Answer to Question #144211 in Physics for me

Question

a wheel turning with angular velocity of 60π rad/s is brought to rest with a constant acceleration. it makes 60 rotations before it stops .calculate its angular velocity.

Answer 1

If the wheel makes 60 rotations before it stops then it coveres angular distance

$\varphi = 60\times 2\pi =120\space rad$

before it stops. On the other hand, the angular distance covered in time $t$ is:

$\varphi =\omega_0t -\dfrac{\varepsilon t^2}{2}$

where $\omega_0 = 60\pi \space rad/s$ is the initial angular velocity and $\varepsilon$ is the constant angular acceleration. Expressing $\varepsilon$, obtain:

$\varepsilon = \dfrac{2}{t^2}(\omega_0t-\varphi)$

The angular velocity in any given moment of time is:

$\omega(t) = \omega_0-\varepsilon t = \omega_0 - \dfrac{2}{t}(\omega_0t-\varphi) = \dfrac{2\varphi}{t} - \omega_0$

Substituting values, obtain:

$\omega(t) = \dfrac{2\varphi}{t} - \omega_0 = \dfrac{2\cdot 120\pi}{t} - 60\pi = 240\pi\dfrac{1}{t} - 60\pi\space [rad/s]$

Answer. Angular velocity at any given moment of time (in seconds) is $\omega(t) = 240\pi\dfrac{1}{t} - 60\pi\space [rad/s]$