We've been instructed to use energy concepts, so let's define the top of the ramp as our initial position and the bottom of the ramp as our final position. Then identify the energies that are present at each of those positions.

We've been told that this is a sphere that's rolling, so we can use the known moment-of-inertia for a sphere:

So this is the velocity of the sphere at the bottom of the ramp. Note that neither the mass nor the radius of the rolling sphere factor into the final velocity of the sphere at the bottom. Interesting!

Assuming that the sphere is accelerating constantly (and it is, as we'll soon see), we can use kinematics to identify its acceleration as it rolls.

This problem is asking the same questions as the previous one, but we're using a force-and-torque analysis instead of energy. That, of course, requires a free-body diagram for the rolling object, shown here.

Because the ball is accelerating down the ramp, we choose to tilt our axes, so positive-x is down the ramp. Forces acting in that direction include the force-parallel due to gravity and the force of friction between the ball and the ramp, which both inhibits the motion of the ball down the ramp and acts as a torque around the axis of rotation, causing the ball to start to rotate.

You might be tempted to substitute in f_static = μF_Normal, but recall that μ_static represents a "maximum stickiness" that is possible; the f_static in the problem at this point is not necessarily that maximum value. At this point we don't know what that friction force actually is, nor what the acceleration of the sphere is. We'll need to turn to another strategy to get additional information.

Substituting this into the previous equation:

This matches the result from our energy-based analysis above.

From here, it's simple enough to use this acceleration value with kinematics to identify the velocity of the ball at the bottom of the ramp, a result which also matches that arrived at above.

For a given ramp angle θ a certain amount of "stickiness" is required for the sphere to be able to roll. Let's calculate exactly how much.

Using information from our calculations above:

If the actual μ_static is smaller than this value, the sphere will be slipping as it rolls, losing some energy to frictional heat losses. As a result, it will have less total K by the time it reaches the bottom of the ramp. Its rotational speed will be less (why? Do a τ = Iα analysis!) but its translational speed will be greater (why? Do an F_net = ma analysis!).

One strategy for solving this problem consists of repeating the acceleration calculation that we just performed on the sphere, applying that analysis to the hoop (I_hoop = MR²), and the solid cylinder (I_cylinder = (1/2) MR²).

We've already determined in our analysis of the sphere above that the mass and the radius of the rolling object are not a factor in its acceleration.

1. The magnitude of the angular momentum can be calculated using the values given in the problem: The direction of the angular momentum, using the coordinate system shown in the diagram above, is determined using the Right Hand Rule. That direction is "counterclockwise" as we view the situation, but a more precise answer would be "out of the page" or "in the positive-z direction."
2. First find the angular velocity of the particle, then use that relationship to calculate L. Notice that this result is a rotational analog of the relationship p = mv.

The angular momentum is originally calculated as follows:

What about three seconds later? Let's first identify where the particle is three seconds later... ... and then calculate the "new" angular momentum L.

Note that this value is the same as the value we had before. The angular momentum of the particle relative to the origin remains unchanged.

This is similar to the way the linear momentum of a moving particle remains unchanged... unless, of course, a net force acts on it, providing an impulse to change its momentum.

Using the angular momentum for a disk:

1. To find L as a function of ω we simply need to identify the moment of inertia of the system, because Let's do that then: Thus, we end up with:
2. To get the angular acceleration, let's use a torque analysis:

1. I_total = I_disk + I_student
   I_total = (1/2)(100)(2)² + (60)(2)² = 440 kg m²
2. ω = 4.09 rad/s
3. K_i = 880 J, K_f = 1.8e3 J.
4. As student walks across the platform, they are doing work on the system, which adds energy.
   Follow-up question: If the student started closer to the middle of the platform and walked to the edge, would they still be doing work on the system, adding energy to it? Or would the system be doing work on them, and thus "losing" mechanical energy?