Conservation of Mechanical Energy — Question 4

Question 4

Completing a loop requires more than reaching its height

Neglect friction and air resistance, take $g=10$ m/s$^2$, and treat all stated guides as fixed. A small bead slides on the inside of a smooth vertical circular track of radius R. It starts from rest at height h above the bottom on a connecting track. The contact force can push inward but cannot pull.

1. Find speed squared at the top in terms of h and R.

2. Derive the minimum release height for continuous contact through the loop.

3. Why is the energy-only threshold h=2R insufficient?

Question 4 - Solution

Completing a loop requires more than reaching its height

(a) Use the top height of 2R.

Conserve energy between release and the top, whose height is $2R$: $$mgh=\tfrac12mv_t^2+2mgR\Rightarrow\boxed{v_t^2=2g(h-2R)}.$$

(b) Impose nonnegative contact force.

At the top, inward radial dynamics require $$mg+N=\frac{mv_t^2}{R},\qquad N\geq0\Rightarrow v_t^2\geq gR.$$ $$2g(h-2R)\geq gR\Rightarrow\boxed{h_{\min}=\frac52R}.$$ More generally, at angle $\theta$ from the bottom, $N/(mg)=2h/R-2+3\cos\theta$, minimized at the top.

(c) Check dynamics at the proposed threshold.

The energy-only threshold $h=2R$ gives $$v_t=0\Rightarrow N=-mg<0.$$ The track cannot pull outward on the bead. Loss of contact occurs before the assumed full circular trajectory is completed.