Irodov 1.58: When A Spin Axis Itself Precesses – Amit & Deepali Sinha Foundation

Problem Irodov 1.58

A solid body spins about its own horizontal axis $AB$ at a constant rate $\omega_0=0.50$ rad/s. Starting at $t=0$ , the axis $AB$ itself begins to turn about a fixed vertical axis through the same point, with constant angular acceleration $\beta_0=0.10$ rad/s $^2$ (so the axis’s own turning rate about the vertical grows from zero at that constant rate), while remaining horizontal throughout. Find the modulus of the body’s total angular velocity and the modulus of its total angular acceleration at $t=3.5$ s.

The two perpendicular contributions to the body's angular velocity at a general instant: ω_0 fixed along the body's own axis AB, and Ω along the vertical, growing from zero. Since AB stays horizontal, the two are perpendicular at every instant, and ω is their vector sum. — The two perpendicular contributions to the body’s angular velocity at a general instant: $\vect\omega_0$ fixed along the body’s own axis $AB$ , and $\vect\Omega$ along the vertical, growing from zero. Since $AB$ stays horizontal, the two are perpendicular at every instant, and $\vect\omega$ is their vector sum.

Given

$\omega_0=0.50$ rad/s, constant, about $AB$
$\beta_0=0.10$ rad/s $^2$ , constant precession of $AB$ about the vertical, starting from rest at $t=0$
$t=3.5$ s

Find

\omega(t)

and

\beta(t)

t=3.5

Solution

Step 1The two angular velocity contributions, and their sum

The precession starts from rest with constant angular acceleration $\beta_0$ , so its rate at time $t$ is $\Omega(t)=\beta_0t$ , directed vertically; the spin $\vect\omega_0$ stays along $AB$ with constant modulus. The total angular velocity is their vector sum, and since $\vect\omega_0\perp\vect\Omega$ ,

\vect\omega(t) = \vect\omega_0+\vect\Omega(t), \qquad \omega(t) = \sqrt{\omega_0^2+\Omega(t)^2} = \sqrt{\omega_0^2+(\beta_0t)^2}.

\omega = \omega_0\sqrt{1+(\beta_0t/\omega_0)^2} = 0.6\ \text{rad/s}

(1)

Step 2How the spin term itself changes, as

AB

precesses

Writing $\vect\omega=\omega_0\hat{\vect a}(t)+\vect\Omega(t)$ with $\hat{\vect a}$ the (rotating) unit vector along $AB$ , the derivative splits into the two pieces below.

\frac{d\vect\omega}{dt} = \omega_0\frac{d\hat{\vect a}}{dt}+\frac{d\vect\Omega}{dt} = \omega_0\bigl(\vect\Omega\times\hat{\vect a}\bigr)+\vect\beta_0, \tag{2}

where $\vect\beta_0=d\vect\Omega/dt$ is vertical, constant, modulus $\beta_0$ . The first term has modulus $\omega_0\Omega(t)$ and, being $\vect\Omega\times\hat{\vect a}$ with $\vect\Omega$ vertical and $\hat{\vect a}$ horizontal, is itself horizontal and perpendicular to $AB$ .

Step 3The two contributions to

\vect\beta

are perpendicular, so their moduli combine directly

The two pieces of : β_0, the growth of the vertical piece, and the axial mark at the right, the term of modulus ω_0Ω made by AB's own direction sweeping around with the precession, pointing into the page. — The two pieces of (2): $\vect\beta_0$ , the growth of the vertical piece, and the axial mark at the right, the term of modulus $\omega_0\Omega$ made by $AB$ ’s own direction sweeping around with the precession, pointing *into* the page.

$\vect\beta_0$ (vertical) and $\omega_0\vect\Omega\times\hat{\vect a}$ (horizontal, perpendicular to $AB$ ) are mutually perpendicular at every instant, so the modulus of their sum is the root of the sum of squares. With $AB$ along $+x$ and the vertical along $+y$ as in the first figure, $\vect\Omega\times\hat{\vect a}$ lies along $-z$ , into the plane $\vect\beta_0$ and $\vect\omega_0$ span.

\beta(t) = \sqrt{\beta_0^2+\bigl(\omega_0\Omega(t)\bigr)^2} = \sqrt{\beta_0^2+(\omega_0\beta_0t)^2} = \beta_0\sqrt{1+\omega_0^2t^2}.

\beta = \beta_0\sqrt{1+\omega_0^2t^2} = 0.2\ \text{rad/s}^2

(3)

Check

At $t=0$ , $\Omega=0$ so $\omega(0)=\omega_0=0.50$ rad/s exactly (the body has not begun precessing yet), while $\beta(0)=\beta_0=0.10$ rad/s $^2$ : the precession’s own angular acceleration is present from the first instant even though $\Omega$ itself starts at zero, since $\vect\beta_0=d\vect\Omega/dt$ does not require $\Omega$ to be nonzero. Both formulas reduce correctly. Dimensionally, $\beta_0t/\omega_0$ and $\omega_0t$ are both dimensionless, as required inside the square roots.

Remark

The two answers share one structure, $\omega_0\sqrt{1+(\ldots)^2}$ against $\beta_0\sqrt{1+(\ldots)^2}$ , because the whole problem is one perpendicular vector composition, first for the velocity ( $\vect\omega_0$ against $\vect\Omega$ ) and then for its own rate of change ( $\vect\beta_0$ against $\omega_0\vect\Omega$ ): whenever a spin axis of fixed strength is itself carried around by a second, perpendicular rotation, both the spin’s total rate and the total rate’s own rate of change come out as this same Pythagorean combination.