Irodov 1.46: Averages of a Cubic Rotation Law – Amit & Deepali Sinha Foundation

Problem Irodov 1.46

A solid body turns about a fixed axis with rotation angle $\varphi(t) = at - bt^3$ , where $a = 6.0$ rad/s and $b = 2.0$ rad/s $^3$ are constants and $t$ is measured from $t=0$ . The body comes momentarily to rest at some later instant. Find

the mean angular velocity and the mean angular acceleration, both averaged over the interval from $t=0$ until the body stops
the angular acceleration at the instant the body stops

A solid body turning about a stationary axis. The angular velocity ω curls in the positive sense of the rotation angle φ, which is the sense in which φ=at-bt^3 is measured. — A solid body turning about a stationary axis. The angular velocity $\vect\omega$ curls in the positive sense of the rotation angle $\varphi$ , which is the sense in which $\varphi=at-bt^3$ is measured.

Given

$\varphi(t) = at - bt^3$
$a = 6.0$ rad/s
$b = 2.0$ rad/s $^3$

Find

(a)

\langle\omega\rangle

\langle\beta\rangle

over

[0,t_1]

, (b)

\beta

at the stop

t_1

Solution

Step 1Angular velocity and acceleration from the rotation law

Differentiating the given $\varphi(t)$ once gives the angular velocity and a second time gives the angular acceleration.

\begin{aligned} \omega(t) &= \frac{d\varphi}{dt} = a - 3bt^2 \\ \beta(t) &= \frac{d\omega}{dt} = -6bt \end{aligned} \tag{1}

The same body with the angular acceleration this step produces. For every t>0, β=-6bt is negative while the body is still turning forward, so β curls opposite to ω: the rotation is slowing. — The same body with the angular acceleration this step produces. For every $t>0$ , $\beta=-6bt$ is negative while the body is still turning forward, so $\vect\beta$ curls opposite to $\vect\omega$ : the rotation is slowing.

Step 2The stopping instant

The body stops when $\omega=0$ ; solving the $\omega(t)$ of (1) for the positive root gives that instant $t_1$ .

0 = a - 3bt_1^2 \;\Longrightarrow\; t_1^2 = \frac{a}{3b} \tag{2}

t_1 = \sqrt{\dfrac{a}{3b}}

(3)

Step 3Mean angular velocity over

[0,t_1]

The angle swept by $t_1$ follows from $\varphi(t)$ , rewriting $t_1^3$ as $t_1\cdot t_1^2$ and substituting (2).

\begin{align*}\varphi(t_1) &= at_1 - bt_1^3 \\ &= at_1 - b\,t_1\cdot t_1^2 \\ &= at_1 - bt_1\cdot\frac{a}{3b} \\ &= at_1 - \frac{a}{3}t_1 \\ &= \frac{2a}{3}t_1\end{align*}

Dividing by $t_1$ gives the mean angular velocity.

\langle\omega\rangle = \frac{\varphi(t_1)}{t_1} = \frac{2a}{3}

\langle\omega\rangle = \dfrac{2a}{3} = 4.0 \text{ rad/s}

(4)

Step 4Mean angular acceleration over

[0,t_1]

The mean angular acceleration is the total change in $\omega$ divided by the elapsed time; $\omega$ falls from $a$ at $t=0$ to $0$ at $t_1$ , and substituting (3) for $t_1$ gives the closed form.

\begin{align*}\langle\beta\rangle &= \frac{\omega(t_1)-\omega(0)}{t_1} = \frac{0-a}{t_1} \\ &= -\frac{a}{\sqrt{a/(3b)}} \\ &= -\sqrt{3ab}\end{align*}

\langle\beta\rangle = -\sqrt{3ab} = -6.0 \text{ rad/s}^2

(5)

Step 5Angular acceleration at the stop

The $\beta(t)$ of (1) evaluated at $t_1$ (3) gives the instantaneous deceleration at the moment the body stops.

\begin{align*}\beta(t_1) &= -6bt_1 \\ &= -6b\sqrt{\frac{a}{3b}}\end{align*}

\beta(t_1) = -2\sqrt{3ab} = -12 \text{ rad/s}^2

(6)

The book’s own key quotes $\langle\beta\rangle$ and $\beta(t_1)$ as the magnitudes $\sqrt{3ab}$ and $2\sqrt{3ab}$ ; the minus signs above are the same numbers, carrying the direction of the deceleration relative to the positive sense chosen for $\omega$ , not a disagreement with the corpus.

Check

Dimensionally $[a][b]=(\text{rad/s})(\text{rad/s}^3)=\text{rad}^2/\text{s}^4$ , so $\sqrt{3ab}$ has units rad/s $^2$ , correct for an angular acceleration; and both $\langle\beta\rangle$ and $\beta(t_1)$ come out negative, consistent with a rotation that is slowing down, $\omega$ falling from $a>0$ toward $0$ . Numerically $t_1=\sqrt{6.0/6.0}=1.0$ s, and $\beta(t_1)=-6(2.0)(1.0)=-12$ rad/s $^2$ , matching the boxed result.

Remark

The $\beta(t)=-6bt$ of (1) is linear in $t$ , and a linear function’s time average over any interval always equals the mean of its endpoint values, $(\beta(0)+\beta(t_1))/2 = (0+(-12))/2 = -6$ rad/s $^2$ , exactly $\langle\beta\rangle$ . That endpoint-mean property is not exclusive to linear functions, but it is what makes $\langle\beta\rangle$ so easy to read off here. $\omega(t)$ , by contrast, is quadratic in $t$ and shares no such symmetry about the midpoint of $[0,t_1]$ , so the same shortcut fails for it: the endpoint mean $(\omega(0)+\omega(t_1))/2 = (6.0+0)/2 = 3.0$ rad/s disagrees with the true $\langle\omega\rangle = 4.0$ rad/s found in Step 3.