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Acceleration of the center of mass

  1. Oct 14, 2011 #1
    1. The problem statement, all variables and given/known data
    A body of mass m, moment of inertia I, radius R is rolling (without slipping) upon the action of force F applied in horizontal direction at a distance x above center of mass (com).
    Prove that acceleration of com (a) = F[(1+(x/R)]/[m(1+(I/mR2)]
    2. Relevant equations

    F - frictional force = mg
    F.x - (Frictional force).R = Ia/R

    3. The attempt at a solution

    I first decided to eliminate frictional force from equation but its not helping
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 14, 2011 #2
    At any instant in time if your object is not slipping on the surface that it is rolling on, there is something called the instantaneous center. It is at the point where it touches the ground at any instant. What you need to do is sum torques (moments) about this point and set it equal to I*alpha plus a*m*R, where a is acceleration, m is mass, and R is radius. Alpha equals a/R.

    Remember, you have two types of acceleration here. One is linear motion while the other is rotational speed. The answer drops right out after some algebraic manipulation.

    Let's see your work.
     
  4. Oct 14, 2011 #3
    If it rolls without slipping then why does it matter where the force is applied as long as it is in the same direction?
     
  5. Oct 14, 2011 #4
    I m not getting the purpose behind doing that. I need motion equations to solve this thing mathematically.
     
  6. Oct 14, 2011 #5
    You have two types of acceleration that cause inertial torques. One is the translational acceleration of the object moving horizontally. Since it must rotate to do this, there is another inertial torque caused by the rotation. There is acceleration in two forms. One is rotational. The other is translational. Both must be accounted for.

    If it were a frictionless block merely sliding, you would only have F=ma which accounts for translational acceleration. This has rotation as well.

    I * alpha + m * a * R represents the torque inhibiting motion. F * (R + x) is the torque trying to cause motion. Equate them. Recall that alpha = a/R.
     
  7. Oct 14, 2011 #6
    The amount of torque due to the force depends on the moment arm of the force. If it were applied at the top the torque at the instantaneous center would be F * R * 2. At the center, the torque is F * R. Below the center, the torque reverses.
     
  8. Oct 14, 2011 #7
    The torque will be equal. If all the force is applied at the top of the disk lets say, then there would only be "rolling". If it is applied to the centre of the disk there will only be translational momentum. Except that it rolls so all of that translational force will be converted to torque by the friction.
     
  9. Oct 14, 2011 #8
    Another way you can look at this problem is by using the parallel axis theorem where you would move the rotational inertia from the center where it is I to the contact point where it becomes I + m * R^2. Then the equation reads:

    F * (x + R) = (I + m * R * R) * alpha

    alpha = a/R

    F * (x + R) = (I + m * R * R) * a/R

    F * (x + R) = (I/R + m * R) * a

    Solve for acceleration 'a'
     
  10. Oct 14, 2011 #9
    Hmm okay, but this is where I'm stuck.

    No matter where the force is applied, there is no energy lost in the system right? So regardless of the value of x there should be 0 translational energy, and some number of rotational energy. How can two different x values yield different accelerations?
     
  11. Oct 14, 2011 #10
    No energy is lost in the system. But the work done by the force causes the object to rotate as well as move in a line. So there would be two components of kinetic energy. One is rotational KE; the other is translational KE. Think of a thrown football. It spins and it translates. The kinetic energy is the sum of the two kinds of motion. This is what you have here except the football is traveling sideways rather than point first.

    However in this problem we are not concerned with energy. Since nothing is said about distance moved, work/energy principles are not a consideration. We must stick with Newton's Second Law where F=ma and its rotational equivalent, T=I*alpha. The torque contributes to both rotation as well as translational motion.
     
  12. Oct 14, 2011 #11
    the rotational and translational motions are the same thing though aren't they? If it rolls without slipping than all of its rotational motion is translational motion.
     
  13. Oct 14, 2011 #12
    With no slipping, the cylinder must move if it is rotating. Go back to my previous comment about a block of mass M on a frictionless table. If you push on the block with a force, F, then the acceleration of the block is F/M. Now lets change the block to a cylinder that has the same mass as the block just mentioned. It also is on a frictionless table so it slides and does not rotate. With a force F, its acceleration would be F/M also.

    But now let's consider friction where the cylinder has to rotate as well as just move across the table. The acceleration cannot be the same (F/M) because you are getting two types of motion for the same input. You'd be getting something for nothing. The translational acceleration must be less than the previous two cases because you are also accelerating it angularly.
     
  14. Oct 14, 2011 #13
    Thanks a lot, I got it !!!
     
  15. Oct 14, 2011 #14
    Good for you!
     
  16. Oct 14, 2011 #15
    I think I understand now. Basically, if I were to push a cylinder on a surface with a ton of friction it would roll. By my previous way of thinking, it would roll at the same speed at which it would translate given no friction. If this friction-full surface turns into a frictionless surface, then by my old hypothesis it would be translating at the same speed as the frictioneless surface push, and also spinning. So i basically added a bunch of energy. Thanks.
     
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