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Cylinder over a platform; force on the platform

  1. Jan 18, 2012 #1
    1. The problem statement, all variables and given/known data
    I have a cylinder of mass M and radius R, it stands over a platform of mass m; the platform stands over a no-friction floor. There is a given force F on the platform, directed horizontally on the right. The coefficient of static friction between the cylinder and the platform is μ. The cylinder has a pure roll motion relatively to the platform. I want to know
    a) the accelerations of the platform and of the cylinder.
    b) The value of the friction force
    c) The maximum force F to be applied to the platform for having a pure roll of the cylinder

    I can't set the equations of motion properly, can someone help me? Thanks
     
  2. jcsd
  3. Jan 18, 2012 #2
    I would recommend drawing a picture. Identify the forces acting on each object(i.e. the platform and cylinder), and then apply Newton's second law.
    [itex]F_{net}=ma=F_{1}+F_{2}+...[/itex]
    My technique is to consider all the forces acting on an object that (a)don't "touch" it, like gravity or electrical forces, and (b) forces that do touch it, like a bat hitting a ball or a frictional force. That helps me to make sure I don't miss a force.
    In the problem, I see that the platform experiences gravity, the normal force from the floor, the given force F and the frictional force from the cylinder.
    Hope this helps to get you started.
     
  4. Jan 19, 2012 #3
    Thanks, but i've already done this things. I am convinced that i'm doing somethig wrong.
    Because my equations for the platoform are
    F - F[itex]_{fric}[/itex] = m a[itex]_{plat}[/itex]
    N = m*g + N[itex]_{1}[/itex]

    where N[itex]_{1}[/itex] is the normal reaction on the cylinder, and N is the normale reaction of floor to the platform.

    Now my problem, the equations for the cylinder
    F[itex]_{fric}[/itex] = M a[itex]_{cyl}[/itex]
    F[itex]_{fric}[/itex] * R = torque = I * [itex]\alpha[/itex]
    The exercise says the the motion is a pure roll, so that [itex]\alpha[/itex] = a[itex]_{cyl}[/itex] / R
    I is 1/2MR², so I get from the second equation
    F[itex]_{fric}[/itex] = 1/2Ma[itex]_{cyl}[/itex]

    How it's possible? the other one says F[itex]_{fric}[/itex] = Ma[itex]_{cyl}[/itex] ???
     
  5. Jan 19, 2012 #4
    You are getting ahead of yourself a little bit. Parts (a) and (b) will not involve rotational issues, just Newton's Second Law.
    Your equations of motion for the platform look good with a subtle tweak. Specifically, [itex]N_{1}[/itex]is the force of the platform on the cylinder, the force of the cylinder on the platform is Mg.
    Now you need to find [itex]F_{fric}[/itex]. Do you know the equation for that?
    Your first equation for the horizontal motion of the cylinder is correct. The problem I think is the statement "pure roll". The original post said the "The cylinder has a pure roll motion relatively to the platform." This is a way of saying the cylinder does not slip on the platform. Hope this helps.
     
  6. Jan 19, 2012 #5
    I'm so sorry but I can't figure it out.

    I think that having a pure roll RELATIVELY to the platform, just means that its center of mass isn't moving, it only rotates about the CM. But for a[itex]_{cyl}[/itex] to be 0, F[itex]_{fric}[/itex] also has to be 0, and this is impossible.

    thank you for help, i forget it before :)
     
  7. Jan 19, 2012 #6
    I will provide a summary of what's been done so far;
    The platform:
    Horizontal forces.
    [itex]F-F_{fric}=ma_{plat}[/itex]
    Vertical forces.
    [itex]N=mg+N_{1}[/itex]
    as I stated in post 4, this [itex]N_{1}[/itex]is really just Mg. One way to think of it is that the normal force of the floor must support the weight of both the platform and the cylinder.
    Now we have,
    [itex]N=mg+Mg=(m+M)g[/itex]
    Also in post 4 I mentioned the equation for frictional force, its general form is F=Nμ.
    Do you see how you can use this to find [itex]a_{plat}[/itex]? A similar analysis will give you the acceleration of the cylinder and the force due to friction will be found with the frictional force equation.
     
  8. Jan 19, 2012 #7
    I used to know that μ[itex]_{s}[/itex]N is the maximum frictional force when it is a STATIC frictional force, but F[itex]_{fric}[/itex] can be everything, from 0 to μN. The kinetic frictional force is always μ[itex]_{k}[/itex]N.
    If the cylinder moves in pure-roll, the force must be a static friction, doesn't it?
     
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