Irodov 1.136: Loop Break-Off Height Fixes Apex Speed

Problem Irodov 1.136

A small body $A$ starts from rest at height $h$ above the ground on a smooth (frictionless) groove. The groove descends and joins smoothly into a vertical half-circle of radius $h/2$ whose lowest point touches the ground, so the top of the half-circle sits at height $h$ , level with the starting point. The body slides down and partway around the inside of the half-circle before it leaves the track and continues afterward as a free projectile. Find the body’s speed at the highest point of that subsequent trajectory.

From rest at A, height h, the body slides into the half-circle of radius h/2, leaves the track at angle θ from the upward vertical with velocity v, and follows the dashed arc to the top, where v_top is horizontal. — From rest at $A$ , height $h$ , the body slides into the half-circle of radius $h/2$ , leaves the track at angle $\theta$ from the upward vertical with velocity $\vect v$ , and follows the dashed arc to the top, where $\vect v_{\mathrm{top}}$ is horizontal.

Given

$h$
loop radius $h/2$ , frictionless, released from rest

Find

the speed at the highest point of the trajectory after break-off

Solution

Step 1Height on the loop as a function of the break-off angle

Let $r=h/2$ be the loop’s radius and let $\theta$ be measured from the upward vertical through the center, as in the figure. The center sits at height $r$ above the ground, so a point on the rim at angle $\theta$ sits at height

y(\theta) = r + r\cos\theta.

Step 2Speed on the loop from energy conservation

The track is frictionless, so the body’s mechanical energy is conserved from its release at height $h$ to any point at height $y(\theta)$ .

\begin{align*}\frac{1}{2}mv^2 &= mg\bigl[h-y(\theta)\bigr] \\ &= mg\bigl[h-r-r\cos\theta\bigr]\end{align*}

v^2 = 2g(h-r-r\cos\theta).

Step 3The break-off condition

On the inside of the loop the track can only push, never pull, so the body leaves it exactly where the required normal force $N$ first reaches zero. Newton’s second law along the radius (positive toward the center) reads

N + mg\cos\theta = \frac{mv^2}{r}.

Setting $N=0$ isolates the break-off speed.

v^2 = gr\cos\theta. \tag{1}

Step 4Solving for the break-off angle

\begin{aligned} gr\cos\theta &= 2g(h-r-r\cos\theta) \\ r\cos\theta &= 2h-2r-2r\cos\theta \\ 3r\cos\theta &= 2(h-r) \end{aligned} \tag{2}

With $r=h/2$ , the right side is $2(h-h/2)=h$ and $3r=3h/2$ .

\cos\theta = \dfrac{2}{3}

(3)

Step 5Speed and its components at break-off

Substituting (3) into (1) gives the speed there.

\begin{align*}v^2 &= gr\cos\theta \\ &= g\left(\frac{h}{2}\right)\left(\frac{2}{3}\right)\end{align*}

v = \sqrt{\dfrac{gh}{3}}

(4)

Because the velocity is tangent to the circle and the radius to the break-off point makes angle $\theta$ with the vertical, the tangent (the velocity direction) makes the same angle $\theta$ with the horizontal. With $\cos\theta=2/3$ and $\sin\theta=\sqrt{5}/3$ from (3), the horizontal and vertical components of $\vect v$ are

v_x = v\cos\theta = \frac{2}{3}v, \qquad v_y = v\sin\theta = \frac{\sqrt5}{3}v.

Step 6Speed at the highest point of the flight

After break-off the body is a free projectile: gravity changes $v_y$ but never $v_x$ , and the trajectory’s highest point is exactly where the vertical component vanishes, leaving only $v_x$ .

\begin{align*}v_{\mathrm{top}} &= v_x \\ &= \frac{2}{3}v\end{align*}

Substituting the break-off speed (4),

v_{\mathrm{top}} = \dfrac{2}{3}\sqrt{\dfrac{gh}{3}}

(5)

Check

Dimensionally $\sqrt{gh}$ is a speed, as required. Also $v_{\mathrm{top}}<v$ (the break-off speed), consistent with the body still having to rise further after break-off before leveling off, and $\cos\theta=2/3\in(0,1)$ confirms a genuine break-off angle exists strictly between the side of the loop and its top.

Remark

The break-off angle $\cos\theta=2/3$ does not depend on $h$ at all: the relation $3r\cos\theta=2(h-r)$ in (2) becomes a pure number only once the given proportion $r=h/2$ is substituted, and that substitution cancels $h$ completely because every length in the problem is measured in units of $h$ . Doubling $h$ (and so $r$ ) therefore multiplies every speed in the problem by the same factor $\sqrt2$ , since $v^2\propto gr$ throughout, but leaves the break-off angle, and so the geometry of where the body leaves the track, unchanged.