Uniform Circular Motion — Question 6

Question 6

A constant-speed vertical loop needs tangential control

Take $g=10$ m/s$^2$. A 0.50 kg bead follows the inside of a smooth vertical circular track of radius 2 m at constant speed 5 m/s. An external drive acts only tangentially to maintain that speed. The track can push inward but cannot pull outward.

1. Find the track normal forces at the top and bottom.

2. Find the minimum constant speed that preserves contact around the full loop.

3. At either side point, find the normal force and the drive-force magnitude; explain why a tangential drive is needed.

Question 6 - Solution

A constant-speed vertical loop needs tangential control

(a) Point inward separately at top and bottom.

The constant radial demand is $mv^2/r=.50(25)/2=6.25$ N. Choose inward separately at each point: $$N_{\mathrm{top}}+mg=6.25\Longrightarrow\boxed{N_{\mathrm{top}}=1.25\ \mathrm N},$$ $$N_{\mathrm{bottom}}-mg=6.25\Longrightarrow\boxed{N_{\mathrm{bottom}}=11.25\ \mathrm N}.$$

(b) Test the weakest contact point.

If $\phi$ is measured from the top, the normal is $$N(\phi)=\frac{mv^2}{r}-mg\cos\phi.$$ Its minimum occurs at the top. Nonnegative contact force everywhere requires $$\boxed{v\geq\sqrt{gr}=\sqrt{20}=4.472\ \mathrm{m/s}}.$$ Equality is marginal contact at the top.

(c) Separate perpendicular components.

At either side, gravity is purely tangential: $$\boxed{N=\frac{mv^2}{r}=6.25\ \mathrm N},\qquad F_{{\mathrm{drive}},t}+F_{g,t}=ma_t=0.$$ Thus $\boxed{|F_{{\mathrm{drive}},t}|=mg=5\ \mathrm N}$, directed upward along the tangent. The drive cancels the tangential gravitational force; without it the speed would vary.