Chapter 2 Homework

Homework H2.E - Sp24

17 Comments
Problem statement
Solution video

DISCUSSION THREAD

DISCUSSION

Note that with the disk not slipping on the outer cable, and with section BC of that cable being at rest, the velocity of point C on the disk has zero velocity. This means that the disk rolls without slipping on the cable, which is the same is it were rolling without slipping on a fixed vertical wall.

With the above in mind, consider the following:

  • Let M represent the point on the left side of the pulley from which the outer cable leaves contact with the disk. Note that M is moving vertically upward with a speed of vA.
  • Write a rigid body kinematics equation relating the motion of points C and M: vM = vC + ω x rM/C . That equation will produce an expression for ω, the angular speed of the disk.
  • Then, you can use vN = vC + ω x rN/C and  vH = vC + ω x rH/C to find the velocities of point N and H on the disk. Since the inner cable does not slip on the disk, then vN = vD and vH = vQ.
  • Also,  vE = vC + ω x rE/C.

Any questions?? Please ask/answer questions regarding this homework problem through the "Leave a Comment" link above.

17 thoughts on “Homework H2.E - Sp24”

  1. Do we assume that point O is moving, and if so, what is a good way to find its velocity and acceleration?

    1. Because B is fixed, C is also not moving, meaning O must be moving. Something to keep in mind is that there is only one ω since every point is on the same rigid body. And the problem statement does not ask for velocity

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    2. Cable AB is a fixed length. Think about what happens when one end of the cable is pulled upwards. O will have velocity as the disk rotates clockwise and moves upwards with no slipping. You can find it by using the velocity equation that we have been using throughout this semester with either point C or A. However, you should not need to know that to solve any part of the problem because you can use the velocity equation with any arbitrary point, not just the center.

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  2. Does it make sense for the velocities of blocks B and Q to be positive? I'm unsure of the implications of both being positive at the same time. I assume that it has to do with the fact that after the system passes this instant, that points N and H are higher than point C. Is this assumption correct?

    1. I was wondering the same thing as we look at the figure, we find the disk to be rolling clockwise. Intuitively, it would make sense to see block D have a positive velocity upwards, however, block Q would make sense to have a negative velocity as the disk is moving clockwise and the block would be falling downwards.

      1. At first, I was thinking the same thing, but it is important to remember that the entire disk is moving in the positive j direction. If you keep this in mind, even though block Q may be falling downwards, the net motion of block Q could still be positive.

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  3. The best way for me to think about the string pulling the disk was imagining the disk was rolling and point C was tangent to the non-slip ground, making the velocity of that point 0. Point O will have to be moving, so I wouldn't recommend using that as your reference point. Also, the rotation of the body is clockwise, so make sure to pay attention to your signs.

  4. It's helpful when working on this problem to find an equation for the velocity of A as observed by C rather than solving for the values at the center of the disk, or at point O. Essentially, you don't have to use two equations to find omega, as the velocity of point C is zero. This would leave only one unknown in your first velocity equations.

  5. When finding rA/C should we just be using the square root of the diameter plus AM? Or does that not work since AM is unkown?

    1. Keep in mind that the rigid body kinematics equation can be applied to only a pair of points whose distance between them is constant. The distance between A and C is not constant. Therefore, you cannot use the rigid body equation between A and C.

      However, note that the velocity of A and M are the same (where M is defined in the above DISCUSSION). And, points C and M are on the same rigid body.

      1. How are we able to find rA/C if the length of AM is not defined? I don't fully understand how to utilize the velocity of A and M to solve for rA/C

        1. As noted above, you cannot write a rigid body kinematics equation between points A and C since the distance between these points changes with time. The suggestion above is that you use a rigid body kinematics equation between C and M. Then, use the fact that A and M have the same velocity.

          Does that make sense?

    1. Yes, in the hint they told us that the velocity of C is 0, and since B is fixed, its velocity is 0 as well.

  7. Wouldn't the velocity of D and Q be the same value in opposite directions since they are connected? When I run the calculations like in the Discussion portion, I get vN=3/4va and vH=1/4v_a? Thanks

    1. What you say would be true if the disk were rotating about O. Because of the no-slip condition at C, the disk is actually rotating about C. In that case D and Q are both moving upward with different speeds.

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