Homework H2.E - Sp23

Problem statement
Solution video


We encourage you to interact with your colleagues here in conversations about this homework problem.

Discussion and hints

From the simulation results above, we see that point A travels on a cycloidal path, with the velocity vector for A being tangent to this path, as expected. In addition, the acceleration of A points inward to the path (again, expected).  The angle between v and a is initially obtuse, implying that A is initially decreasing in speed. At some point, this angle because acute indicating that the speed of A begins to increase. In fact, this rate of change of speed becomes very large as A approaches the surface on which the disk rolls. Do you know why?

The velocity analysis is a straight-forward application of our rigid body kinematics equations where we write a velocity equation for each rotating member:

vA = vC + ωdisk x rA/C =ωdisk x rA/C
vA = vB + ω
AB x rA/B = vBi + ωAB x rA/B

From these, you can solve for ωdisk and ωAB.

Applying the same procedure to acceleration:

aA = aCαdisk x rA/C - ωdisk2 rA/C
aA = aB + α
AB x rA/B - ωABrA/B

produces too few equations for the number of unknowns. It is recommended that you also use the following equation in your acceleration solution:

aO = aC + αdisk x rO/C - ωdisk2 rO/C


32 thoughts on “Homework H2.E - Sp23”

    1. I believe that a_c != 0. Instead a_o = 0, which can be used to calculate a_c using equation 3, leaving 2 equations and 2 unknowns if the other two equations.

          1. I believe that ao is zero in the j direction since it is moving on a horizontal surface and ac is zero in the i direction since it is the contact point. This would mean that the j component of aoj = 0 and ac = acj. You can then find acj in terms of omega disk and R. You can then plug this in for ac in a prior acceleration equation. This is at least how I thought to go about it.

              1. Emily: Please review pp. 91-92 in the lecture book. This gives a brief review on rolling without slipping. That provides an explanation regarding your question.

    1. Why do you say that alpha_disk = 0? The acceleration equation that you mention relating the motion of O and C has three unknowns: a_C, a_O and alpha_disk. You cannot solve for these three unknowns with only the two scalar equations that come from your acceleration equation. As suggested in the discussion above, you need to also relate the acceleration of points C and A.

      1. I think I am still a bit confused. Now we have four unknowns in three equations. Is there an assumption we are suppose to make about a_O? Is it only in one direction, not both?

        1. Not sure of your confusion. You have three unknowns. And you have three equations. You can solve these for the three unknowns.

          Point O moves along a straight path (rho = infinity, or 1/rho = 0) parallel to the surface on which the disk rolls. Therefore, it does not have a component perpendicular to the path. The acceleration of O is in only one direction. Please note that all of this is NOT an assumption, it is a consequence of the motion.

  1. Do we need a 3rd equation for the velocities to find omega_ab and omega_disk? Because we have two unknowns and only 2 equations, or am I missing something?

        1. I had the same mistake originally but splitting the equations into i and j components allowed me to isolate one of the variables into all known or easily calculated values.

        2. Once you have the equations relating A with B and C, you can set the i^ and j^ components of each equation together, leaving you with two unknowns (the angular velocities) and two equations to solve them with.

    1. aO is zero only under special circumstances: when O is traveling at a constant speed along a straight-line path (which, in turn, corresponds to zero angular acceleration of the no-slip disk). Here, although O is traveling along a straight path, it is not given that O has a constant speed.

      In general, one should avoid making any assumptions on motion for any problem. Assumptions (such as zero acceleration) are NOT needed since you can use the kinematics equations in order to determine the result.

    1. For a disk rolling without slipping on a stationary surface, the acceleration of the no-slip point C is in the direction of the center of the disk O and has a value of r*omega^2. You do not need to memorize this - just work it out: write out the acceleration equation relating points O and C.

  2. Is it just me or was anybody else's acceleration solution a super long equation? I think I got it right (positive alpha disk, negative alpha link), but I was just curious if that's how it was for other people.

    1. Yes, this is what I had. I found the acceleration for O in terms of C, then went from O to A, then from A to B, and using the knowledge that B has no acceleration, I worked my way backwards all the way to O. This process took a while (and I am not quite sure that I did it correctly.) but I understand the basic idea of it.

    1. Yes. B is traveling along a straight-line path at a constant speed. This says that both v_dot = 0 and rho = infinity, making both the tangential and normal components of a_B = 0.

    1. It might help to use the 3x3 matix techique for cross products in order to avoid sign errors. In my experience it helps me avoid careless mistakes

Leave a Reply