Irodov 1.47: When Acceleration Tips Away from Velocity

Problem Irodov 1.47

A solid body begins rotating from rest about a fixed axis, with angular acceleration growing in time as $\beta(t) = at$ , where $a = 2.0\times10^{-2}$ rad/s $^3$ . Consider any point of the body at some fixed distance from the axis; that point’s total (linear) acceleration vector makes some angle with its velocity vector, an angle that changes as the body spins up. Find the time after the start of rotation at which this angle equals $\alpha = 60^\circ$ .

A point P at radius r on the rotating body. Its velocity v is tangential; its total acceleration w is the resultant of the tangential w_τ and the centripetal w_n, and makes the angle α with v. — A point $P$ at radius $r$ on the rotating body. Its velocity $\vect v$ is tangential; its total acceleration $\vect w$ is the resultant of the tangential $w_\tau$ and the centripetal $w_n$ , and makes the angle $\alpha$ with $\vect v$ .

Given

$\beta(t) = at$
$a = 2.0\times10^{-2}$ rad/s $^3$
$\omega(0)=0$

Find

t

such that

\alpha=60^\circ

Solution

Step 1Angular velocity from the growing angular acceleration

Integrating the given $\beta(t)=at$ from rest gives the angular velocity at any later time.

\omega(t) = \int_0^t a t'\,dt' = \frac{at^2}{2}

Step 2The two acceleration components at radius

r

For a point at radius $r$ , the tangential component follows the angular acceleration and the normal (centripetal) component follows the square of the angular velocity found in Step 1.

\begin{align*}w_\tau &= r\beta = rat \\ w_n &= r\omega^2 = r\left(\frac{at^2}{2}\right)^2 = \frac{ra^2t^4}{4}\end{align*}

Step 3The angle between the total acceleration and the velocity

The velocity is purely tangential, so the angle $\alpha$ between the total acceleration $\vect w = w_\tau\vect\tau + w_n(-\hat{\vect r})$ and $\vect v$ equals the angle the total acceleration makes with the tangential direction, whose tangent is the ratio of the normal to the tangential component.

\tan\alpha = \frac{w_n}{w_\tau} = \frac{ra^2t^4/4}{rat}

The radius $r$ cancels, leaving

\tan\alpha = \frac{at^3}{4} \tag{1}

Step 4Solving for the time

Rearranging (1) for $t$ gives the time at which the acceleration reaches the stated angle.

t^3 = \frac{4\tan\alpha}{a}

t = \sqrt[3]{\dfrac{4\tan\alpha}{a}} = 7.0 \text{ s}

(2)

Check

Dimensionally $\tan\alpha$ is pure number and $a$ has units rad/s $^3$ , so $4\tan\alpha/a$ has units s $^3$ and its cube root is a time, correct. Numerically, $4\tan60^\circ/a = 4(1.732)/(2.0\times10^{-2}) = 346$ , and $346^{1/3}\approx7.0$ s, matching the boxed value. The radius $r$ never entered the final answer, consistent with the problem asking about “an arbitrary point.”

Remark

Relation (1) grows without bound as $t\to\infty$ , so $\alpha\to90^\circ$ eventually: the longer a body spins up under a steadily increasing angular acceleration, the more its points’ accelerations tip away from the direction of motion and toward the center, because $\omega$ (and hence $w_n\propto\omega^2$ ) keeps climbing long after $\beta$ has grown only linearly. A body spun up this way can never keep its acceleration close to the direction of travel indefinitely.