[Physics] the friction between cylinder and wall (ground)


A hollow cylinder (radius $R$) is rolling against the wall at angular speed $\omega$. The coefficient of friction between the cylinder and the wall(ground) is $\mu$. After how many rotations the cylinder will stop rotating?

So I figured I need to find the time taken for cylinder to stop moving, and that would be
\beta=-\omega/t => t = -\omega/\beta
Where $\beta$ is angular acceleration, which is known from torques:
2*F_f*R = I*\beta
That's where I got stuck… How do I know the friction? I'm familiar with such equation:
F_f = \mu*F_n
How do I find the normal force? Does it have anything to do with centripetal force?

Best Answer

Hint: Look at the following diagram, and then solve the equations:

enter image description here

Or just notice that $F_w$ and $F_f$ do not depend on $\omega$, then use the Work-Energy principle.

Solution: $$\theta = \frac{R \omega^2 \left( 1+ \mu^2\right)}{2 g \mu (1+\mu)}$$

step by step solution:

Horizontal forces:$$F_f-N_w=0$$ Vertical forces: $$F_w+N_f-mg=0$$ We also know: $$F_f=\mu N_f$$ and $$F_w=\mu N_w \\$$ Solving them we find: $$ \\ F_f=\frac{mg\mu}{1+\mu^2} \\ F_w=\frac{mg\mu^2}{1+\mu^2} \\ \Rightarrow \left(F_f+F_w \right)R \theta = \frac{1}{2}I \omega^2 \\ \Rightarrow \theta = \frac{R \omega^2 \left( 1+ \mu^2\right)}{2 g \mu (1+\mu)} $$

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