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Homework Help: Rotating inclined cylinder

  1. Feb 24, 2010 #1
    1. The problem statement, all variables and given/known data

    We have an inclined cylinder, something like a funnel, which rotates around its symmetrical axis. A mass (m) resides on the wall of this cylinder and rotates with the cylinder. Now the angular velocity of the cylinder increases. Whether the newton laws (or else) can anticipate what will happen? (Don't use intuition or your previous experiments)

    2. Relevant equations
    6bb871a3f5ace1ec265a7e0e82f5c5382g.jpg


    3. The attempt at a solution
    I know that the mass remains somewhere on the inclined wall. Now it may go up the wall or may go down or remain at the same place.
    First case: The mass goes up: r (rotation radius) increases, w (angular velocity) increase so it requires a greater centripetal force (mrw^2) and the normal force (N) will provide this force so N will increase.
    Second case: the mass goes down: r decreases, w increases so the centripetal force (mrw^2) can increase, decrease or remain the same consequently the normal force (N) can increase, decrease or remain constant.
    Third case: The mass remains at its place: r constant, w increases so mrw^2 becomes greater. Normal force increases to provide this increase.
    I think nothing theoretically prevents these cases to occur.
     
    Last edited: Feb 24, 2010
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  3. Feb 24, 2010 #2

    rl.bhat

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    If theta is the angle made by the inclined wall with the horizontal, mg*sin(theta) is the force acting on the body along the inclined plane in the down ward direction. And this force in constant. Now the centripetal force is m*w^2*r acts towards towards the axis of rotation. Its one component is normal reaction and the other component m*w^2*r*cos(theta) is along the inclined plane in the upward direction. It changes with w. These two components will decide the motion of the body.
     
  4. Feb 24, 2010 #3
    thanks. but don't forget that centripetal force is not a real force. That is, when you draw free diagram of forces you don't include such a force. In free diagram you just enter Normal force, Gravity and friction (if exists). Then you set the sum of these vectors to m*a which a is the acceleration and you know that is toward the axis.
    f4fa6370349c457f4277db7a3831f1232g.jpg
    free diagram
    In the above diagram the forces are shown in red and acceleration in green.
    Newton's second law:
    ∑ F = ma
    mg + N = ma (mg,N,a are vectors)
    magnitude of a = mrw^2
     
  5. Feb 24, 2010 #4

    rl.bhat

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    Can you identify the forces winch keep the body at rest on the rotating inclined plane?
     
  6. Feb 24, 2010 #5
    yes
    the N vector can be shown by the sum of two components:
    Nsin(theta)*i + Ncos(theta)*j
    which i and j are unit vectors of cartesian coordination.
    we can also write mg like:
    -mg*j
    also the vector representation of acceleration is:
    -a*i
    Then we write the sum of forces:
    F = ma (F and a are vectors)
    N + mg = ma (N,mg,a are vectors)
    Nsin(theta)*i + Ncos(theta)*j + (-mg)*j = -a*i
    then we have
    Ncos(theta) - mg = 0
    Nsin(theta) = -a = rw^2
    then:
    N = mg/cos(theta)
    we can see that this is Nsin(theta) that creates the centripetal acceleration.
    from above equation we can conclude that N is constant (independent of the position of the mass on the funnel plate)
    N is constant so the left hand side of the equation:
    Nsin(theta) = rw^2
    is constant. Now if w increases for being constant the r must decrease. So we conclude that the mass will goes down which does not seem reasonable.
     
  7. Feb 24, 2010 #6

    rl.bhat

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    The block is moving along the inclined surface of the funnel. It never leaves the surface of the funnel. Now identify the components of the forces acting along the inclined surface of the funnel. If the funnel is not rotating, the block placed on the inclined surface will slide down. Which force causes this slide? The rotation causes the additional force acting on the block. What is that force? What is its direction?
     
    Last edited: Feb 24, 2010
  8. Feb 25, 2010 #7
    evidently this is m*g*sin(theta) and its direction is parallel to the inclined surface of the funnel. Any way we have only two forces mg and N.
     
  9. Feb 25, 2010 #8

    rl.bhat

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    To keep the body at rest on the inclined plane, there must be a force equal and opposite to m*g*sin(theta). This force must depend on the angular velocity.
    When you are sitting in a moving car and the car is taking a sharp left turn, you are thrown toward right due to inertia. Similarly the block in the rotating funnel will be accelerated away from the axis. Its one component will contribute to N and the other component is along the inclined plane in the upward direction.
     
