Find the maximum value of ratio of coefficients of friction?

fardeen_gen · #1 $Old$ May 19th, 2009, 11:01 AM

A sphere of mass m is placed on a rough block of mass 2m which in turn is placed on a rough inclined plane. Friction coefficient between block and sphere is $\mu_{2}$ and that between block and the inclined plane is $\mu_{1}$ . Find the maximum value of $\frac{\mu_{1}}{\mu_{2}}$ to ensure pure rolling at block and sphere surface.

the_doc · #2 $Old$ June 6th, 2009, 11:28 AM

This question, as it is currently stated, makes no sense!

No statement is made about whether the block is at the point of slipping or is slipping and so to maximise the ratio one can assume that $\mu_1$ is sufficiently large that the block remains where it is. In this case there is only a fixed condition on the minimum of $\mu_2$ and as we can set $\mu_1$ as large as we like then there cannot be any maximum.

fardeen_gen · #3 $Old$ June 6th, 2009, 09:37 PM

It made no sense to me too

Thats why I posted it. Are we supposed to assume something?

luobo · #4 $Old$ August 8th, 2009, 06:56 PM

Probably we are supposed to assume this:

Find the maximum value of $\frac{\mu_1}{\mu_2}$ to ensure pure rolling at block and sphere surface, and sliding at block and plane surface.

If so, here is the solution (just tentative, you are very welcome to point out if there is error):

For the spherical ball,
$f_2$ =frictional force
$N_2$ =contact force
$I$ =moment of inertia $, =kmr^2$
$\beta$ =angular acceleration
$a_2$ =translational acceleration along the surface

$mg\sin\theta-f_2=ma_2$ (1)
$mg\cos\theta-N_2=0$ (2)
$f_2 r=kmr^2\beta$ (3)
For pure rolling:
$a_2=\beta r$ (4)
$f_2\leq \mu_2 N_2$ (5)

From (1)~(5),
$f_2=\frac{k}{1+k}mg\sin\theta$ (6)
$N_2=mg\cos\theta$ (7)
$\mu_2\geq\frac{k}{1+k}\tan\theta$ (8)

For the block,
$f_1$ =frictional force along the block-plane interface
$N_1$ =contact force along the block-plane interface
$M$ =mass of block, $=2m$

$Mg\sin\theta+f_2-f_1\geq 0$ (9)
$Mg\cos\theta+N_2-N_1=0$ (10)
$f_1\geq\mu_1 N_1$ (11)

From (6)(7) and (9)~(11),
$\mu_1\leq\frac{M+\frac{k}{1+k}m}{M+m}\tan\theta$ (12)

From (8) and (12),
$\frac{\mu_1}{\mu_2}\leq$ $\frac{(1+k)M+km}{kM+km}=1+\frac{1}{k}\,\frac{M}{M+ m}$ (13)
Let $M=2m$ , so
$\max\frac{\mu_1}{\mu_2}=1+\frac{2}{3k}$ (14)

Note $k=\frac{2}{5}$ for solid sphere, $=\frac{2}{3}$ for spherical shell, $=\frac{1}{2}$ for solid cylinder, $=1$ for cylindrical shell,. So finally
$\max \frac{\mu_1}{\mu_2}=\frac{8}{3}$ for solid sphere.
$\max \frac{\mu_1}{\mu_2}=2$ for spherical shell.
$\max \frac{\mu_1}{\mu_2}=\frac{7}{3}$ for solid cylinder.
$\max \frac{\mu_1}{\mu_2}=\frac{5}{3}$ for cylindrical shell.

I am also wondering if there exists another possibility, i.e. the block goes up along the plane (rather than go down). If so, equation (9) needs to be changed. Interested persons might investigate that possibility.

Luobo