Irodov 1.56: Two Vectors, Seventeen Degrees Apart – Amit & Deepali Sinha Foundation

Problem Irodov 1.56

A rigid body rotates with an angular velocity vector $\vect\omega(t) = at\,\vect i + bt^2\,\vect j$ , where $a = 0.50$ rad/s $^2$ , $b = 0.060$ rad/s $^3$ , and $\vect i$ , $\vect j$ are the unit vectors along the $x$ and $y$ axes. Find:

the magnitudes of the angular velocity and the angular acceleration at the instant $t = 10.0$ s
the angle between the angular velocity vector and the angular acceleration vector at that same instant

The angular velocity at a general instant, drawn at an arbitrary pose, with the two components the statement gives: at along the x axis and bt^2 along the y axis. Both grow with time, at different rates. — The angular velocity at a general instant, drawn at an arbitrary pose, with the two components the statement gives: $at$ along the $x$ axis and $bt^2$ along the $y$ axis. Both grow with time, at different rates.

Given

$\vect\omega(t) = at\,\vect{i}+bt^2\vect{j}$
$a=0.50$ rad/s $^2$
$b=0.060$ rad/s $^3$
$t=10.0$ s

Find

(a)

\omega

and

\beta=|\dot{\vect\omega}|

t

, (b) the angle between

\vect\omega

and

\vect\beta

Solution

Step 1Read off the angular velocity components

At the given instant $\vect\omega$ has components $\omega_x=at$ , $\omega_y=bt^2$ , so its modulus is the root sum of their squares,

\omega = \sqrt{\omega_x^2+\omega_y^2} = \sqrt{a^2t^2+b^2t^4}.

Factoring $a^2t^2$ out from under the root,

\omega = at\sqrt{1+(bt/a)^2}

(1)

Step 2Differentiate for the angular acceleration

The angular acceleration is $\vect\beta=d\vect\omega/dt$ , with components $\beta_x=a$ (a genuine constant) and $\beta_y=2bt$ (which still grows with $t$ ), so its modulus is

\beta = \sqrt{\beta_x^2+\beta_y^2} = \sqrt{a^2+4b^2t^2}.

Factoring $a^2$ out from under the root,

\beta = a\sqrt{1+(2bt/a)^2}

(2)

Step 3Evaluate both at

t=10.0

\omega_x = at = 5.0\text{ rad/s},\qquad \omega_y = bt^2 = 6.0\text{ rad/s}, \tag{3}

\beta_x = a = 0.50\text{ rad/s}^2,\qquad \beta_y = 2bt = 1.2\text{ rad/s}^2. \tag{4}

\begin{gathered}\omega = \sqrt{5.0^2+6.0^2}\text{ rad/s} \approx 8\text{ rad/s}\\ \beta = \sqrt{0.50^2+1.2^2}\text{ rad/s}^2 = 1.3\text{ rad/s}^2\end{gathered}

(5)

Step 4The angle between

\vect\omega

and

\vect\beta

\cos\theta = \frac{\vect\omega\cdot\vect\beta}{\omega\beta} = \frac{\omega_x\beta_x+\omega_y\beta_y}{\omega\beta} = \frac{5.0\times0.50+6.0\times1.2}{7.81\times1.3}. \tag{6}

\theta = \arccos\dfrac{\omega_x\beta_x+\omega_y\beta_y}{\omega\beta} \approx 17^\circ

(7)

The two vectors at t=10.0 s, drawn at the directions their components fix (though not at their true relative lengths): ω at (ω_y/ω_x)50.2° from the x axis and β at (β_y/β_x)67.4°. The wedge between them is the angle θ boxed above. — The two vectors at $t=10.0$ s, drawn at the directions their components fix (though not at their true relative lengths): $\vect\omega$ at $\arctan(\omega_y/\omega_x)\approx50.2^\circ$ from the $x$ axis and $\vect\beta$ at $\arctan(\beta_y/\beta_x)\approx67.4^\circ$ . The wedge between them is the angle $\theta$ boxed above.

Check

The two direction angles found straight from the components, $\arctan(6.0/5.0)\approx50.2^\circ$ for $\vect\omega$ and $\arctan(1.2/0.50)\approx67.4^\circ$ for $\vect\beta$ , differ by $17.2^\circ$ , matching the dot-product answer independently: the figure’s angle $\theta$ was drawn at these two directions, not guessed.

Remark

Because $\vect\omega(t)$ ’s own direction drifts, $\arctan(bt/a)$ growing with $t$ , this is rotation about an axis that sweeps rather than one fixed in space; $\vect\beta$ is not simply describing how fast the body spins up but also how fast that axis is turning, which is precisely why $\vect\beta$ points off at a different angle from $\vect\omega$ rather than lying along it as it would for spin-up about a fixed axis.