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Cylinder Rotating

  1. Jul 11, 2010 #1
    1. The problem statement, all variables and given/known data

    We have a rigid body that is formed by two coaxial cylinders, one with radius [tex]r[/tex] and another with radius [tex]R[/tex], [tex]r < R[/tex]. There's a winding of string around the cylinder of smaller radius. The string is pulled with a constant force [tex]F[/tex] making an angle [tex]\alpha[/tex] with the horizontal.

    There is no "sliding", only rotation.

    What is the "critical" angle [tex]\alpha[/tex] for which the direction of the movement changes?


    2. Relevant equations

    [tex]\vec{F} = m \vec{a}[/tex]
    [tex]\tau = \vec{F} \times \vec{r}[/tex]
    [tex]\tau = I \vec{\vec{\ddot{\theta}}}[/tex]

    3. The attempt at a solution

    Well, at first I was a bit skeptical since I couldn't quite imagine pulling the string to one side and the cylinder coming to that side. I thought that pulling the string to one side would make the cylinder rotate away always. So, I made the experiment and the direction of the movement did in fact change with the angle.

    I've been thinking and I could only conjecture one thing: when I'm pulling the string with an angle, I can not assume that the force is being applied to the lowest point, yielding a torque [tex]\tau = F r \cos \alpha[/tex], instead I should think that the string is always perpendicular to [tex]\vec{r}[/tex], and that the net torque is [tex]\tau = F r[/tex]. But this lead me nowhere.

    Can you help me? I don't want a solution, I just want a little help to "see the physics" of the problem.

    Regards
    Krappy
     

    Attached Files:

  2. jcsd
  3. Jul 11, 2010 #2
    Because the instantaneous motion of the body can be described as rotating about the axis going through the point of contact (i.e. the lowest point), you can apply the equation Torque = I.alpha for that axis. And what's the torque about the axis? :smile:
     
  4. Jul 11, 2010 #3

    ehild

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    Homework Helper
    Gold Member

    The torque is defined with respect to a rotation axis. If the axis is the same as the symmetry axis of the cylinder, the torque of F is really Fr. But there is one more force that have torque: static friction. The angular acceleration will depend on the sum of both.

    As you do not know anything about either the magnitude and the direction of the force of friction except that it is parallel to the ground, it is more convenient to calculate the torque with respect to the instantaneous axis of rotation, the line where the cylinder touches the ground. In this case, the torque of friction is zero, and easy to see, in what direction will the cylinder rotate if you change the direction of F and what is the angle when F has zero torque.

    ehild
     
  5. Jul 11, 2010 #4
    EDIT

    Should I assume that the force is always perpendicular to the vector that goes from the axis of symmetry to the point of contact between the rope and the body?

    Thanks for the answers.
     
    Last edited: Jul 11, 2010
  6. Jul 11, 2010 #5
    yes of course because the tangent of a circle is always perpendicular to its radius
     
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