Irodov 1.108: The Push That Frees the Body Early – Amit & Deepali Sinha Foundation

Problem Irodov 1.108

A small body rests at the top of a smooth (frictionless) sphere of radius $R$ . The sphere is then given a constant horizontal acceleration $w_0$ , and the body, initially at rest relative to the sphere, begins to slide over its surface under gravity and the sphere’s own acceleration. Let $\theta$ be the angle, at the sphere’s center, between the true vertical and the radius to the body’s current position. Find:

the body’s speed $v$ relative to the sphere at the instant it leaves the surface
the angle $\theta_0$ at that instant, and its numerical value when $w_0=g$

The setup: a small body on the smooth sphere of radius R, at the angle θ measured at the center O from the true vertical, while the sphere itself is driven horizontally with the constant acceleration w_0. — The setup: a small body on the smooth sphere of radius $R$ , at the angle $\theta$ measured at the center $O$ from the true vertical, while the sphere itself is driven horizontally with the constant acceleration $\vect w_0$ .

Given

$R$
$w_0$

Find

(a)

v

at break-off, (b)

\theta_0

, and its value at

w_0=g

Solution

Step 1Working in the sphere’s own frame

The free body this step arrives at, drawn at a general angle θ in the sphere's own frame: the radial normal N, the true weight m g (downward) and the pseudo-force -m w_0 (horizontal, opposite the sphere's acceleration) are concurrent at the point-like body. — The free body this step arrives at, drawn at a general angle $\theta$ in the sphere’s own frame: the radial normal $\vect N$ , the true weight $m\vect g$ (downward) and the pseudo-force $-m\vect w_0$ (horizontal, opposite the sphere’s acceleration) are concurrent at the point-like body.

In this frame the body starts from rest and all three of these forces act on it at every angle $\theta$ it reaches.

Step 2Radial (normal) equation of motion at angle

\theta

What this step's projection amounts to: in the sphere's frame, true gravity g and the pseudo-force per unit mass - w_0 add to a single effective field g ', tilted by δ=(w_0/g) from the true vertical. — What this step’s projection amounts to: in the sphere’s frame, true gravity $\vect g$ and the pseudo-force per unit mass $-\vect w_0$ add to a single effective field $\vect g\,'$ , tilted by $\delta=\arctan(w_0/g)$ from the true vertical.

Projecting $m\vect g$ and the pseudo-force $-m\vect w_0$ onto the outward radial direction at angle $\theta$ (which is $-\cos\theta$ times $g$ for gravity, since it is inward, and $+\sin\theta$ times $w_0$ for the pseudo-force, since part of it points outward once the body is off to the side) and equating the outward sum to the required centripetal (inward) force,

N - mg\cos\theta + mw_0\sin\theta = -\frac{mv^2}{R}

Solving for $N$ ,

N = mg\cos\theta - mw_0\sin\theta - \dfrac{mv^2}{R}

(1)

Step 3Speed from work done by both forces

Over the angle $0$ to $\theta$ , gravity does work $mgR(1-\cos\theta)$ (the height dropped) and the horizontal pseudo-force does work $mw_0R\sin\theta$ (the horizontal distance moved in its own direction, $R\sin\theta$ ). Equating total work to kinetic energy gained,

\frac{1}{2}mv^2 = mgR(1-\cos\theta) + mw_0R\sin\theta

v^2 = 2R\bigl[g(1-\cos\theta)+w_0\sin\theta\bigr]

(2)

Step 4The break-off condition

Setting $N=0$ in (1) and using (2) for $v^2$ ,

\begin{align*}mg\cos\theta_0-mw_0\sin\theta_0 &= \frac{m}{R}\cdot 2R\bigl[g(1-\cos\theta_0)+w_0\sin\theta_0\bigr] \\ g\cos\theta_0-w_0\sin\theta_0 &= 2g(1-\cos\theta_0)+2w_0\sin\theta_0 \\ 3g\cos\theta_0-3w_0\sin\theta_0 &= 2g\end{align*}

g\cos\theta_0-w_0\sin\theta_0 = \dfrac{2g}{3}

(3)

