Classical-Mechanics · Unit 18 · Video 5 · Interactive Practice

Letting Go at Forty-Eight Degrees: When the Sphere Stops Pushing

IKey Formulas

Formula	Name	What you need
$N - mg\cos\theta = -m\dfrac{v^2}{R}$	Radial component of Newton's Second Law	Outward radial axis; the acceleration is centripetal, $v^2/R$ inward
$N(\theta_1) = 0$	Loss-of-contact condition	A surface can push, but it can never pull
$\tfrac{1}{2}mv^2 + mgR\cos\theta = mgR$	Conservation of mechanical energy, $U = 0$ at the centre	Frictionless surface, released from rest at the top, $W_N = 0$
$\cos\theta_1 = \dfrac{2}{3}, \qquad v_1 = \sqrt{\dfrac{2gR}{3}}$	Separation angle and separation speed	Both equations imposed at the same instant

Key Insight: Eliminating $v^2$ between the two equations gives $N(\theta) = mg\,(3\cos\theta - 2)$ — the normal force is a pure function of angle. Energy conservation alone can never locate its zero, because $N$ does no work and so never enters the energy equation at all.

IIVisualization 1 — The Push That Fades

Gravity's radial pull is spent two ways: the turn the object needs, and whatever the surface still supplies.

Slide 0.0°

The top of the sphere is an unstable equilibrium, so the release nudge must be non-zero yet negligible; a genuinely finite starting speed moves the separation point up the sphere, as Problem 4 shows.

IIIVisualization 2 — Two Equations, One Crossing

Newton fixes the speed contact can survive; energy fixes the speed the fall delivers — they agree once.

Angle θ 24.0°

IVVisualization 3 — The Angle Nothing Can Move

Mass, radius and gravity all set the speed at separation; none of them touches the angle.

Radius R 1.8 m Gravity Mass m 2.0 kg

The cancellation is a fact about spheres, not about domes in general: on a surface whose radius of curvature changes as the object descends, the separation angle changes with it.

VQuiz Questions

Problem 1 · Where Contact Breaks

Given: a small block is released from rest at the very top of a fixed, frictionless sphere of radius $R = 0.50\ \text{m}$ — find the angle $\theta_1$ , measured from the vertical, at which the block leaves the surface.

✅ Correct!

\cos\theta_1 = \tfrac{2}{3}

gives

\theta_1 = 48.19\degree

— and notice that

R = 0.50\ \text{m}

never entered the calculation.

❌ That is the complement.

41.8\degree

is the angle measured from the horizontal;

\theta

is measured from the vertical, and

90\degree - 41.8\degree = 48.2\degree

❌ Check the factor of 2.

\tfrac{1}{2}mv^2 = mgR(1 - \cos\theta)

gives

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

; using

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

instead yields

\cos\theta_1 = \tfrac{1}{2}

and

60\degree

❌ Not quite. Set

N = 0

in the radial equation to get

gR\cos\theta_1 = v_1^2

, then substitute

v_1^2 = 2gR(1 - \cos\theta_1)

from energy conservation.

Show solution

Step 1 — Radial component of Newton's Second Law (outward positive, acceleration centripetal):

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

At the instant of separation the surface has nothing left to give, $N(\theta_1) = 0$ , so

mgR\cos\theta_1 = mv_1^2

Step 2 — Conservation of energy with $U = 0$ at the centre of the sphere. Released from rest at the top, $E_i = mgR$ ; at angle $\theta_1$ , $E_f = \tfrac{1}{2}mv_1^2 + mgR\cos\theta_1$ . The surface is frictionless and $N$ does no work, so $E_f = E_i$ :

\tfrac{1}{2}mv_1^2 + mgR\cos\theta_1 = mgR

Step 3 — Combine. Substituting $mv_1^2 = mgR\cos\theta_1$ into the kinetic term:

\tfrac{1}{2}mgR\cos\theta_1 + mgR\cos\theta_1 = mgR \;\Longrightarrow\; \tfrac{3}{2}\cos\theta_1 = 1

\cos\theta_1 = \frac{2}{3} \;\Longrightarrow\; \theta_1 = \cos^{-1}\!\left(\frac{2}{3}\right) = 48.19\degree

Every term carried the same factor $mgR$ , so $m$ , $g$ and $R$ all cancelled: the radius given in the problem is a red herring.

Problem 2 · Why Energy Is Not Enough

Given: the block on the frictionless sphere, with mechanical energy perfectly conserved throughout the slide — identify why conservation of energy alone cannot locate the separation angle $\theta_1$ .

✅ Correct! Energy is blind to

N

. The radial component of Newton's Second Law is the tool that can see it, and

N(\theta_1) = 0

supplies the second equation — two equations for the two unknowns

\theta_1

and

v_1

❌ $N$ does no work at all. The displacement is tangent to the sphere and

\vec{N}

is radial, so

\vec{N} \cdot d\vec{r} = 0

at every instant — the work is zero, not merely unknown.

❌ Mechanical energy is conserved here. The surface is frictionless, air resistance is neglected and

W_N = 0

, so

W_{nc} = 0

and

\Delta E_m = 0

. The energy equation is valid — it is just insufficient by itself.

❌ The mass is not the obstacle.

m

is a common factor in every term and cancels; the missing information is the normal force, which the energy equation cannot contain.

