Answer to Question #237154 in Physics for lion

Question

A wheel turns with angular speed of 30 rev s^-1 is brought to rest with constant acceleration. It turns 60 rev before it stops. Calculate its angular acceleration in rad s^-2

Answer 1

By definition, the angular acceleration is given as follows:

$$\varepsilon = \dfrac{\omega_f - \omega_i}{t} = \dfrac{ - \omega_i}{t}$$

where $\omega_f = 0rad/s$ is the final angular speed (the wheel is brought to rest), $\omega_i = 2\pi \cdot 30rev/s = 60\pi\space rad/s$ is the inital angular speed, $t$ is time in which it was brought to rest.

The angular distance is given as follows:

$$\varphi = \omega _it + \dfrac{\varepsilon t^2}{2}$$

Expressing time from the first formula and substituting in into the second one, obtain:

$$t = \dfrac{ -\omega_i}{\varepsilon}\\$$ $$\varphi = \omega _i\cdot \dfrac{ -\omega_i}{\varepsilon} + \dfrac{\varepsilon\omega_i^2}{2\varepsilon^2} = -\dfrac{\omega_i^2}{2\varepsilon}$$

Expressing acceleration and substituting $\varphi = 2\pi\cdot 60rev = 120\pi\space rad$, obtain:

$$\varepsilon = -\dfrac{\omega^2}{2\varphi} = -\dfrac{(60\pi\space rad/s)^2}{2\cdot 120\pi \space rad} = 15\pi \space rad/s^2$$

Answer. $15\pi \space rad/s^2$.