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Friction of Rotating Object

  1. Jan 3, 2008 #1
    If a cylinder is rotating with the circular end pressed against the ground, how can the work done by friction be calculated?
     
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  3. Jan 3, 2008 #2

    stewartcs

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  4. Jan 3, 2008 #3
    ____
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    I mean the cylinder is rotating this way. With the force going downward and the circular bottom face in contact with the ground. Heat will be produced and work will have to be done to turn the cylinder so there must be a force of friction.

    Drawing is so difficult
     
  5. Jan 3, 2008 #4
    [tex]M=\frac{2}{3} \mu_s PR[/tex]

    [tex]\mu_s[/tex] -coeff. static friction
    R-Radius
    P-Axial load
     
  6. Jan 3, 2008 #5

    FredGarvin

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    If the system is steady state, then the work would be:

    [tex]w = \tau \theta[/tex]

    This would imply that you can measure the torque and the angular displacement.
     
  7. Jan 3, 2008 #6

    arildno

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    Essentially, you perform an integration.

    Consider an infinitisemal slice of the cylinder's bottom with area [tex]da=rdrd\theta[/tex]

    where r is its radial position and theta its angular position.

    Now, let the local normal force be of magnitude dN, then, with [itex]\mu[/itex] being the kinetic friction of coefficient, we have that the local frictional force is:
    [tex]d\vec{F}=-\mu{dN}\vec{i}_{\theta}[/tex]
    where [itex]\vec{i}_{\theta}[/itex] is the local direction vector in the direction of motion.

    Now, the local torque contribution is given by:
    [tex]d\vec{\tau}=\vec{r}\times{d}\vec{\vec{F}}=-rdN\vec{i}_{r}\times\vec{i}_{\theta}=-\mu{r}dN\vec{k}[/tex], where k is the direction vector upwards.

    Now, let the normal force per unit area be some constant n, we have that:
    [tex]d\vec{\tau}=-\mu{n}{r}^{2}drd\theta\vec{k}[/tex]

    The total torque is therefore:
    [tex]\vec\tau=\int_{0}^{2\pi}\int_{0}^{R}d\vec{\tau}=-\frac{2\pi{n}\mu{R}^{3}}{3}\vec{k}=-\frac{2}{3}\mu{R}{n}A\vec{k}=-\frac{2}{3}\mu{R}N\vec{k}[/tex]
    where R is the cylinder's radius, N the net normal force on the cylinders bottom.

    Assuming a moment of inertia I, we therefore have that the angular acceleration is given by:
    [tex]\dot{\omega}=-\frac{2\mu{R}{N}}{3I}[/tex]

    This is then the constant rate by which the angular velocity [itex]\omega[/itex] decreases.

    Setting N=k*mg, and I=s*mR^{2} yields:
    [tex]\dot{\omega}=-\frac{k}{s}\frac{2g\mu}{R}[/tex]
     
    Last edited: Jan 3, 2008
  8. Jan 3, 2008 #7
    Thanks but unfortunately it appears that to solve for work by this method will require a lot of research into angular stuff so I might have to change my experimental method.
     
  9. Jan 3, 2008 #8

    arildno

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    kinetic, kinetic, kinetic!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
     
  10. Jan 3, 2008 #9
    Dear god!!!!!!!!!!!!!! Noooooooooooooooo (slow motion jumping leap with arms extended towards the keyboard to edit post)
     
  11. Jan 18, 2010 #10
    hey, i was doing this the other day, but I used a simpler method. I want to show it,but since I am a new user, i am not sure how to use the symbols. I will try anyway.

    let's start this way, think of a small point on the contacting surface of the cylinder.
    W=Fx, where work done W on the point is the product of F is the friction force and x is the distance travelled.

    And we all know F=[tex]\mu[/tex][tex]\Delta[/tex]mg, where [tex]\Delta[/tex]m is the supported mass on the point.

    x=r[tex]\theta[/tex], where r the distance from the point to the rotating axis and [tex]\theta[/tex] is the angular displacement.

    then, W=[tex]\mu\Delta[/tex]mgr[tex]\theta[/tex]
    and, W=[tex]\mu[/tex]A[tex]\rho[/tex]gr[tex]\theta[/tex], where [tex]\rho[/tex] is the area density, and A is area of the point, but now we will change it to the area sum of all the points with the same distance as the original point we were using. (which is a circle).

    now, the circle has an area of A= 2[tex]\Pi[/tex]r [tex]\Delta[/tex]r,
    [tex]\rho[/tex]= M/([tex]\Pi[/tex][R]^{}[/2]), where M is the total mass of the cylinder and R is the radius of the contact surface.

