Irodov 1.126: Constant Push Behind a Quadratic Pull – Amit & Deepali Sinha Foundation

Problem Irodov 1.126

A particle of mass $m$ moves along a circular path of radius $R$ , starting from rest, with its normal (centripetal) acceleration growing as $w_n=at^2$ , where $a$ is a given constant. Find the power $P(t)$ developed by the combined action of all the forces on the particle as a function of time, and the mean power $\langle P\rangle$ of that same combination averaged over the first $t$ seconds of the motion.

The particle of mass m on the circle of radius R, with the resultant force F resolved into normal F_n and tangential F_τ parts; φ is its angle with the velocity v. — The particle of mass $m$ on the circle of radius $R$ , with the resultant force $\vect F$ resolved into normal $F_n$ and tangential $F_\tau$ parts; $\varphi$ is its angle with the velocity $\vect v$ .

Given

$m$
$R$
$w_n=at^2$

Find

P(t)

\langle P\rangle

over the first

t

seconds

Solution

Step 1Speed from the normal acceleration

The normal acceleration of circular motion is $w_n=v^2/R$ ; equating this to the given $w_n=at^2$ and taking the positive root (the particle starts from rest, so $v\ge0$ ) gives $v(t)$ .

\frac{v^2}{R} = at^2 \;\Longrightarrow\; v^2 = aRt^2

v(t) = \sqrt{aR}\,t

(1)

Step 2Tangential force from the growth of speed

The tangential acceleration is the time derivative of (1), a constant; Newton’s second law along the tangent then gives the tangential force $F_\tau$ , the only component of the resultant that can do work.

w_\tau = \frac{dv}{dt} = \sqrt{aR}

F_\tau = m\sqrt{aR}

(2)

Step 3Instantaneous power

Only the tangential force does work, since the normal force is always perpendicular to $\vect v$ , so the power is $F_\tau$ (2) times the speed (1).

P(t) = F_\tau\,v(t) = m\sqrt{aR}\cdot\sqrt{aR}\,t

P(t) = maRt

(3)

Step 4Mean power over the first

t

seconds

The power (3) is linear in time and starts from zero, so averaging it over $[0,t]$ integrates a straight line through the origin, which halves its endpoint value.

\langle P\rangle = \frac1t\int_0^t maR\tau\,d\tau = \frac{maR}{t}\left[\frac{\tau^2}{2}\right]_0^t

\langle P\rangle = \dfrac{maRt}{2}

(4)

Check

Dimensionally, $w_n=at^2$ forces $[a]=\mathrm{m/s^4}$ , so $[m][a][R][t]=\mathrm{kg}\cdot(\mathrm{m/s^4})\cdot\mathrm m\cdot\mathrm s=\mathrm{kg\,m^2/s^3}=\mathrm W$ , a power, as required. At $t=0$ the particle is at rest and both $P(0)=0$ and $\langle P\rangle\to0$ as $t\to0$ , consistent with no work yet done; and since $P(t)$ is linear starting from the origin, its average over $[0,t]$ is exactly half its value at $t$ , matching $\langle P\rangle=P(t)/2$ read off the two boxed results.

Remark

The normal acceleration was given growing as $t^2$ , which looks like the more complicated ingredient, yet it never entered $F_\tau$ or $P(t)$ at all: $F_n$ only feeds the wasted, work-free part of the resultant force, so however fast $w_n$ grows, the entire power comes from $F_\tau$ alone, which here happens to be constant because $w_n=at^2$ was exactly the profile that makes $v$ linear in $t$ . A different growth law for $w_n$ would still leave the normal force powerless, but $F_\tau$ and $P(t)$ would no longer be so simple.