Chapter 5 Homework

Homework H5.K – Sp26

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Problem statement
Solution video

DISCUSSION THREAD

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Discussion

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The animation shown above demonstrates the relationship between the translation velocity of the disk’s center A and the angular velocity of the disk. Recall that the disk is initially projected to the right with an initial velocity of A and a counterclockwise rotation rate; that is, the disk is given a “backspin” on release.

As the disk slides, the friction with the floor does two things: it slows down the translational motion of the disk, and exerts a clockwise moment on the disk reducing the rotational rate of the disk. At some point the rotation rate changes from counterclockwise (positive) to clockwise (negative). And shortly after that, slipping ceases to exist between the disk and the floor. Rolling without slipping starts at the point where the plots of velocity of A and angular velocity of the disk go “flat”.

 

HINT: As always, we should follow the four-step plan for solving this problem.
STEP 1: FBD. Here, draw a free body diagram of the disk.
STEP 2: Kinetics (here, linear and angular impulse momentum equations).
STEP 3: Kinematics. Slipping ceases when the velocity of the contact point on the disk with the ground goes to zero. At that point: vA = ω x rA/C, where C is the point on the disk in contact with the ground.
STEP 4: Solve

 

15 thoughts on “Homework H5.K – Sp26”

      1. I am confused why the FBD would remain the same between the states. Does the frictional force acting at the point of contact with the ground simply always act in the opposite direction of movement of the center of mass of the disk?

        1. While I think frictional force does in fact change directions, I don’t think it matters. This is because we are looking at the velocity at the point where friction flips, and the time it takes until the friction flips. Thus, we only need the frictional force when the disk is spinning counterclockwise.

  2. In this problem I found it easiest to set up equations for angular and linear momentums, then solve for time first. To do this I solved for w2 in terms of time from angular momentum and similarly put v2 in terms of time by relating that and angular velocity w2. To finish the problem I solved for time in the linear momentum equation then plugged it back in to get v2.

    1. If you take the moment about A, do friction and the normal force have to be taken accounted for? Should I take the moment about C, the point of contact?

      1. I am pretty sure friction would need to be accounted for but normal force goes through point A so should not be in the moment equation. Also, since point C is slipping, it is neither fixed nor the center of mass of the disk and cannot be used

      2. For a moment around A, the friction force does create a moment the force is acting perpendicular to the radius vector. But the normal force does not, since the force is directed through point A. Knowing this, I found it much easier to take it about A when solving.

  3. Could someone reexplain the parallel axis theorem we have to use for this case with the bar. I think I might have messed up my equation and forgot how to apply it here.

    1. I did not use the parallel axis theorem for this problem, but in general this is how it works: You start at the center of gravity, and find the polar moment of inertia at the center of gravity. Then you take the point you want the moment to be about, and add the mass (distance between the points)^2 to the original moment. So the general formula for the polar moment about any point A would be (I_G) + (mass * (distance_AG)^2)

    1. I believe friction opposes the intended motion. I like to ignore rotation when I think about it and imagine a case where there’s no friction. In that scenario the disk would want to slide to the right (no rotation because there is no friction), so friction is acting to the left

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