Problem statementSolution video |

**DISCUSSION THREAD**

Any questions??

**DISCUSSION**

With the bar welded to the disk, the two bodies move as a single rigid body. That is, they have the same angular velocity as the disk rolls without slipping. For position 2 shown below, the velocities of O (center of mass of the disk) and G (center of mass of the bar) are in opposite directions and appear to have nearly equal speeds. Do they have *exactly* the same speed? (Consider the location of the instant center for the disk as it rolls.)

As always, we should follow the four-step plan for solving this problem.

* STEP 1: FBD. *We will be using the work/energy equation to solve this problem since we are looking for a change in speed for a change in position. Based on earlier recommendations, we will make the choice of our system BIG, including the disk and bar together.

*(here, work/energy). The total kinetic energy of the system shown in your FBD above is that of the bar + disk. For the disk, it is recommended that you choose the no-slip point as your reference point for the KE. For the bar, you could use either C or G for the reference point for its KE. Be reminded that your choice of reference points dictates the mass moments of inertia that are to be used for the rigid bodies. Be sure to identify the datum line for the gravitational potential energy, and use this in writing down this potential. Is energy conserved?*

**STEP 2: Kinetics***As described above, the IC for the bar+disk is the no-slip point C.*

**STEP 3: Kinematics.**

**STEP 4: Solve**

How are we supposed to relate the moment of inertia of the center of gravity of the disk to the no-slip point at C ?

You can use the parallel axis theorem for the disk about point C using the radius R

Would the kinetics equation be vg = omega ( (L/2) - R) since C is the IC?

Sorry I meant kinematics oops