Irodov 1.87: Gravity Outruns the Normal Force – Amit & Deepali Sinha Foundation

Problem Irodov 1.87

A small body $A$ starts from rest at the very top of a fixed, frictionless sphere of radius $R$ and slides down over its outer surface under gravity. Let $\theta$ be the angle between the vertical (through the sphere’s center $O$ and the top) and the radius to the body’s current position. Find the angle $\theta$ at which the body leaves the surface, and its speed at that instant.

The body at colatitude θ from the crown, with the normal N, the weight m g, and the weight's down-slope (tangential) component mgθ that speeds the body up as it descends. The drawn colatitude is illustrative. — The body at colatitude $\theta$ from the crown, with the normal $\vect N$ , the weight $m\vect g$ , and the weight’s down-slope (tangential) component $mg\sin\theta$ that speeds the body up as it descends. The drawn colatitude is illustrative.

Given

$R$
frictionless sphere
starts from rest at the top

Find

the break-off angle

\theta

and the break-off speed

v

Solution

Step 1Speed as a function of

\theta

, from energy conservation

Measured from the top, the body’s height drop at colatitude $\theta$ is $R-R\cos\theta$ . With no friction, mechanical energy is conserved from the start.

\frac12 mv^2 = mgR(1-\cos\theta)

v^2 = 2gR(1-\cos\theta)

(1)

Step 2The radial equation and the break-off condition

Along the radius, toward the center, gravity’s component $mg\cos\theta$ supplies both the centripetal acceleration and whatever the normal force $N$ does not.

mg\cos\theta - N = \frac{mv^2}{R}

The body is in contact only while $N\ge0$ ; it leaves exactly when $N$ first reaches zero, at which point the radial equation reduces to gravity alone providing the centripetal pull.

v^2 = gR\cos\theta

(2)

Step 3Solve for the break-off angle

Equating the speed from energy conservation (1) with the break-off condition (2) eliminates $v$ and $R$ together, leaving one equation in $\theta$ alone.

\begin{aligned} 2gR(1-\cos\theta) &= gR\cos\theta \\ 2(1-\cos\theta) &= \cos\theta \\ 2 &= 3\cos\theta \end{aligned} \tag{3}

\cos\theta = \dfrac23, \quad \theta = \arccos\dfrac23 \approx 48.2^\circ

(4)

The body drawn at the colatitude this step has just found. The normal force is absent from this figure because it has fallen to zero exactly here: gravity is the only force left, and its radial component alone supplies the centripetal pull.

Step 4The break-off speed

Substituting $\cos\theta=2/3$ into (2) gives the speed at the instant of separation.

v^2 = gR\cdot\frac23

v = \sqrt{\dfrac{2gR}{3}}

(5)

Check

At the very top, $\theta=0$ and $v=0$ , and the radial equation gives $N=mg$ , the full weight, exactly as a body at rest on top of a fixed sphere requires. As $\theta$ grows from $0$ , (1) shows $v^2$ increasing while (2)’s required $v^2=gR\cos\theta$ decreases, so the two curves must cross once on the way down, which is the single break-off angle found above.

Remark

Both sides of (3) carry the same factor $gR$ , which cancels before the angle is found: the break-off angle $\arccos(2/3)$ does not depend on the sphere’s size or on $g$ at all, only on the shape of the surface (a sphere). A body sliding down a cylinder of the same radius, released from rest at the top under otherwise identical conditions, leaves at exactly the same angle, because nothing in this derivation used the sphere’s curvature in the third dimension.