1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Paradox of rigid body?

  1. Aug 22, 2009 #1
    There is a rod treating as rigid body, the rod which mass is m and lenghth is R rotate about one end of itself with the angular acceleration [tex]\alpha [/tex]. Apply a tangential external force f on another end to make the rod rotate. Now divide the rod into elements of mass. For each mass element dm with the distance r from the fixed end, it acted by an tangential component of resultant force dF which satisfy:

    [tex]dF=dm\cdot a_{t} = dm\cdot \alpha r =\lambda dr \cdot \alpha r[/tex]([tex]\lambda[/tex] is the linear density)

    [tex]\alpha =\frac{M}{I}=\frac{fR}{\frac{1}{3}mR^{2}}[/tex] (M is the moment of force,I is the moment of interia about the fixed end)

    So the tangential component of resultant force F of the rod will be:

    [tex]F = \int_{0}^{R}\lambda dr \cdot \alpha r=\frac{m}{R}\frac{fR}{\frac{1}{3}mR^{2}}\int_{0}^{R}r\cdot dr=\frac{3}{2}f[/tex]

    As we see, F is not equal to f, is this a paradox? what is wrong in this argument?
     
  2. jcsd
  3. Aug 22, 2009 #2

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    There is a constraint force at the fixed end of the bar that is keeping the fixed end fixed, and you are ignoring this.
     
  4. Aug 22, 2009 #3
    I think ,at the fixed end ,the tangential component of resultant force dF is zero according to [tex]dF =\lambda dr \cdot \alpha r[/tex],which have included the constraint force.
     
  5. Aug 22, 2009 #4

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Try again.

    If there were no constraint force the bar as a whole would go flying off.
     
  6. Aug 22, 2009 #5

    Mapes

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Isn't this http://en.wikipedia.org/wiki/Carroll%27s_paradox" [Broken]? If you sum the moments around the free end, you'll see that there must be a nonzero component of force perpendicular to the bar at its fixed end, as D H says. Otherwise the bar wouldn't begin to rotate.
     
    Last edited by a moderator: May 4, 2017
  7. Aug 22, 2009 #6

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    This is not Carroll's paradox. Carroll's paradox describes an unreal situation. The OP describes something that is real.
     
  8. Aug 23, 2009 #7

    Mapes

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Ah, right you are; I spoke (wrote) too soon.
     
  9. Aug 23, 2009 #8

    rcgldr

    User Avatar
    Homework Helper

    Note that the tangental force at the end of the rod is partially opposed by a force at the hinged end of the rod, and I don't see where this opposing force at the hinged end of the rod is taken into account in the original post.
     
  10. Aug 24, 2009 #9
    Yep, I have realized that I ignored the constraint force, but I still do not understand how to calculate the constraint force f',it is f'=3f/2-f=f/2, isn't it? Is there another way to calculate it ? I only learn a few thing about constraint force before. Constraint force can adjust to the condition automatically, is that right?
    Thanks!
     
  11. Aug 24, 2009 #10

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    You have:
    [tex]\frac{dF}{dr}=\lambda\alpha{r}=\frac{3f\lambda}{Rm}r[/tex]

    This yields:
    [tex]F(r)=\frac{3f\lambda{r}^{2}}{2mR}+C[/tex]
    where C is an undetermined constant.
    Using the relation F(R)=f, we get:
    [tex]C=f-\frac{3}{2}fR\frac{\lambda}{m}=-\frac{1}{2}f[/tex]
    and that is the constraint force at the hinge.

    Your fallacy lies in assuming that constraint force to be 0.
     
  12. Aug 24, 2009 #11

    Mapes

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You could sum the moments around the center of mass and apply

    [tex]M_C=\dot{H}_C=I_C\alpha=ml^2\alpha/12[/tex].
     
  13. Aug 27, 2009 #12

    I am afraid I could not agree with you, because under your assumption,there will be F(0)=C Which is not equal to zero at the fixed end, and this will make the rod fly off, so it sounds impossible. I would like to think:

    [tex]f+C=F=\frac{3f\lambda R^{2}}{2mR}=\frac{3}{2}f [/tex]

    and therefore: [tex]C=\frac{1}{2}f [/tex]

    Is that right ?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Paradox of rigid body?
  1. Rotation of rigid body (Replies: 4)

  2. Rotation of Rigid body (Replies: 2)

Loading...