Irodov 1.85: Two Parts Share One Critical Angle – Amit & Deepali Sinha Foundation

Problem Irodov 1.85

A small sphere of mass $m$ hangs from a thread of length $l$ from a fixed pivot. It is pulled aside until the thread is horizontal (a $90^\circ$ deflection from the vertical) and released from rest, so it swings down under gravity alone. Let $\theta$ be the thread’s angle from the vertical at any later instant. Find

the sphere’s total acceleration and the thread’s tension, each as a function of $\theta$ ;
the tension at the instant the vertical component of the sphere’s velocity is largest;
the angle $\theta$ at which the sphere’s total acceleration vector points horizontally.

The sphere at a general angle θ from the vertical, with the tension T, the weight m g, and the arc still ahead of it down to the lowest point of the swing. The drawn deflection is illustrative. — The sphere at a general angle $\theta$ from the vertical, with the tension $\vect T$ , the weight $m\vect g$ , and the arc still ahead of it down to the lowest point of the swing. The drawn deflection is illustrative.

Given

$m$
released from rest with the thread horizontal

Find

(a)

w(\theta)

and

T(\theta)

; (b)

T

when the vertical velocity is greatest; (c)

\theta

where

\vect w

is horizontal

Solution

Step 1Speed at angle

\theta

from energy conservation

Measured from the pivot, the sphere’s depth below it is $l\cos\theta$ ; at release ( $\theta=90^\circ$ ) that depth is zero, so falling to angle $\theta$ drops it a height $l\cos\theta$ . No friction acts at the pivot, so mechanical energy is conserved from release.

\frac12 mv^2 = mg\,l\cos\theta

v^2 = 2gl\cos\theta

(1)

Step 2Tangential and normal acceleration, and the total (part a)

The tangential acceleration is gravity’s component along the direction of travel, $w_\tau=g\sin\theta$ ; the normal (centripetal) acceleration is $v^2/l$ , and (1) removes $l$ from it entirely.

\begin{align*}w_\tau &= g\sin\theta \\ w_n &= \frac{v^2}{l} = 2g\cos\theta\end{align*}

The two are perpendicular, so the total acceleration’s modulus is their Pythagorean sum.

w = \sqrt{w_\tau^2+w_n^2} = \sqrt{g^2\sin^2\theta+4g^2\cos^2\theta}

w = g\sqrt{1+3\cos^2\theta}

(2)

Step 3Tension from Newton’s second law along the thread (part a)

Along the thread, toward the pivot, the tension and the radial component of gravity combine to supply the centripetal acceleration $v^2/l$ .

T - mg\cos\theta = \frac{mv^2}{l}

Substituting (1) for $v^2/l=2g\cos\theta$ ,

T = mg\cos\theta + 2mg\cos\theta.

T = 3mg\cos\theta

(3)

Step 4The angle where the total acceleration is horizontal (part c)

With the pivot at the origin and $y$ measured upward, the sphere sits at $(l\sin\theta,-l\cos\theta)$ , so the outward radial direction is $\hat r=(\sin\theta,-\cos\theta)$ and the direction of travel is $\hat t=(-\cos\theta,-\sin\theta)$ (it reduces to straight down at release, $\theta=90^\circ$ , as it must). The acceleration is $w_\tau\hat t$ plus $w_n$ toward the pivot, i.e. along $-\hat r$ .

\begin{align*}\vect w &= w_\tau\hat t - w_n\hat r \\ &= g\sin\theta(-\cos\theta,-\sin\theta) - 2g\cos\theta(\sin\theta,-\cos\theta) \\ &= \bigl(-3g\sin\theta\cos\theta,\ g(2\cos^2\theta-\sin^2\theta)\bigr)\end{align*}

$\vect w$ is horizontal exactly when its vertical component vanishes.

2\cos^2\theta - \sin^2\theta = 2\cos^2\theta-(1-\cos^2\theta) = 3\cos^2\theta - 1 = 0

\cos\theta = \dfrac{1}{\sqrt3}

(4)

\theta = \arccos\dfrac{1}{\sqrt3} \approx 54.7^\circ

(5)

The acceleration at the angle this step finds, decomposed into its tangential part w_τ (along the direction of travel) and its normal part w_n (toward the pivot O). The resultant w lies exactly horizontal. — The acceleration at the angle this step finds, decomposed into its tangential part $w_\tau$ (along the direction of travel) and its normal part $w_n$ (toward the pivot $O$ ). The resultant $\vect w$ lies exactly horizontal.

Step 5Tension when the vertical velocity peaks (part b)

The vertical velocity is $v_y=dy/dt$ , so its own time derivative is exactly the vertical component of the acceleration, $w_y$ ; a vanishing $w_y$ marks a stationary point of $v_y$ , but that alone does not yet say it is a maximum. It is one here: $v_y=v\sin\theta$ is continuous, equals zero at both ends of the swing ( $\theta=90^\circ$ , released from rest, and $\theta=0^\circ$ , where the velocity is purely horizontal), and is strictly positive in between, while (4) is its only stationary point in that open interval; a continuous function with this shape can only be turning from increasing to decreasing at its one interior stationary point, so that point is the maximum, not a minimum or inflection. Parts (b) and (c) therefore share one critical angle. Evaluating (3) there,

T = 3mg\cos\theta = 3mg\cdot\frac{1}{\sqrt3}.

T = \sqrt3\,mg

(6)

Check

At release, $\theta=90^\circ$ : $w=g\sqrt{1+0}=g$ and $T=0$ , consistent with a sphere momentarily at rest on a horizontal thread, where gravity is entirely tangential and there is no speed yet to need a centripetal pull. At the bottom, $\theta=0$ : $w=g\sqrt{1+3}=2g$ and $T=3mg$ , the familiar bottom-of-the-swing tension $mg+mv^2/l$ with $v^2=2gl$ .

Remark

The identity used in Step 5, that a maximum of one coordinate’s velocity occurs exactly where that coordinate’s acceleration vanishes, is just $dv_y/dt=w_y=0$ at an extremum; it has nothing to do with circular motion specifically; the circle only supplies the particular $w(\theta)$ that makes the vanishing condition solvable in closed form.