Step 5Part (a): the break-off speed

Setting $N=0$ in (1) and solving for $v^2$ ,

\begin{align*}0 &= mg\cos\theta_0 - mw_0\sin\theta_0 - \frac{mv^2}{R} \\ v^2 &= R\bigl(g\cos\theta_0-w_0\sin\theta_0\bigr)\end{align*}

Substituting (3)’s value for that combination,

v^2 = R\cdot\frac{2g}{3}

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

(4)

Step 6Part (b): solving for

\theta_0

Writing $\eta=w_0/g$ and isolating the $\sin\theta_0$ term in (3), then squaring and using $\sin^2\theta_0=1-\cos^2\theta_0$ ,

\begin{align*}g\cos\theta_0-\frac{2g}{3} &= w_0\sin\theta_0 \\ g^2\Bigl(\cos\theta_0-\frac{2}{3}\Bigr)^2 &= w_0^2\bigl(1-\cos^2\theta_0\bigr)\end{align*}

Dividing by $g^2$ , expanding, and collecting powers of $\cos\theta_0$ ,

(1+\eta^2)\cos^2\theta_0 - \frac{4}{3}\cos\theta_0 + \frac{4}{9}-\eta^2 = 0

Multiplying through by $9$ clears the fractions and leaves a quadratic in $\cos\theta_0$ with integer coefficients.

9(1+\eta^2)\cos^2\theta_0 - 12\cos\theta_0 + \bigl(4-9\eta^2\bigr) = 0

Its discriminant collapses, because the constant $144$ and the $36\times4$ coming out of the product cancel:

\begin{align*}\Delta &= 12^2 - 4\cdot 9(1+\eta^2)\bigl(4-9\eta^2\bigr) \\ &= 144 - 36\bigl(4-9\eta^2+4\eta^2-9\eta^4\bigr) \\ &= 36\eta^2\bigl(5+9\eta^2\bigr)\end{align*}

The quadratic formula then gives the two roots, with $\sqrt{\Delta}=6\eta\sqrt{5+9\eta^2}$ .

\cos\theta_0 = \frac{12 \pm 6\eta\sqrt{5+9\eta^2}}{18(1+\eta^2)} = \frac{2 \pm \eta\sqrt{5+9\eta^2}}{3(1+\eta^2)}

The unsquared relation decides which sign is physical. Rearranging (3) and dividing by $g$ ,

\cos\theta_0-\frac{2}{3} = \eta\sin\theta_0

The body breaks off in the upper hemisphere, where $\sin\theta_0\ge0$ , and $\eta=w_0/g$ is nonnegative, so the right side cannot be negative and $\cos\theta_0\ge2/3$ is forced. Only the plus sign keeps that inequality, so it is the root to take,

\cos\theta_0 = \dfrac{2+\eta\sqrt{5+9\eta^2}}{3(1+\eta^2)}, \qquad \eta=\dfrac{w_0}{g}

(5)

Putting $w_0=g$ , so that $\eta=1$ , into the boxed formula and evaluating,

\begin{align*}\cos\theta_0 &= \frac{2+\sqrt{5+9}}{3(1+1)} = \frac{2+\sqrt{14}}{6} \\ &\approx \frac{2+3.742}{6} \approx 0.957\end{align*}

\theta_0 \approx 17^\circ \quad (w_0=g)

(6)

Check

As $\eta\to0$ (no horizontal push), (3) reduces to $\cos\theta_0=2/3$ , the textbook break-off angle for a body sliding off a stationary smooth sphere under gravity alone, exactly what the boxed formula gives at $\eta=0$ . Dimensionally, $v=\sqrt{2gR/3}$ has units of speed, and it involves no $w_0$ at all, consistent with the elegant cancellation in Step 5.

Remark

Part (a)’s answer is independent of $w_0$ only because the break-off speed is read off the same linear combination $g\cos\theta_0-w_0\sin\theta_0$ that the break-off condition fixes at $2g/3$ ; the horizontal push changes where the body leaves the sphere (through $\theta_0$ , which does depend on $w_0$ , as part (b) shows) but, remarkably, not the speed it has when it does. It is the same $\sqrt{2gR/3}$ found for a body sliding off a stationary smooth sphere under gravity alone ( $w_0=0$ , where $\theta_0=\arccos(2/3)$ ), even though the sphere here is accelerating and the body leaves from a different point.