6.05f - Leaving a circular path

When the circular model ends

A particle slides over the outside of a smooth hemisphere. Does it leave the surface when its speed becomes zero, when its weight vanishes, or when something else happens? Neither of those is the right condition. The surface stops controlling the motion when the required normal contact force reaches zero and would need to reverse direction to preserve the circle.

The sign restriction comes from the object providing the force. A smooth external surface pushes the particle away from its centre; it cannot pull the particle in. A light string pulls towards its fixed end; it cannot push away. A rigid rod can do either, so a negative force in an assumed inward direction can mean compression rather than loss of the path.

Zero force is a boundary condition, not a command to replace every other quantity by zero. At departure the position is continuous and the velocity is unchanged in that instant: it is tangent to the circle. Acceleration becomes the free-fall acceleration. There is no outward kick and no new impulse merely because contact ends.

Check the direction at this handover: confusing velocity with radial acceleration would launch the particle along the wrong line.

Leaving the outside of a surface

For the outside of a sphere or circular arc, choose an angle from the upward vertical. This makes the height and the inward component of weight easy to read.

Let the particle start at the top with horizontal speed $u$, and let $\phi$ increase down the right-hand side of a smooth sphere of radius $r$. Its height above the centre is $r\cos\phi$; it has fallen $r(1-\cos\phi)$. Energy gives

and hence $v^2=u^2+2gr(1-\cos\phi)$. Meanwhile weight has inward component $mg\cos\phi$, but the reaction $R$ acts outwards. Thus $mg\cos\phi-R=mv^2/r$.

[DIAGRAM: asset_name: Leaving the outside of a smooth sphere; asset_slug: h245_6_05_surface_departure; caption: At departure the outward reaction is zero, and the particle continues with a tangent velocity; description: The upper semicircle of a sphere is drawn with centre O and a departure point P on its upper right side. The radius OP makes angle phi with the upward vertical. Height r cos phi is marked above O. At an earlier point a short outward normal reaction arrow and downward weight arrow illustrate the signs. At P the reaction is labelled zero, a blue tangent velocity points down and right, and a clay projectile trajectory continues outside the circle to a horizontal plane through O. A side panel states mg cos phi - R = mv squared over r and R = 0 at departure]

At the top, contact requires $R=mg-mu^2/r\ge0$. As the particle descends, its speed increases while the inward component of weight decreases; eventually weight alone is just enough to turn it at that instant. Thereafter the surface would need to pull, and the circular model ends. A particle exactly at rest at the exact highest point is an equilibrium of this ideal model; a slide from near the top can be considered as a limit, but no unmentioned push is needed in the non-zero-speed examples below.

Change only the initial speed to a larger value. Less descent is needed before gravity alone can no longer supply the required inward force.

After departure, the value $v^2/r$ is no longer a required radial acceleration for the particle: it is no longer following that circle. Carry the launch velocity into a new free-fall model.

Change the launch speed once more and keep both equations visible.

Slack strings and complete circles

For a string, measure $\theta$ from the downward vertical as in the preceding lesson. The force that can fail is inward tension, so the radial equation has the opposite constraint-force direction from the external-surface example.

For initial speed $u$ at the bottom and string length $r$, energy and radial dynamics give

Combining them gives $T=(m/r)(u^2-2gr+3gr\cos\theta)$. On the first ascent, tension decreases as $\theta$ increases. To locate slackening, solve $T=0$ and check that the speed is real and the point is reached before a reversal.

[DIAGRAM: asset_name: A string reaches zero tension above the centre; asset_slug: h245_6_05_string_slack; caption: At the first slackening point the tangent velocity is still non-zero, and gravity alone provides the instantaneous inward acceleration; description: A vertical circle of radius r with centre O and a point P in the upper right quadrant at theta = 120 degrees from the downward vertical. The string OP is blue and its tension at P is labelled T = 0. A tangent velocity arrow points up and left. A downward weight arrow from P is resolved conceptually into an inward component and a down-tangent component. A short dashed projectile curve bends inside the former circle after P. Text gives cos theta = -1/2 and v squared = gr/2 for the check example, with the circular arc beyond P greyed and labelled no longer the path]