Show solution

The normal force is perpendicular to the surface, and the block's displacement is tangent to it. Therefore

dW_N = \vec{N} \cdot d\vec{r} = 0 \quad\text{at every instant}\;\Longrightarrow\; W_N = 0

so $N$ is absent from

\tfrac{1}{2}mv^2 + mgR\cos\theta = mgR

That single equation relates $v$ to $\theta$ , but "loss of contact" is a statement about $N$ , which it cannot express. The radial component of Newton's Second Law does contain $N$ :

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

Eliminating $v^2$ between the two gives $N(\theta) = mg\,(3\cos\theta - 2)$ , and setting it to zero locates $\theta_1$ . Neither tool answers the question alone; together they pin down both unknowns at the single instant of separation.

Problem 3 · Halfway Down, Still Touching

Given: the same frictionless sphere with $R = 1.20\ \text{m}$ and $g = 9.8\ \text{m/s}^2$ , the block released from rest at the top — find its speed and the normal force acting on it at $\theta = 30\degree$ , while it is still in contact.

Speed at 30 degrees

Normal force at 30 degrees

✅ Correct! The fall has delivered

1.78\ \text{m/s}

, and the surface is still pushing with about

60\%

of the block's weight — contact is a long way from breaking.

❌ Check the energy step. The block drops

R(1 - \cos 30\degree)

, so

\tfrac{1}{2}mv^2 = mgR(1 - \cos 30\degree)

with

1 - \cos 30\degree = 0.134

, giving

v^2 = 2(9.8)(1.20)(0.134)

❌ Check the radial equation.

N = mg\cos\theta - m v^2/R

, and here

mv^2/R = 2mg(1 - \cos 30\degree) = 0.268\,mg

— the centripetal demand, not the answer itself.

Show solution

Step 1 — Speed from energy conservation. The height above the centre falls from $R$ to $R\cos\theta$ :

v^2 = 2gR(1 - \cos\theta) = 2(9.8)(1.20)(1 - 0.8660) = 3.151\ \text{m}^2/\text{s}^2

v = 1.78\ \text{m/s}

Step 2 — Normal force from the radial equation. With $N - mg\cos\theta = -mv^2/R$ ,

N = mg\cos\theta - m\frac{v^2}{R} = mg\cos\theta - 2mg(1 - \cos\theta) = mg\,(3\cos\theta - 2)

N = mg\,[3(0.8660) - 2] = 0.598\,mg

Check the budget. Gravity's radial pull is $0.866\,mg$ ; the turn consumes $mv^2/R = 0.268\,mg$ ; the surface supplies the remaining $0.598\,mg$ . The three numbers satisfy $0.866 = 0.268 + 0.598$ , and $N$ reaches zero only when $\cos\theta = \tfrac{2}{3}$ .

Problem 4 · A Push at the Top

Given: the same frictionless sphere, but the block is launched from the top with a small initial speed $v_0$ satisfying $v_0^2 = \tfrac{1}{3}gR$ instead of being released from rest — find the new separation angle $\theta_1$ .

✅ Correct!

\cos\theta_1 = \tfrac{2}{3} + \dfrac{v_0^2}{3gR} = \tfrac{2}{3} + \tfrac{1}{9} = \tfrac{7}{9}

, so the extra speed pushes separation up the sphere, to

38.9\degree

❌ The universal angle assumes release from rest. What cancelled was

mgR

, not the initial kinetic energy: a non-zero

v_0

adds

v_0^2/(3gR)

\cos\theta_1

❌ Check the direction of the effect. Starting faster means the block reaches the critical speed

v^2 = gR\cos\theta

sooner — at a smaller

\theta

, nearer the top, not further down.

❌ Watch the factor of 2. Clearing the

\tfrac{1}{2}

from

\tfrac{1}{2}mv_0^2

doubles that term too: energy gives

v^2 = v_0^2 + 2gR(1 - \cos\theta)

, not

\tfrac{1}{2}v_0^2 + 2gR(1 - \cos\theta)

❌ Not quite. Rebuild the energy equation with the extra initial kinetic energy, then set it equal to the

N = 0

requirement

v^2 = gR\cos\theta

Show solution

Step 1 — Energy with a head start. $E_i = \tfrac{1}{2}mv_0^2 + mgR$ , and $E_f = \tfrac{1}{2}mv^2 + mgR\cos\theta$ . Setting $E_f = E_i$ ,

v^2 = v_0^2 + 2gR(1 - \cos\theta)

Step 2 — The separation condition is unchanged. $N = 0$ still requires $v_1^2 = gR\cos\theta_1$ . Substituting,

gR\cos\theta_1 = v_0^2 + 2gR - 2gR\cos\theta_1 \;\Longrightarrow\; 3gR\cos\theta_1 = 2gR + v_0^2

\cos\theta_1 = \frac{2}{3} + \frac{v_0^2}{3gR}

Step 3 — Substitute $v_0^2 = \tfrac{1}{3}gR$ :

\cos\theta_1 = \frac{2}{3} + \frac{1}{9} = \frac{7}{9} = 0.7778 \;\Longrightarrow\; \theta_1 = 38.9\degree

Setting $v_0 = 0$ recovers $\cos\theta_1 = \tfrac{2}{3}$ , so the familiar $48.2\degree$ is the largest possible separation angle for this sphere: any head start only makes the block let go sooner.

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