Detachment of body from hemisphere: exact formulas

Detachment of a Body from a Surface: From the Classic Hemisphere to Generalized Mechanics Problems

A body of mass m is placed at the top of a smooth hemisphere of radius R in a gravitational field g. Initial horizontal velocity is v₀. We need to find the angle φ at which the body detaches from the surface (N=0).

We neglect the radius and shape of the body. Centripetal acceleration is directed toward the center O.

High School Method via Newton's Second Law

Newton's second law in the radial projection:

ma_c = mg\cos\phi - N

where a_c = v²/R. At detachment N=0:

\frac{v^2}{R} = g\cos\phi

Law of conservation of energy:

\frac{mv_0^2}{2} + mgR = \frac{mv^2}{2} + mgR(1 - \cos\phi)

Simplification:

v_0^2 + 2gR = 3gR\cos\phi

\phi = \arccos\left(\frac{v_0^2 + 2gR}{3gR}\right)

If v₀ > √(gR), detachment occurs immediately (φ=0).

Deriving the Energy Law Without Assumptions

Distance traveled s = φR, velocity v = R dφ/dt, tangential acceleration dv/dt = g sin φ.

Separation of variables:

v dv = gR \sin\phi d\phi

Integration with initial conditions leads to the same law of conservation of energy:

\frac{mv_0^2}{2} + mgR = \frac{mv^2}{2} + mgR(1 - \cos\phi)

Then — substitution and solution as in the first method.

Generalization to an Arbitrary Convex Surface

Consider a surface y(x) with y''(x) < 0, vertex at (0, H), y(0)=H, y'(0)=0. Detachment at (x, y), N=0.

Radius of curvature R at the point, angle φ between the normal and vertical:

\frac{v^2}{R} = g\cos\phi

Transition to h = y - y_{O'} (O' — instantaneous center):

v^2 = gh = g(y - y_{O'})

Energy law:

\frac{mv_0^2}{2} + mgH = \frac{mv^2}{2} + mgy

Substitution gives a Cauchy differential equation:

(v_0^2 + 2g(H - y)) y'' + g(1 + (y')^2) = 0

Solution (Wolfram Alpha):

y = H - \frac{g x^2}{2 v_0^2}

The detachment point is the second intersection of y(x) with the parabola y = H - (g x²)/(2 v₀²). No intersection — detachment at the vertex.

Accounting for Sliding Friction Force

F_fric = μN. Equations in radial and tangential projections:

ma_c = mg\cos\phi - N

m \frac{dv}{dt} = -mg\sin\phi + \mu N

Expression for N and substitution:

2v \frac{dv}{d\phi} = g(\mu \cos\phi - \sin\phi) - 2\mu v^2

Linear non-homogeneous 1st order equation in v² (Bernoulli method):

v^2(\phi) = \frac{2gR}{\mu^2 + 1} \left[ (2\mu^2 + 1)\cos\phi - \mu \sin\phi \right] + \left( v_0^2 - 2gR \frac{1 + 2\mu^2}{1 + 4\mu^2} \right) e^{-2\mu \phi}

At N=0: v² = gR cos φ. Substitution gives a transcendental equation for φ (solved numerically, 0 < φ < 90°).

• If only φ=0 — immediate detachment.

Extension to Air Resistance

Add F_c = k v². Tangential equation:

\frac{dv}{d\phi} = g(\mu \cos\phi - \sin\phi) - \frac{1}{2} (2\mu v^2 + k v^2)

Structure is analogous, solved by the same method.

Key Takeaways

• For a smooth hemisphere: φ = arccos((v₀² + 2gR)/(3gR)), immediate detachment if v₀ > √(gR).
• Generalization: detachment at the intersection of y(x) with the parabola y = H - (g x²)/(2 v₀²).
• With friction: transcendental equation for φ from explicit v²(φ).
• Differential equations are solved analytically (Bernoulli) or numerically.
• The method is applicable to air resistance without complicating the structure.

— Editorial Team