A stick of mass density $\rho$ per unit length rests on a circle of radius $R$ which makes an angle $\theta$ with horizontal and tangent to the circle at its upper end. Find the friction between them?

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written 8.4 years ago by

anjij • 70

Let $N$ be the normal force between stick and circle and let $F_f$ be the friction between ground and circle (see figure below). Then we immadiatly see that the friction force between stick and circle is also $F_f$, because the torques from the two friction forces on the circle must cancel. Looking at torques on the stick around the point of contact with ground we have $Mgcos{\theta}\displaystyle\frac{L}{2}=NL$ ,where $M$ is the mass of the stick and $L$ length. Therefore $N=mgcos{\theta}$ .Balancing the horizontal force on circle then gives $Nsin{\theta}=F_f+F_f cos{\theta}$. So we have $$ F_f = \frac{Nsin{\theta}}{1+cos{\theta}}=\frac{Mgsin{\theta}cos{\theta}}{2(1+cos{\theta})}$$

But $M=\rho L$ and from the figure we have $L=\displaystyle\frac{R}{tan{\frac{\theta}{2}}}$ and also using the identity $tan{\theta}=\displaystyle\frac{sin{\theta}}{1+cos{\theta}}$ , we obtain $$F_f=\frac{1}{2}\rho gRcos{\theta}$$

In the limit $\theta\rightarrow \frac{\pi}{2}$ , $F_f$ approaches 0, which makes sense. In the limit $\theta\rightarrow$ 0 $F_f$ approaches $\frac{\rho gR}{2}$ , which is not so obvious.

Figure- Stick resting on circle

ADD COMMENT EDIT

That circle isn't in horiz equilibrium because Nsin@, Fcos@ and F all point to the left.

N (on the circle) points to the circle's centre, equal & opposite to N (on the stick). F (on the circle) is equal & opposite to F (on the stick).

Note the 'N' in your diagram is infact N (on the stick), which was used in taking moments.

Also tan@ doesn't equal sin@/(1 + cos@), this re-arranges to tan@ + sin@ = sin@, which is wrong. Its tan(@/2) = sin@/(1 + cos@).

7.4 years ago by nima.alyson • 0