Irodov 1.128: Centrifugal Work Only Reads the Radius

Problem Irodov 1.128

In a reference frame rotating about a fixed axis with constant angular velocity $\omega$ , a small body of mass $m$ moves from a point $1$ to a point $2$ along some arbitrary path in that frame, where point 1 is a distance $r_1$ from the rotation axis and point 2 is a distance $r_2$ from it. Find the work done by the centrifugal force of inertia during this transfer, given $m=0.10$ kg, $\omega=5.0$ rad/s, $r_1=30$ cm and $r_2=50$ cm.

A path in the rotating frame from point 1 (radius r_1) to point 2 (radius r_2), about the axis O turning at ω. The outward centrifugal force is drawn at each end; the path dips inside r_1 at P. — A path in the rotating frame from point $1$ (radius $r_1$ ) to point $2$ (radius $r_2$ ), about the axis $O$ turning at $\omega$ . The outward centrifugal force is drawn at each end; the path dips inside $r_1$ at $P$ .

Given

$m$
$\omega$
$r_1$
$r_2$

Find

A_{cf}

, the work done by the centrifugal force from point 1 to point 2

Solution

Step 1The centrifugal force as a function of radius

In the rotating frame, the centrifugal force of inertia on the body at distance $r$ from the rotation axis points radially outward from that axis with magnitude proportional to $r$ .

\vect F_{cf}(r) = m\omega^2 r\,\hat{\vect r}

(1)

Step 2Work as a path integral, reduced to a radial one

The work done as the body moves along an element $d\vect r$ of an arbitrary path is $\vect F_{cf}\cdot d\vect r$ using (1); since $\vect F_{cf}$ is always directed along $\hat{\vect r}$ , only the radial component of the displacement, $dr$ , ever contributes, so the line integral collapses to an ordinary integral in $r$ alone from $r_1$ to $r_2$ .

A_{cf} = \int_1^2 \vect F_{cf}\cdot d\vect r = \int_{r_1}^{r_2} m\omega^2 r\,dr

(2)

Step 3Evaluating the integral

Carrying out the integration in (2) gives the work in terms of the endpoint radii alone.

A_{cf} = \left.\frac{m\omega^2 r^2}{2}\right|_{r_1}^{r_2}

A_{cf} = \dfrac{m\omega^2}{2}\bigl(r_2^2-r_1^2\bigr) = 0.20\ \mathrm J

(3)

Check

Dimensionally, $[m][\omega^2][r^2] = (\mathrm{kg})(\mathrm{s^{-2}})(\mathrm{m^2}) = \mathrm J$ , as required. Numerically, $0.5\times0.10\times5.0^2\times(0.50^2-0.30^2) = 0.5\times0.10\times25\times0.16 = 0.20$ J, matching the boxed formula. As $r_2\to r_1$ the work correctly vanishes, whatever path connects two points at the same radius.

Remark

Step 2 is the same argument that makes gravity’s work path independent: a force directed radially outward from a fixed axis (not a point), with magnitude a function of the axial distance $r$ alone, is still conservative, because its work along any path depends only on how $r$ changes and never on the displacement along the axis or around it, exactly as for a true central force, with potential $U(r) = -\int F(r)\,dr$ . Here $U_{cf}(r) = -m\omega^2r^2/2$ , the familiar centrifugal potential, and $A_{cf}=U_{cf}(r_1)-U_{cf}(r_2)$ reproduces the boxed result without ever integrating along the actual path.