The Forces of Rolling — Question 1

Question 1

Static friction accelerates spin on a downhill ramp

A solid disk rolls from rest down a $30^\circ$ incline without slip. Let its mass be $M$, radius $R$, and $I=MR^2/2$; use $g=9.8$.

1. Derive its acceleration and friction force.

2. Find the minimum static friction coefficient.

Question 1 - Solution

Static friction accelerates spin on a downhill ramp

(a) Derivation and calculation

Take downhill translation positive and let $f$ denote uphill friction. The equations are $$Mg\sin\theta-f=Ma,\qquad fR=I\frac aR.$$ For $I=MR^2/2$, $f=Ma/2$, hence $$\boxed{a=\frac{2g\sin\theta}{3}=\frac{2(9.8)(.5)}3=3.267\ \mathrm{m/s^2}},$$ $$\boxed{f=\frac{Mg\sin\theta}{3}=Mg/6\quad\text{uphill}}.$$

(b) Result and interpretation

Normal balance gives $N=Mg\cos\theta$. The no-slip solution is admissible only if $$f\le\mu_sN\Rightarrow\mu_s\ge\frac{f}{N}=\frac13\tan\theta.$$ Thus $$\boxed{\mu_{s,\min}=\frac1{3\sqrt3}=.1925}.$$ Static friction takes the value required by the dynamics and reaches $\mu_sN$ only at its limiting case.

Physical check

The uphill friction both reduces translational acceleration and supplies the torque needed for rolling.