Uniform Circular Motion — Question 7

Question 7

A crest and a dip require opposite normal-force corrections

Take $g=10$ m/s$^2$. A 50 kg rider moves at 10 m/s through two separate smooth road segments, each with local curvature radius 20 m: the top of a convex crest and the bottom of a concave dip. Treat the rider and vehicle support as a particle model with vertical road normal at these points.

1. Find the normal force on the rider at the crest and at the dip.

2. Find the crest speed at which the normal force first reaches zero.

3. Why can the road not enforce the circular crest path above that speed without a restraint?

Question 7 - Solution

A crest and a dip require opposite normal-force corrections

(a) Choose the local inward direction.

The local inward acceleration has magnitude $v^2/r=100/20=5$ m/s$^2$. At the crest inward is downward; at the dip it is upward: $$\boxed{N_{\mathrm{crest}}=mg-\frac{mv^2}{r}=500-250=250\ \mathrm N},$$ $$\boxed{N_{\mathrm{dip}}=mg+\frac{mv^2}{r}=500+250=750\ \mathrm N}.$$

(b) Find the contact boundary.

The crest contact boundary satisfies $$N_{\mathrm{crest}}=m\left(g-\frac{v^2}{r}\right)=0.$$ $$\boxed{v_c=\sqrt{gr}=\sqrt{200}=10\sqrt2=14.142\ \mathrm{m/s}}.$$

(c) Reject a negative contact force.

For $v>v_c$, the prescribed-path equation would demand $$N=m\left(g-\frac{v^2}{r}\right)<0.$$ An ordinary road below the rider supplies only $N\geq0$. Without a restraint, contact is lost and the rider follows a different trajectory; the negative formal normal is not a physical contact force.