  10. Feb 25, 2010 #9
    yes you are right. But pay attention that from an inertial observer's point of view (for example someone stands on the ground) there are only two forces: N and mg. Ok you may say "so, how the block can remain at its place without sliding" and the answer is: "For someone outside the funnel i.e. someone on the ground the block is not moveless, it has a movement with acceleration towards the center which we call centripetal acceleration"
    Now we consider a non-inertial observer; "someone resides on the funnel". This observer senses a force outward the center of rotation which we call centrifugal force. this force exists only for non-inertial observer i.e. someone on the funnel and so is an imaginary force.
    Anyway
    but you didn't answer my question, with increase of angular velocity what happens? the block goes up, down or ....?
     
  11. Feb 26, 2010 #10

    rl.bhat

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    The object is not glued to the funnel. To keep it in the circular motion, some one must push it towards the center continuously. Due to inertia, the object in the rotating funnel has the tendency to move in a straight line with velocity V. But the walls of the rotating funnel pushes it with some force towards the center. If the velocity of the block is constant, the magnitude of this force is m*w^2*r. Resole this force into two components. m*w^2*r*cos(theta) is along the inclined plane in the upward direction and m*w^2*r*sin(theta) perpendicular to the surface. As w increases, m*w^2*r*cos(theta) also increases. If this force is greater than m*g*sin(theta), the block will go up. Other wise it will go down.
     
  12. Feb 26, 2010 #11
    OK but "some force towards the center" is not a force that comes magically when the mass rotates. This force is simply the sum of two forces; gravity and normal force. you may ask the normal and gravity forces exist when the block does not rotate. Yes but the magnitude of N is different when the block rotate so the wall of the funnel senses a greater pressure when it rotates.
    And about the main question I think if there is not friction between the funnel and the block, when the velocity of the funnel increases the velocity of the block will not change because there is no means to transfer this increase.
     
  13. Feb 26, 2010 #12

    rl.bhat

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    When the funnel is not rotating, the block will be at the bottom of the inclined wall of the funnel. When the funnel starts rotating, do you mean that the block remains at rest at the bottom?
    You have mentioned that "the magnitude of N is different when the block rotate so the wall of the funnel senses a greater pressure when it rotates." Which force exerts the pressure on the wall of the funnel when it is rotating? Precisely one component that force itself pushes the block along the inclined wall of the funnel. The block will occupy a stable orbit when m*w^2*r*cos(theta) = m*g*sin(theta). Again you have mentioned that "the force on the block is simply the sum of two forces". Where is the reaction to this force?
     
  14. Feb 27, 2010 #13
    you said"
    "When the funnel starts rotating, do you mean that the block remains at rest at the bottom?"
    my answer: yes I think so. If there is not any friction then the block will remain at the bottom of the funnel.
     
  15. Feb 27, 2010 #14

    rl.bhat

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    In that case how is it possible to keep the block inside the funnel as shown in the figure?
    Are you keeping the block in the rotating funnel?
    If you keep a coin on a frictionless disc and rotate it, do you think that coin remains at rest?
     
  16. Feb 27, 2010 #15

    Cleonis

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    Let me first discuss another case: not a cone but a surface with parabolic cross section. A liquid spinning around a vertical axis assumes a parabolic shape. (Which is why the mirrors for telescopes are cast inside a rotating oven. It's also the operating principle of Mercury mirror telescopes.)

    When a surface has a parabolic shape then at every distance to the center of rotation the slope of the surface provides the required centripetal force. You can create a parabolic surface, and when you later spin that at exactly the original angular velocity, then an object will remain at rest wrt the cone at every distance to the central axis.

    Now to the case of the cone, as submitted by the original poster.
    For every angular velocity there will be exactly one distance to the central axis where the object on the cone will experience precisely the required amount of centripetal force. Any lower and it will slump down, any higher and it will climb up and slide out.
    So just the case of the object in the cone is a knife-edge situation. You can set up an equilibrium state, but it's unstable equilibrium.

    So what will happen if you increase the rotation rate of the cone? Assume that some force keeps the angular velocity of the object in step with the angular velocity of the cone, while allowing free motion in radial direction.
    If the angular velocity increases then the centripetal force arising from the slope will not be strong enough. The object will climb up, eventually dropping out of the cone.

    Note the extra assumption: that at all times a force keeps the objects angular velocity in step with the cone's angular velocity. You get a different scenario if only a single impulse is given to the object, making only the initial angular velocity the same. As the object climbs up the slope it loses angular velocity again.
     
    Last edited: Feb 27, 2010
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