Coefficient of kinetic friction on ramp

  • #1
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A block and disk with equal masses are released from the same height on a 30° incline. The block slides down the ramp while the disk rolls. If they both reach the bottom at the same time, what must be the coefficient of kinetic (sliding) friction?


ANSWER: mk = (1/3)tanX = (1/3)tan(30°) = 0.192

my question is... how did they derive the equation:
mk = (1/3)tanX

all I got was 2/3 g sin X

[tex]\mu m g cos\theta=ma[/tex]
[tex]\mu m g cos\theta=m*\frac{2}{3} g sin\theta[/tex]
[tex]\mu=\frac{2}{3}tan \theta[/tex]
 

Answers and Replies

  • #2
Andrew Mason
Science Advisor
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UrbanXrisis said:
A block and disk with equal masses are released from the same height on a 30° incline. The block slides down the ramp while the disk rolls. If they both reach the bottom at the same time, what must be the coefficient of kinetic (sliding) friction? ...

[tex]\mu m g cos\theta=ma[/tex]
[tex]\mu m g cos\theta=m*\frac{2}{3} g sin\theta[/tex]
[tex]\mu=\frac{2}{3}tan \theta[/tex]
If they reach the bottom at the same time, then:

[tex]t_{block} = \sqrt{2s/a_{block}} = t_{disk} = \sqrt{2s/a_{disk}}[/tex] so the two accelerations must be equal.

The acceleration of the centre of mass of the disk is given by:

[tex]\tau = I\alpha = \frac{3}{2}mr^2a_{disk}/r = mgsin(30)r[/tex]

This is the tricky part. You have to use the parallel axis theorem to get the moment of inertia about a point on the rim: [itex]I_{rim} = I_{cm} + mr^2[/itex]

So:
(1)[tex]a_{disk} = \frac{2}{3}gsin(30)[/tex]

For the block:

[tex]F = ma_{block} = mgsin(30) - \mu_kmgcos(30)[/tex]

(2)[tex]a_{block} = g(sin(30) - \mu_kcos(30))[/tex]

From (1) and (2):

[tex]sin(30)(1 - \frac{2}{3})= \mu_kcos(30)[/tex]

[tex]\mu_k = \frac{1}{3}tan(30)[/tex]

AM
 
Last edited:
  • #3
1,789
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I suppose you have this figured out already but still...

For the disk, it is not really necessary to refer to the rim. If you do so however, you will take the torque contribution from mgsin\theta only and not from the friction force. So I guess for this problem, Andrew's method is better. You can also set up the torque equation about the mass center of the disk, taking torque due to friction only. But you would have to solve another equation (in particular, for translation of the mass center) to eliminate the frictional force. Remember, frictional force is in general not written as a function of the friction coefficients in case of general plane motion (ie rotation + translation) and definitely not in case of rolling. But for the block which does not rotate, frictional force is written as a function of the coefficient of kinetic friction.

Cheers
Vivek
 

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