Irodov 1.86: Where Tangential Meets Centripetal in Magnitude

Problem Irodov 1.86

A ball on a thread of length $l$ swings back and forth in a vertical plane, momentarily at rest at each extreme of the swing, deflected by angle $\theta$ from the vertical there. Find $\theta$ given that the ball’s acceleration has the same modulus at the extreme position as it does at the lowest point of the swing.

The ball at an extreme of its swing, deflected by θ from the vertical and momentarily at rest. The gray arc is the path down to the lowest point, the second position the problem compares. The drawn deflection is illustrative. — The ball at an extreme of its swing, deflected by $\theta$ from the vertical and momentarily at rest. The gray arc is the path down to the lowest point, the second position the problem compares. The drawn deflection is illustrative.

Given

swings between $\pm\theta$ , momentarily at rest at each extreme

Find

the angle

\theta

Solution

Step 1Acceleration at the extreme: purely tangential

The same extreme position with the acceleration this step deduces there. Because the ball has no speed at an extreme, w carries no centripetal part at all and lies along the arc, pointing toward the lowest point. — The same extreme position with the acceleration this step deduces there. Because the ball has no speed at an extreme, $\vect w$ carries no centripetal part at all and lies along the arc, pointing toward the lowest point.

At an extreme the ball is momentarily at rest, so the centripetal term $v^2/l$ vanishes and only gravity’s tangential component remains.

w_{\mathrm{ext}} = g\sin\theta

Step 2Acceleration at the bottom: purely centripetal

At the lowest point the thread is vertical, so gravity is entirely radial and contributes no tangential acceleration there; only the centripetal term survives. Energy conservation from the extreme (height $l\cos\theta$ below the pivot) to the bottom (height $l$ below the pivot) gives the speed there.

\begin{align*}mgl(1-\cos\theta) &= \frac12 mv_{\mathrm{bot}}^2 \\ v_{\mathrm{bot}}^2 &= 2gl(1-\cos\theta)\end{align*}

w_{\mathrm{bot}} = \frac{v_{\mathrm{bot}}^2}{l} = 2g(1-\cos\theta)

Step 3Equate the two and solve for

\theta

Setting $w_{\mathrm{ext}}=w_{\mathrm{bot}}$ and canceling the common factor $g$ leaves one equation in $\theta$ alone, with $l$ already gone from both sides.

\sin\theta = 2(1-\cos\theta)

Writing $t=\tan(\theta/2)$ and using the standard half-angle identities turns this into a polynomial in $t$ .

\sin\theta = \frac{2t}{1+t^2}, \qquad \cos\theta = \frac{1-t^2}{1+t^2}

\begin{align*}\frac{2t}{1+t^2} &= 2\left(1-\frac{1-t^2}{1+t^2}\right) \\ 2t &= 2(1+t^2)-2(1-t^2) \\ 2t &= 4t^2\end{align*}

The root $t=0$ is the trivial equilibrium; the swinging solution is $t=1/2$ .

\tan\dfrac{\theta}{2} = \dfrac12

(1)

\theta = 2\arctan\dfrac12 \approx 53.1^\circ

(2)

Check

At $\tan(\theta/2)=1/2$ , the half-angle identities give $\sin\theta=2(1/2)/(1+1/4)=0.8$ and $\cos\theta=(1-1/4)/(1+1/4)=0.6$ , a $3$ - $4$ - $5$ triangle in disguise. Substituting back: $\sin\theta=0.8$ and $2(1-\cos\theta)=2(0.4)=0.8$ , so the two accelerations do agree.

Remark

Neither $g$ nor $l$ survives into this particular answer, but that is a fact about comparing two quantities of the same kind, not a general property of pendulum swings: an acceleration here always carries a bare factor of $g$ (as in $w_{\mathrm{ext}}=g\sin\theta$ above, or 1.85’s $w=g\sqrt{1+3\cos^2\theta}$ ), so equating two accelerations cancels that common $g$ , and $l$ had already left $w_{\mathrm{bot}}$ on its own once $v_{\mathrm{bot}}^2=2gl(1-\cos\theta)$ was substituted. A tension, by contrast, carries a bare factor of $mg$ (1.85’s $T=3mg\cos\theta$ ), so it does not cancel against an acceleration or drop out unless two tensions are compared against each other.