Irodov 1.44: Radius Cancels, Speed and Time Remain – Amit & Deepali Sinha Foundation

Problem Irodov 1.44

A wheel of radius $R$ turns about a fixed axis, its rotation angle obeying $\varphi=at^2$ with $a=0.20$ rad/s $^2$ . At the instant $t=2.5$ s, a point $A$ on the rim has linear speed $v=0.65$ m/s. Find the magnitude of $A$ ’s total acceleration at that instant.

Point A on the rim of the wheel of radius R, centered at O. With the spin ω growing, A's total acceleration w has a tangential part w_τ along v and a normal part w_n toward O. — Point $A$ on the rim of the wheel of radius $R$ , centered at $O$ . With the spin $\vect\omega$ growing, $A$ ’s total acceleration $\vect w$ has a tangential part $w_\tau$ along $\vect v$ and a normal part $w_n$ toward $O$ .

Given

$\varphi=at^2$
$a=0.20\text{ rad/s}^2$
$v=0.65\text{ m/s at }t=2.5\text{ s}$

Find

w

t=2.5

Solution

Step 1Angular velocity and angular acceleration from

\varphi(t)

Differentiating the rotation law once gives the wheel’s angular velocity, and again gives its angular acceleration.

\begin{aligned} \omega &= \frac{d\varphi}{dt} = 2at \\ \beta &= \frac{d\omega}{dt} = 2a \end{aligned} \tag{1}

Both are exact at every instant; $\beta$ is constant, but $\omega$ grows linearly with $t$ .

Step 2The wheel’s radius, eliminated rather than given

v = \omega R \;\Longrightarrow\; R = \frac{v}{\omega} = \frac{v}{2at} \tag{2}

Step 3The tangential and normal acceleration of point

A

The rim’s tangential acceleration is $\beta R$ and its normal (centripetal) acceleration is $v^2/R$ ; substituting (2) into each removes the unknown radius entirely.

\begin{align*}w_\tau &= \beta R = 2a\cdot\frac{v}{2at} \\ w_n &= \frac{v^2}{R} = v^2\cdot\frac{2at}{v}\end{align*}

w_\tau = \frac{v}{t}

(3)

w_n = 2avt

(4)

Step 4Total acceleration

The two parts are perpendicular, so $w$ is their root sum of squares; substituting (3) and (4) and factoring out $v/t$ leaves a compact closed form.

\begin{align*}w &= \sqrt{w_\tau^2+w_n^2} = \sqrt{\left(\frac{v}{t}\right)^2+(2avt)^2} \\ &= \frac{v}{t}\sqrt{1+(2at^2)^2}\end{align*}

w = \frac{v}{t}\sqrt{1+4a^2t^4} = 0.70\text{ m/s}^2

(5)

Check

Dimensionally $[v/t]=\text{m/s}^2$ , an acceleration, and $at^2$ is a pure angle (rad/s $^2$ times s $^2$ ), so $2at^2$ is dimensionless, exactly what the sum under the square root needs. As a further check on the closed form itself, its small- $t$ behavior is sensible: the correction term $(2at^2)^2$ becomes negligible beside the leading $1$ as $t\to0$ , so $w\to v/t$ , meaning the tangential piece would dominate at early times, consistent with $w_\tau$ staying fixed while $w_n$ grows only quadratically in $t$ (Step 3).

Remark

The wheel’s actual radius never had to be found: it enters $w_\tau=\beta R$ and $w_n=v^2/R$ in a way that cancels completely once both are rewritten through $v$ and $\omega$ alone. Whenever a rim quantity and an angular quantity are both given for the same point, expect $R$ to be eliminable rather than needed, since $v=\omega R$ is exactly the bridge between the two descriptions.