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More angular stuff

  1. Mar 28, 2005 #1
    No one in my class seems to get this:

    A 4.8-kg uniform cylinder rolls without slipping down a plane at 26.3-degree angle above horizontal. (a) What is the translational acceleration of the cylinder? (b) What is the magnitude of the frictional force?

    Now I know that the without slipping part is a crucial part of the problem that dictates what forumula should be used. I don't know what formula to use exactly though, because none of them seem to fit.
     
  2. jcsd
  3. Mar 28, 2005 #2
    Translational motion is talking about motion external to the actual object. In this case, translational motion means the motion of the cylinder neglecting any spinning or rotating effects.

    Drawing a force diagram, identify the forces acting on the cylinder, and find the magnitude of this force at the angle given. Divide the force by mass.

    For part b, recall the frictional force is [tex] F_{friction} = \mu N [/tex] where N is the normal force [tex] mgcos(\theta) [/tex]. Other than that I've got nothing for part b.
     
  4. Mar 28, 2005 #3
    Ok, I understand
     
    Last edited: Mar 28, 2005
  5. Mar 28, 2005 #4
    I really think the original poster meant to say tangential acceleration.

    Suppose the object with rotational inertia I, rolls down from an incline. We need to apply the two versions of Newton's second law. We have gravity, friction and a normal force. Also, for the torque, it is important to realize that the friction-force-vector is perpendicular to the radius at the point where the object touches the ground. Suppose that the x-axis is directed upwards, so the object rolls in opposite x-direction.

    The object moves opposite the x-direction, so a (tangential acceleration along the x-axis) will be negative

    The direction of our rotation is clockwise so the torque and the angular acceleration (alpha) should be negative.

    [tex]F_{friction} -Mgsin( \theta ) = Ma[/tex]
    [tex]\tau = -F_{friction}R = I \alpha[/tex]

    and

    [tex]alpha = \frac{a}{R}[/tex] where a is the tangential acceleration

    Now just solve the above equations for a and your problem is okelidokeli...

    You should get [tex]a = \frac{-gsin( \theta )}{1 + \frac{I}{MR^2}}[/tex]

    R is the radius of the object, M is the mass

    regards
    marlon
     
  6. Mar 28, 2005 #5
    ...But I wasn't given a radius....
     
  7. Mar 28, 2005 #6
    Hmm, ok then, i gave too much info...Forget about the rotation of the object. Just keep on working with the first version of Newton's second law...sorry

    regards
    marlon
     
  8. Mar 28, 2005 #7
    Ok, I'll try that. Thank you.
     
  9. Mar 28, 2005 #8
    I'm so sorry... but I still am not getting the right answer. I don't know what I am doing wrong.
     
  10. Mar 28, 2005 #9
    For part a)

    [tex] Mass = 4.8kg [/tex]

    [tex] Acceleration = 9.8 ms^-2*sin(26.3) = 4.34 ms^-2 [/tex]

    [tex] Force = Mass x Acceleration [/tex]

    That should be the answer to a. If you had more information I could help you with b, but I cant think of any way to find the frictional force without a coefficient of friction.
     
  11. Mar 28, 2005 #10
    That is not the correct answer I tried that the very first time and the computer told me I was wrong. I don't understand, it should be simple, considering it is only the translational accl.
     
  12. Mar 28, 2005 #11
    x-axis is pointing up the incline

    you are forgetting the normal force.

    [tex] \mu N - mgsin( \theta) = ma[/tex]

    [tex]N -mgcos( \theta ) = 0[/tex]

    solve the last equation for N and plug it in the first one that you solve for a

    marlon
     
    Last edited: Mar 28, 2005
  13. Mar 28, 2005 #12
    Perhaps I'm not understanding the original problem. Is it not a standard rolling down ramp problem? Is the not slipping acting as a restriction to the acceleration?
     
  14. Mar 28, 2005 #13
    I think my Professor isn't the best at creating these questions... because I don't have a coeff. of friction, right?
     
  15. Mar 28, 2005 #14
    Marlon, the funny thing is that these questions were made by the Physics department chair at the University of Kansas, and my whole class doesn't seem to understand.
     
  16. Mar 28, 2005 #15
    Yes, you have two unknowns, your resultant acceleration and your friction coefficient.
     
  17. Mar 28, 2005 #16
    You definetly should not forget about rolling.
    Use
    [tex]a = \frac{-gsin( \theta )}{1 + \frac{I}{MR^2}}[/tex]

    Do you know I for uniform cylinder?
     
  18. Mar 28, 2005 #17
    Yes, that was my first hint too but indeed you CAN treat the object not as a solid object but as a 'point-particle' , as is often done...

    marlon
     
  19. Mar 28, 2005 #18
    Marlon's 4th post was absolutely right (I suppose).
    For uniform cylinder I=(M*R^2)/2
    Thus: a=(2g/3)*Sin(theta)
     
  20. Mar 28, 2005 #19
    That is indeed how i would solve it too

    marlon
     
  21. Mar 28, 2005 #20

    Integral

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    The only force acting to cause rotation is the frictional force so we have a torque [itex] \tau [/itex] applied.

    From the kenimatics of rotation we have

    [tex] \tau = I_{cm} \alpha [/tex]

    The torque due to friction is

    [tex] \tau = f R [/tex]

    For a cylinder

    [tex] I_{cm} = \frac 1 2 MR^2 [/tex]

    combining these and using [itex] \alpha = \frac a R [/itex] we get

    [tex] f = \frac 1 2 Ma [/tex]

    Now applying Newtons Law to the motion down the incline we have
    [tex] Mg \sin( \theta ) - f = M a [/tex]

    we can now solve for a

    [tex] a = \frac 2 3 g sin ( \theta ) [/tex]

    Of course [itex] \theta [/itex] is the angle of your incline.
     
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