The question concerns a wheel rolling without slipping along a circular disk and involves calculating the angular motion and angular momentum of a system containing point masses. Let's analyze and answer both parts step by step.

Part (a): How many rotations does the wheel make?

The wheel with radius $r$ rolls without slipping on the edge of a circular disk with circumference $6\pi r$.

Key Concept:

When a wheel rolls without slipping:

• The distance traveled along the path equals the arc length the wheel covers.
• The circumference of the wheel is $2\pi r$, meaning it covers this distance in one full rotation.

Given:

• The circular disk's circumference $C_{\text{disk}} = 6\pi r$,
• Wheel's circumference $C_{\text{wheel}} = 2\pi r$,

The number of rotations $N$ the wheel makes is given by: $N = \frac{\text{Distance traveled (disk circumference)}}{\text{Wheel circumference}}.$

Substitute the values: $N = \frac{6\pi r}{2\pi r} = 3.$

Important Clarification: Although the wheel travels 3 times its own circumference, it undergoes an additional rotation due to rolling along a curved path. This is a result of the "geometric phase" effect. Therefore, the total rotations are:

$N_{\text{total}} = 3 + 1 = 4 \, \text{rotations}.$

Part (b): Calculate the angular momentum $L$ of the wheel.

Given Values:

• Mass of each point: $m = 10 \, \text{kg}$,
• Radius of the wheel: $r = 0.1 \, \text{m}$,
• Two point masses are located at the edge of the wheel (radius $r$).

The wheel rolls without slipping with one complete rotation per second. The angular velocity $\omega$ is therefore:

$\omega = 2\pi \, \text{rad/s}.$

Moment of Inertia:

For point masses on a rotating wheel: $I = \sum m_i r_i^2.$

Since there are two identical point masses ($m_1 = m_2 = 10 \, \text{kg}$) at radius $r = 0.1 \, \text{m}$: $I = 2 \cdot m r^2.$

Substitute the values: $I = 2 \cdot 10 \cdot (0.1)^2 = 2 \cdot 10 \cdot 0.01 = 0.2 \, \text{kg·m}^2.$

Angular Momentum:

The angular momentum $L$ is given by: $L = I \omega.$

Substitute $I = 0.2 \, \text{kg·m}^2$ and $\omega = 2\pi \, \text{rad/s}$: $L = 0.2 \cdot 2\pi = 0.4\pi \, \text{kg·m}^2/\text{s}.$

Numerical Value:

Approximating $\pi \approx 3.14$: $L \approx 0.4 \cdot 3.14 = 1.256 \, \text{kg·m}^2/\text{s}.$

Final Answers:

1. Part (a): The wheel makes 4 rotations when it rolls once around the circular disk.
2. Part (b): The angular momentum of the wheel is approximately: $L = 1.256 \, \text{kg·m}^2/\text{s}.$

Do you need a more detailed breakdown or additional explanations? Here are some related questions:

1. Why does the wheel undergo an additional rotation while rolling on a curved path?
2. How would the angular momentum change if the radius of the wheel were doubled?
3. What is the physical explanation for the "geometric phase" effect in part (a)?
4. How would you calculate the angular momentum if the wheel had a continuous mass distribution?
5. How does rolling without slipping differ from pure rotation or pure translation?

Tip: When analyzing rotational motion, always consider both the geometric path and the rotational phase to avoid underestimating rotations!