Irodov 1.185: Torque Stays Fixed While Momentum Turns

Problem Irodov 1.185

A particle’s angular momentum about a fixed point $O$ is given at every time $t\geq0$ by $\vect M(t) = \vect a + \vect b\,t^2$ , where $\vect a$ and $\vect b$ are constant vectors with $\vect a$ perpendicular to $\vect b$ . Find the moment of force (torque) $\vect N$ about $O$ at the particular instant when the angle between $\vect N$ and $\vect M$ is $45^\circ$ .

The two constant vectors of the statement, a and b, fixed in space and at right angles to each other, both drawn from the point O about which the angular momentum is taken. — The two constant vectors of the statement, $\vect a$ and $\vect b$ , fixed in space and at right angles to each other, both drawn from the point $O$ about which the angular momentum is taken.

Given

$\vect a$
$\vect b$
$\vect a\perp\vect b$
angle $(\vect N,\vect M)=45^\circ$

Find

\vect N

Solution

Step 1Torque is the rate of change of angular momentum

By definition the torque about $O$ is the time derivative of the angular momentum about $O$ ; since $\vect a$ is constant, only the second term contributes.

\begin{align*}\vect N = \frac{d\vect M}{dt} &= \frac{d\vect a}{dt} + \frac{d(\vect b\,t^2)}{dt} \\ &= 0 + 2\vect b\,t\end{align*}

\vect N = 2\vect b\,t

(1)

Step 2Resolving

\vect M

along and across the torque’s direction

Because $\vect a\perp\vect b$ , the vector $\vect a$ is entirely perpendicular to $\vect N$ ’s direction while $\vect b\,t^2$ is entirely parallel to it, so the angle $\theta$ between $\vect M$ and $\vect N$ satisfies

\tan\theta = \frac{|\vect a|}{|\vect b\,t^2|} = \frac{a}{bt^2}. \tag{2}

Step 3The instant at which

\theta=45^\circ

The two components of M, the fixed a and the growing bt^2, are equal here, so M makes 45° with each. N is drawn offset below the axis it shares with bt^2, joined by the dashed leader. — The two components of $\vect M$ , the fixed $\vect a$ and the growing $bt^2$ , are equal here, so $\vect M$ makes $45^\circ$ with each. $\vect N$ is drawn offset below the axis it shares with $bt^2$ , joined by the dashed leader.

Setting $\theta=45^\circ$ in (2), so $\tan\theta=1$ , and solving for $t$ gives

\begin{align*}1 &= \frac{a}{bt^2} \\ t^2 &= \frac{a}{b}\end{align*}

t = \sqrt{\dfrac{a}{b}}

(3)

Step 4The torque at that instant

Substituting (3) into (1) gives $\vect N$ at the moment the angle reaches $45^\circ$ . It keeps the constant direction of $\vect b$ , as (1) already showed it must, and only its size has been fixed by the condition.

\vect N = 2\vect b\sqrt{\dfrac{a}{b}}

(4)

Check

Dimensionally, the boxed vector has magnitude $2b\sqrt{a/b}=2\sqrt{ab}$ , and since $[a]$ is angular momentum ( $\mathrm{kg\,m^2/s}$ ) and $[b]=[a]/\mathrm{s^2}$ , the product $[a][b]=(\mathrm{kg\,m^2/s^2})^2$ , whose square root is a torque, as required. Also, $\tan\theta=a/(bt^2)$ falls monotonically from $+\infty$ at $t=0$ to $0$ as $t\to\infty$ , so it equals $1$ at exactly one time, matching the problem’s request for “the” instant rather than a family of instants.

Remark

Although $\vect N$ never changes direction, always pointing along the fixed $\vect b$ , the angular momentum $\vect M$ continuously swings from the direction of $\vect a$ at $t=0$ toward the direction of $\vect b$ as $t\to\infty$ : a torque of unchanging direction can still steer the angular momentum it produces through a full quarter turn, because $\vect M$ carries the memory of its initial value $\vect a$ even as $\vect N$ keeps adding momentum along a single fixed axis.