfrac{\omega_0^2}{\varepsilon}$$

Expressing $\varepsilon$ from here, get:

$$\varepsilon = \dfrac{3}{2}\dfrac{\omega_0^2}{\Delta\varphi} = \dfrac{3}{2}\cdot \dfrac{6.6^2}{64\pi} \approx 0.32 rad/s^2$$

Substitutin $\varepsilon$ to the secong equation, get:

$$t = \dfrac{\omega_0}{\varepsilon}= \dfrac{2}{3}\dfrac{\Delta\varphi}{\omega_0} = \dfrac{2}{3}\cdot \dfrac{64\pi}{6.6} \approx 20.3 s$$

c) Solving the first equation with $\Delta\varphi = 16\cdot 2\pi = 32\pi\space rad$, obtain:

$$\Delta\varphi = \omega_0 t + \dfrac{\varepsilon t^2}{2}$$ $$0.16t^2 + 3.6t - 32\cdot 3.14 = 0$$ $$0.16t^2 + 3.6t - 100.5 = 0$$

The solution is:

$$t = 11.8s$$

Answer. a) Time to stop: 20.3 s, b) angular acceleration: 0.32 rad/s^2, c) time to perform 16 revolutions: 11.8 s.