Amit & Deepali Sinha Foundation

Rotation That Never Quite Finishes

Irodov 1.49DifficultyUniqueness

Problem Irodov 1.49

A solid body rotates about a fixed axis, and its angular velocity depends on the angle φ\varphi already turned through as ω(φ)=ω0−aφ\omega(\varphi) = \omega_0 - a\varphi, where ω0\omega_0 and aa are given positive constants. The angle is measured from φ=0\varphi=0 at t=0t=0. Find, as functions of time,

  1. the rotation angle φ(t)\varphi(t)

  2. the angular velocity ω(t)\omega(t)

The body after turning through an angle φ from its starting position (dashed reference), with its current angular velocity ω; the marked point on the rim tracks how far it has turned.

The body after turning through an angle φ\varphi from its starting position (dashed reference), with its current angular velocity ω\vect\omega; the marked point on the rim tracks how far it has turned.

Given
  • ω(φ)=ω0−aφ\omega(\varphi) = \omega_0-a\varphi
  • φ(0)=0\varphi(0)=0
Find
(a) φ(t)\varphi(t), (b) ω(t)\omega(t)

Solution

Step 1A separable equation for φ(t)\varphi(t)

Writing ω=dφ/dt\omega=d\varphi/dt in the given relation and separating variables,

dφdt=ω0−aφ  ⟹  dφω0−aφ=dt\frac{d\varphi}{dt} = \omega_0-a\varphi \;\Longrightarrow\; \frac{d\varphi}{\omega_0-a\varphi} = dt
Step 2Integrating from the stated initial condition

Integrating the left side from φ=0\varphi=0 and the right side from t=0t=0,

∫0φdφ′ω0−aφ′=∫0tdt′−1a[ln⁡(ω0−aφ′)]0φ=tln⁡(ω0−aφω0)=−at\begin{align*}\int_0^\varphi\frac{d\varphi'}{\omega_0-a\varphi'} &= \int_0^t dt' \\ -\frac{1}{a}\Bigl[\ln(\omega_0-a\varphi')\Bigr]_0^\varphi &= t \\ \ln\left(\frac{\omega_0-a\varphi}{\omega_0}\right) &= -at\end{align*}

Exponentiating and solving for φ\varphi,

ω0−aφ=ω0e−at\omega_0-a\varphi = \omega_0 e^{-at}
φ(t)=ω0a(1−e−at)\varphi(t) = \dfrac{\omega_0}{a}\left(1-e^{-at}\right)
(1)
Step 3The angular velocity from the same result

Differentiating (1) directly gives ω(t)=dφ/dt\omega(t)=d\varphi/dt.

ω(t)=dφdt=ω0a(ae−at)\omega(t) = \frac{d\varphi}{dt} = \frac{\omega_0}{a}\left(ae^{-at}\right)
ω(t)=ω0e−at\omega(t) = \omega_0 e^{-at}
(2)
Check

At t=0t=0, (1) gives φ(0)=(ω0/a)(1−1)=0\varphi(0)=(\omega_0/a)(1-1)=0, matching the stated initial condition, and (2) gives ω(0)=ω0\omega(0)=\omega_0, matching the given ω=ω0−aφ\omega=\omega_0-a\varphi at φ=0\varphi=0. Also, substituting φ(t)\varphi(t) and ω(t)\omega(t) back into the original relation ω=ω0−aφ\omega=\omega_0-a\varphi gives ω0e−at=ω0−a⋅ω0a(1−e−at)=ω0e−at\omega_0 e^{-at}=\omega_0-a\cdot\dfrac{\omega_0}{a}(1-e^{-at})=\omega_0 e^{-at}, an identity, so both answers satisfy the governing relation exactly.

Remark

Although ω(t)=ω0e−at\omega(t)=\omega_0e^{-at} never reaches zero at any finite time, so the body never truly stops, φ(t)\varphi(t) approaches the finite limit ω0/a\omega_0/a as t→∞t\to\infty: the body sweeps out only a bounded total angle in an unbounded time. The exponential time law is forced by the specific shape of the given relation, ω\omega linear in φ\varphi: a first order linear equation like dφ/dt=ω0−aφd\varphi/dt=\omega_0-a\varphi always relaxes exponentially toward its fixed point φ=ω0/a\varphi=\omega_0/a. More generally, if ω\omega vanished as some power nn of the remaining angle near a limiting angle LL, the arrival time ∫dφ/ω\int d\varphi/\omega converges, and the body actually reaches LL in finite time, whenever ω\omega falls off slower than linearly (n<1n<1); it diverges, giving only an asymptotic approach as here, for n≥1n\ge1. Linear decay is exactly the marginal case sitting on that boundary.