Irodov 1.192: Sideways Push Toward a Right Angle – Amit & Deepali Sinha Foundation

Problem Irodov 1.192

A small ball hangs from a fixed point $O$ on a light, inextensible thread of length $l$ . It is drawn aside until the thread makes an angle $\theta$ with the vertical, and from that position it is launched with initial speed $v_0$ , directed horizontally and perpendicular to the vertical plane that contains the thread (so the ball leaves the plane of its initial displacement). Find the initial speed $v_0$ for which the thread’s maximum angle from the vertical, reached anywhere during the subsequent motion, equals $\pi/2$ .

The launch: the ball of mass m on a thread of length l from the fixed point O, drawn aside to angle θ from the vertical and released with speed v_0 directed out of the page. — The launch: the ball of mass $m$ on a thread of length $l$ from the fixed point $O$ , drawn aside to angle $\theta$ from the vertical and released with speed $v_0$ directed out of the page.

Given

$l$
$\theta$
maximum angle from vertical $=\pi/2$

Find

v_0

Solution

Step 1Two constants of the motion on the sphere

Left, the free body: tension T toward O, weight m g down, the thread of length l at θ. Right, the frame at the ball: z, r, φ, the lever arm lθ out to the axis and the depth lθ below O. — Left, the free body: tension $\vect T$ toward $O$ , weight $m\vect g$ down, the thread of length $l$ at $\theta$ . Right, the frame at the ball: $\hat{\vect z}$ , $\hat{\vect r}$ , $\hat{\vect\varphi}$ , the lever arm $l\sin\theta$ out to the axis and the depth $l\cos\theta$ below $O$ .

Because the thread is inextensible and stays taut (it can pull the ball inward but never push it outward, and the launch only ever swings the ball away from the vertical, never slack), the ball’s distance from $O$ stays fixed at $l$ throughout, so it moves on the sphere of radius $l$ centered on $O$ . Take $\theta$ as the polar angle measured from the downward vertical through $O$ ; the ball’s distance from the vertical axis is then $l\sin\theta$ and its depth below $O$ is $l\cos\theta$ , the two distances drawn in the frame panel. Two quantities are conserved. The first is the mechanical energy, with the potential energy measured from the level of $O$ .

E = \frac{1}{2}mv^2 - mgl\cos\theta \tag{1}

The second is the component of angular momentum about the vertical axis through $O$ , which is the mass times the moment of the ball’s azimuthal velocity about that axis, at the lever arm $l\sin\theta$ .

L_z = m\,(l\sin\theta)^2\,\dot\varphi = m\,(l\sin\theta)\,v_\varphi \tag{2}

Step 2Both conservation laws at the launch point and at the extreme angle

At launch the velocity $v_0$ is purely horizontal and perpendicular to the vertical plane of the thread, that is, purely along $\hat{\vect\varphi}$ , so the whole speed there is azimuthal; likewise at the angle of maximum deviation $\theta_{\max}$ the polar angle is momentarily stationary, $\dot\theta=0$ , so the speed there, $v_{\max}$ , is again purely azimuthal. At both instants, then, $v_\varphi$ is the whole speed, and the second form of (2) can be written down directly.

\begin{align*}L_z &= m\,(l\sin\theta)\,v_0 \\ L_z &= m\,(l\sin\theta_{\max})\,v_{\max}\end{align*}

Equating the two expressions for the conserved $L_z$ ,

v_{\max} = v_0\,\frac{\sin\theta}{\sin\theta_{\max}}

(3)

Step 3Energy conservation between launch and the extreme angle

Writing (1) at the launch point and at $\theta_{\max}$ and equating the two,

\frac{1}{2}mv_0^2 - mgl\cos\theta = \frac{1}{2}mv_{\max}^2 - mgl\cos\theta_{\max}.

Setting $\theta_{\max}=\pi/2$ (so $\sin\theta_{\max}=1$ , $\cos\theta_{\max}=0$ ) and substituting $v_{\max}=v_0\sin\theta$ from (3),

\begin{align*}\frac{1}{2}v_0^2 - gl\cos\theta &= \frac{1}{2}v_0^2\sin^2\theta \\ \frac{1}{2}v_0^2\bigl(1-\sin^2\theta\bigr) &= gl\cos\theta \\ \frac{1}{2}v_0^2\cos^2\theta &= gl\cos\theta.\end{align*}

Dividing through by $\cos\theta$ (nonzero, since $0<\theta<\pi/2$ for the ball to rise further to $\pi/2$ ),

\frac{1}{2}v_0^2\cos\theta = gl.

v_0 = \sqrt{\dfrac{2gl}{\cos\theta}}

(4)

Check

Dimensionally $\sqrt{gl/\cos\theta}$ has units $\sqrt{(\text{m/s}^2)(\text{m})}=\text{m/s}$ , a speed, as required. As $\theta\to0$ (ball started hanging almost straight down) the required launch speed approaches $\sqrt{2gl}$ , matching ordinary energy conservation for a pendulum bob raised from the bottom of its arc to the horizontal, $v_0^2/2=gl$ , exactly the case with no sideways angular momentum complication since $\sin\theta\to0$ makes $L_z\to0$ regardless of $v_0$ .

Remark

As $\theta\to\pi/2$ the required $v_0$ diverges: a ball started already near the horizontal needs an unboundedly large sideways speed to still be climbing (rather than already falling back) by the time it reaches $\theta_{\max}=\pi/2$ , because (3) forces $v_{\max}\to v_0$ there while gravity has almost no height left to absorb into the sideways kinetic energy. The angle $\theta$ therefore cannot be taken past $\pi/2$ in this formula, consistent with $\theta_{\max}=\pi/2$ being the ball’s highest possible point on the sphere below $O$ ’s own level.