Irodov 1.105: The Hodograph Is a Circle – Amit & Deepali Sinha Foundation

Problem Irodov 1.105

A particle of mass $m$ , at rest at $t=0$ , is acted on by a force $\vect F$ of constant magnitude $F$ whose direction rotates within a fixed plane at a constant angular rate $\omega$ . Find:

the particle’s speed $v(t)$
the distance $\Delta s$ the particle covers between two successive instants at which it is momentarily at rest, and its mean speed $\langle v\rangle$ over that same interval

The force F at a general instant, on axes chosen so F points along x at t=0. Having turned through the angle ω t, it has components Fω t along x and Fω t along y. — The force $\vect F$ at a general instant, on axes chosen so $\vect F$ points along $x$ at $t=0$ . Having turned through the angle $\omega t$ , it has components $F\cos\omega t$ along $x$ and $F\sin\omega t$ along $y$ .

Given

$F$
$\omega$
$v(0)=0$

Find

(a)

v(t)

, (b)

\Delta s

and

\langle v\rangle

between successive stops

Solution

Step 1Velocity components from Newton’s second law

Take axes with $\vect F$ along $x$ at $t=0$ , so at a general time $\vect F$ has turned through $\omega t$ and has the components shown in the figure. Newton’s second law integrates directly, component by component, since the particle starts from rest.

\begin{align*}v_x(t) &= \int_0^t \frac{F}{m}\cos\omega t'\,dt' = \frac{F}{m\omega}\sin\omega t \\ v_y(t) &= \int_0^t \frac{F}{m}\sin\omega t'\,dt' = \frac{F}{m\omega}\bigl(1-\cos\omega t\bigr)\end{align*}

Step 2The speed as a function of time

Forming $v^2=v_x^2+v_y^2$ and applying the half-angle identity for $1-\cos\omega t$ turns the sum into a perfect square.

\begin{align*}v^2 &= \left(\frac{F}{m\omega}\right)^2\Bigl[\sin^2\omega t+(1-\cos\omega t)^2\Bigr] \\ &= \left(\frac{F}{m\omega}\right)^2\cdot 2(1-\cos\omega t) \\ &= \left(\frac{2F}{m\omega}\right)^2\sin^2\frac{\omega t}{2}\end{align*}

v(t) = \dfrac{2F}{m\omega}\left|\sin\dfrac{\omega t}{2}\right|

(1)

Step 3The instants of rest

The particle is momentarily at rest whenever the boxed speed vanishes.

\frac{\omega t}{2}=k\pi \quad (k=0,1,2,\dots) \;\Longrightarrow\; t_k=\frac{2k\pi}{\omega}

Consecutive stops $t_k$ and $t_{k+1}$ are therefore separated by a fixed interval

T = \dfrac{2\pi}{\omega}

(2)

Step 4Distance covered between two successive stops

On one interval of length $T$ the argument $\omega t/2$ runs from $0$ to $\pi$ , where $\sin(\omega t/2)\ge0$ , so the absolute value drops and the distance is the plain integral of the boxed speed.

\begin{align*}\Delta s &= \int_0^{T} \frac{2F}{m\omega}\sin\frac{\omega t}{2}\,dt \\ &= \frac{2F}{m\omega}\cdot\frac{2}{\omega}\left[-\cos\frac{\omega t}{2}\right]_0^{T} \\ &= \frac{4F}{m\omega^2}\Bigl[1-\cos\frac{\omega T}{2}\Bigr]\end{align*}

Substituting $\omega T/2=\pi$ from (2), $\cos\pi=-1$ , so the bracket equals $2$ .

\Delta s = \dfrac{8F}{m\omega^2}

(3)

Step 5Mean speed over the same interval

The mean speed is this distance divided by the elapsed time $T$ from (2).

\langle v\rangle = \frac{\Delta s}{T} = \frac{8F/m\omega^2}{2\pi/\omega}

\langle v\rangle = \dfrac{4F}{\pi m\omega}

(4)

Check

Dimensionally $F/m\omega$ is a speed, and every boxed result is built from that combination alone. Part (a)’s peak speed, reached at $\omega t=\pi$ , is $2F/m\omega$ ; the mean speed from the last box is $\langle v\rangle=4F/(\pi m\omega)\approx1.27\,F/(m\omega)$ , numerically below that peak of $2F/(m\omega)$ , as it must be, since a nonnegative quantity that is not constant always averages below its own maximum.

Remark

The hodograph, the curve traced by the tip of $\vect v(t)$ , is exactly a circle: $v_x^2+\bigl(v_y-F/m\omega\bigr)^2=(F/m\omega)^2$ holds at every $t$ , a circle of radius $F/m\omega$ passing through the origin, since $\vect v=\vect 0$ at $t=0$ . A uniformly rotating force of constant magnitude always traces a circular hodograph, for any $F$ , $\omega$ , mass, or initial velocity; that shape is general. What is not general is where the circle sits: its radius is $F/(m\omega)$ , which scales with the force’s magnitude and inversely with its rotation rate, and it passes through the origin here only because the particle starts from rest; a nonzero $v_0$ would translate the same circle away from the origin by $\vect v_0$ , leaving its radius unchanged.