Amit & Deepali Sinha Foundation

An Oscillation Between Two Turning Points

Irodov 1.50DifficultyUniqueness

Problem Irodov 1.50

A solid body begins rotating from rest about a fixed axis. Its angular acceleration along that axis is β(φ)=β0cos⁡φ\beta(\varphi) = \beta_0\cos\varphi, where β0\beta_0 is a constant (the axial component of a fixed vector β0\vect\beta_0) and φ\varphi is the angle turned through from the starting position, φ=0\varphi=0 at the start. Find the angular velocity ω\omega as a function of φ\varphi, and plot that dependence.

Given
  • β(φ)=β0cos⁡φ\beta(\varphi) = \beta_0\cos\varphi
  • ω=0\omega=0 at φ=0\varphi=0
Find
ω(φ)\omega(\varphi), plotted

Solution

Step 1Trading the time derivative for a φ\varphi derivative

Since β=dω/dt\beta=d\omega/dt and dφ/dt=ωd\varphi/dt=\omega,

β=dωdt=dωdφdφdt=ωdωdφ(1) \beta = \frac{d\omega}{dt} = \frac{d\omega}{d\varphi}\frac{d\varphi}{dt} = \omega\frac{d\omega}{d\varphi} \tag{1}

Setting this equal to the given β(φ)\beta(\varphi) and separating variables,

ω dω=β0cos⁡φ dφ\omega\,d\omega = \beta_0\cos\varphi\,d\varphi
Step 2Integrating from rest at φ=0\varphi=0

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

∫0ωω′ dω′=∫0φβ0cos⁡φ′ dφ′\int_0^\omega \omega'\,d\omega' = \int_0^\varphi \beta_0\cos\varphi'\,d\varphi'
ω22=β0sin⁡φ\frac{\omega^2}{2} = \beta_0\sin\varphi

Multiplying through by 2 gives the first integral of the motion, the relation Step 3 leans on:

ω2=2β0sin⁡φ\omega^2 = 2\beta_0\sin\varphi
(2)

Taking the square root,

ω=±2β0sin⁡φ\omega = \pm\sqrt{2\beta_0\sin\varphi}
(3)
Step 3Why the sign is genuinely ±\pm: the body turns around

At φ=π\varphi=\pi the boxed answer (3) gives ω=0\omega=0, but the angular acceleration there is nonzero.

β(π)=β0cos⁡π=−β0≠0\beta(\pi) = \beta_0\cos\pi = -\beta_0 \ne 0

So the body is momentarily at rest, not stopped: it is a turning point, and β\beta immediately drives φ\varphi back down. Since (2) is the same first integral throughout, the return trip retraces the identical curve in φ\varphi with the opposite sign of ω\omega.

ω={+2β0sin⁡φ,φ increasing (the outward swing, 0→π)−2β0sin⁡φ,φ decreasing (the return swing, π→0)\omega = \begin{cases} +\sqrt{2\beta_0\sin\varphi}, & \varphi \text{ increasing (the outward swing, } 0\to\pi\text{)} \\ -\sqrt{2\beta_0\sin\varphi}, & \varphi \text{ decreasing (the return swing, } \pi\to0\text{)} \end{cases}

The body oscillates indefinitely between the two turning points φ=0\varphi=0 and φ=π\varphi=\pi, where β0cos⁡φ\beta_0\cos\varphi has been decelerating it all along the second half of each swing. Plotting the two cases above answers the second half of the question.

Angular velocity ω in multiples of sqrt(β_0) against the angle turned φ. The gold branch is the outward swing, rising from rest to ω_max at φ=π/2 then back to zero; the royal branch below the axis is the return swing.

Angular velocity ω\omega in multiples of β0\sqrt{\beta_0} against the angle turned φ\varphi. The gold branch is the outward swing, rising from rest to ωmax\omega_{\mathrm{max}} at φ=π/2\varphi=\pi/2 then back to zero; the royal branch below the axis is the return swing.

Check

At φ=0\varphi=0 the boxed answer (3) gives ω=0\omega=0, matching the body starting from rest. The extremum of the outward branch sits at φ=π/2\varphi=\pi/2: differentiating (2) gives 2ω dω/dφ=2β0cos⁡φ2\omega\,d\omega/d\varphi=2\beta_0\cos\varphi, so dω/dφ=0d\omega/d\varphi=0 exactly where cos⁡φ=0\cos\varphi=0 (for ω≠0\omega\ne0), which is φ=π/2\varphi=\pi/2, precisely where β=β0cos⁡φ\beta=\beta_0\cos\varphi itself vanishes, as it must at any turning point of ω\omega against φ\varphi. And at both φ=0\varphi=0 and φ=π\varphi=\pi, ω=0\omega=0 while β=±β0≠0\beta=\pm\beta_0\ne0, confirming both are true turning points of the oscillation rather than places the body comes permanently to rest.

Remark

The substitution (1) is the rotational twin of wτ=v dv/dsw_\tau=v\,dv/ds used for translational motion under a position dependent tangential acceleration: both integrate to a first integral, ω2/2\omega^2/2 (or v2/2v^2/2) as a function of position, exactly the way a conservative force’s work integrates to a kinetic energy. That is also why the motion here is periodic rather than one-way: β0cos⁡φ\beta_0\cos\varphi behaves like a restoring influence between the two zeros of sin⁡φ\sin\varphi at φ=0\varphi=0 and φ=π\varphi=\pi, so the body swings between them forever, the rotational analogue of a pendulum confined between two turning points.