    Putting back into the formula, W=Mrg[tex]\theta[/tex][tex]\mu[/tex](2[tex]\Pi[/tex]r[tex]\Delta[/tex]r)/([tex]\Pi[/tex][R]^{}[/2])
    It is simplified to W=Mg[tex]\theta[/tex][tex]\mu[/tex][r]^{}[/2][tex]\Delta[/tex]r/([R]^{}[/2])

    Finally, we can use integration to sum it from 0 to R.
    W=Mg[tex]\theta[/tex][tex]\mu[/tex]/([R]^{}[/2])[tex]\int[/tex][r]^{}[/2][tex]\Delta[/tex]r

    and there's my answer: W= [2]\overline{}[/3][tex]\mu[/tex]MgR[tex]\theta[/tex]
     
  12. Jan 19, 2010 #11
    Let me try this again. This time I think I can do it

    let's start this way, think of a small point on the contacting surface of the cylinder.
    W=Fx, where work done W on the point is the product of F is the friction force and x is the distance travelled.

    And we all know F=[tex]\mu[/tex][tex]\Delta[/tex]mg, where [tex]\Delta[/tex]m is the supported mass on the point.

    x=r[tex]\theta[/tex], where r the distance from the point to the rotating axis and [tex]\theta[/tex] is the angular displacement.

    then, W=[tex]\mu\Delta[/tex]mgr[tex]\theta[/tex]
    and, W=[tex]\mu[/tex]A[tex]\rho[/tex]gr[tex]\theta[/tex], where [tex]\rho[/tex] is the area density, and A is area of the point, but now we will change it to the area sum of all the points with the same distance as the original point we were using. (which is a circle).

    now, the circle has an area of A= 2[tex]\Pi[/tex]r [tex]\Delta[/tex]r, [tex]\rho[/tex]= M/([tex]\Pi[/tex][tex]R^{2}[/tex]), where M is the total mass of the cylinder and R is the radius of the contact surface.

    Putting back into the formula, W=Mrg[tex]\theta[/tex][tex]\mu[/tex](2[tex]\Pi[/tex]r[tex]\Delta[/tex]r)/([tex]\Pi[/tex][tex]R^{2}[/tex])
    It is simplified to W=Mg[tex]\theta[/tex][tex]\mu[/tex][tex]r^{2}[/tex][tex]\Delta[/tex]r/([tex]R^{2}[/tex])

    Finally, we can use integration to sum it from 0 to R.
    W=Mg[tex]\theta[/tex][tex]\mu[/tex]/([tex]R^{2}[/tex])[tex]\int[/tex][tex]r^{2}[/tex][tex]\Delta[/tex]r

    and there's my answer: W= [tex]\frac{2}{3}[/tex][tex]\mu[/tex]MgR[tex]\theta[/tex]
     
  13. Jan 19, 2010 #12
    Please please tell me what you guys think. Sorry that I made a lot of mistakes.
    There were two that I want to rephrase:
    1,where work done W on the point is the product of F the friction force and x the distance travelled.
    2,but now we will change it to the area sum of all the points with the same distance from the rotating axis as the original point we were using. (The points form a circle).
     
  14. Jan 19, 2010 #13
    Here, this time I tried my best.

    let's start this way, think of a small point on the contacting surface of the cylinder.
    [tex]W=Fx[/tex], where work done W on the point is the product of F the friction force and x the distance the point travelled.

    And we all know [tex]F=\mu\Delta mg[/tex], where [tex]\Delta m[/tex] is the supported mass on the point.

    The point moves in circular motion, therefore [tex]x=r \theta[/tex], where r the distance from the point to the rotating axis and [tex]\theta[/tex] is the angular displacement. Arc length equals to the radius times the angle in radians.

    then, [tex]W=\mu\Delta mgr \theta[/tex] This is true because the friction which is constant acts opposite to the motion of the point.
    And [tex]W=\mu\rho Agr\theta[/tex], where [tex]\rho[/tex] is the area density, and A is area of the point, but now we will change it to the area sum of all the points with the same distance to the rotating axis as the original point we were using. (The points will form a circle).

    now, the circle has an area of [tex]A=2\pi r\Delta r[/tex] and [tex]\rho=\frac{M}{\pi R^{2}}[/tex], where M is the total mass of the cylinder and R is the radius of the contact surface.

    Putting back into the formula, [tex]W=\mu\frac{M}{\pi R^{2}}2\pi r\Delta rgr\theta[/tex]
    It is simplified to [tex]W=\frac{2\mu Mg\theta}{R^{2}}r^{2}\Delta r[/tex]

    Finally, we can use integration to sum it from 0 to R.
    [tex]\sum W=\frac{2\mu Mg\theta}{R^{2}}\int^{R}_{0} r^{2}\Delta r[/tex]

    and there's my answer: [tex]W=\frac{2}{3}\mu MgR\theta[/tex]
     
  15. Jan 19, 2010 #14
    There's a direction problem in arildno's calc because he uses the direction of N, in the first eqn. only the magnitude is considered.
     
  16. Jan 19, 2010 #15
    I don't think it's wrong. He can do that. He used [tex]\vec{i}_{\theta}[/tex], which is the direction of motion.
     
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