The Forces of Rolling — Question 9

Question 9

Static friction can transfer energy on an accelerating belt

A disk of mass $M$ and $I=MR^2/2$ starts at rest on a belt whose rightward acceleration is constant $A$. There is sufficient static friction to prevent slip.

1. Find disk center acceleration and angular acceleration.

2. After time $t$, compare frictional power with the rate of increase of disk energy.

Question 9 - Solution

Static friction can transfer energy on an accelerating belt

(a) Derivation and calculation

The moving-surface constraint is $a+R\alpha=A$. With $$f=Ma,\qquad Rf=I\alpha=\tfrac12MR^2\alpha,$$ we obtain $R\alpha=2a$ and therefore $$\boxed{a=A/3},\qquad\boxed{\alpha=2A/(3R)},\qquad\boxed{f=MA/3}.$$ The angular acceleration is counterclockwise.

(b) Result and interpretation

From rest, $V=At/3$ and $\omega=2At/(3R)$. Thus $$K=\tfrac12M(At/3)^2+\tfrac12(MR^2/2)[2At/(3R)]^2 =\frac{MA^2t^2}{6}.$$ The material contact point moves at belt speed $At$, so $$\boxed{\mathcal P_f=f(At)=\frac{MA^2t}{3}=\frac{dK}{dt}}.$$ Static friction can do work when the contacting surface moves in the chosen frame.

Physical check

Translational and rotational contact velocities add to $At/3+2At/3=At$.