A body moves on a circle with angular velocity that depends…
A body moves on a circle with angular velocity that depends on the angle as $\omega(\theta) = \omega_0 - k\theta$, where $\omega_0$ is the initial angular velocity and $k = 2\ \mathrm{s^{-1}}$; $\theta = 0$ at $t = 0$. Find (a) $\theta(t)$; (b) $\omega(t)$; (c) the time after which the angular velocity drops to $\omega_0/2$.
Explanation of Method
Angular velocity that decreases with the angle.
Step 1: (a) Angle as a function of time
Write $\omega = d\theta/dt$, separate the variables and integrate from 0: \[\int_0^\theta \frac{d\theta}{\omega_0 - k\theta} = \int_0^t dt \;\Longrightarrow\; -\frac1k \ln\frac{\omega_0 - k\theta}{\omega_0} = t.\]
\[ \omega_0 - k\theta = \omega_0 e^{-kt}. \]
\[\theta(t) = \frac{\omega_0}{k}\left(1 - e^{-kt}\right)\]
Step 2: (b) Angular velocity as a function of time
The left side of the last equation is exactly $\omega$: $\omega(t) = \omega_0\,e^{-kt}$.
- angular velocity (fraction of ω₀)
- angle (fraction of ω₀/k)
Step 3: (c) Half the initial angular velocity
\[ \omega_0 e^{-kt} = \frac{\omega_0}{2} \;\Longrightarrow\; t = \frac{\ln 2}{k} = \frac{0.693}{2}. \]
\[t \approx 0.35\ \mathrm{s}\]
Check
$ \tfrac{d\theta}{dt} = \tfrac{\omega_0}{k} \cdot k e^{-kt} = \omega_0 e^{-kt} = \omega $, and $ \omega_0 - k\theta = \omega_0 - \omega_0(1 - e^{-kt}) = \omega_0 e^{-kt}\ \checkmark $
Result
\[\boxed{\,\theta(t) = \frac{\omega_0}{k}\left(1 - e^{-kt}\right)\,}\]
\[\boxed{\,\omega(t) = \omega_0\,e^{-kt}\,}\]
\[\boxed{\,t \approx 0.35\ \mathrm{s}\,}\]
Remark
Notes. The body never quite stops: the angle approaches $\omega_0/k$ without reaching it. The constant $k$ has units $\mathrm{s^{-1}}$ (the statement writes rad/s, which is the same thing since the radian is dimensionless). The half-time does not depend on $\omega_0$.