Question

The London Eye (basically a very large Ferris wheel) can be viewed as 32 pods, each with mass \(m_{\mathrm{p}}\), evenly spaced along the edge of a disk with mass \(m_{\mathrm{d}}\) and radius \(R\). Which of the following expressions gives the moment of inertia of the London Eye about the symmetry axis of the disk? a) \(\left(m_{\mathrm{p}}+m_{\mathrm{d}}\right) R^{2}\) b) \(\left(m_{\mathrm{p}}+\frac{1}{2} m_{\mathrm{d}}\right) R^{2}\) c) \(\left(32 m_{\mathrm{p}}+\frac{1}{2} m_{\mathrm{d}}\right) R^{2}\) d) \(\left(32 m_{\mathrm{p}}+m_{\mathrm{d}}\right) R^{2}\) e) \(\left(16 m_{\mathrm{p}}+\frac{1}{2} m_{\mathrm{d}}\right) R^{2}\)

Short answer

Answer: The moment of inertia of the entire system (London Eye) about the symmetry axis of the disk is given by the expression \(\left(32 m_{\mathrm{p}}+\frac{1}{2} m_{\mathrm{d}}\right) R^{2}\).

Step-by-step solution

Moment of Inertia of the Disk

First, we will find the moment of inertia for the disk. The moment of inertia of a solid disk about an axis perpendicular to its plane and passing through its center is given by the following formula: \(I_{\mathrm{d}}=\frac{1}{2} m_{\mathrm{d}} R^{2}\)

Moment of Inertia of One Pod

Next, we will find the moment of inertia for one pod. Since a pod can be considered as a point particle, its moment of inertia can be calculated using the following formula: \(I_{\mathrm{p}}=m_{\mathrm{p}} R^{2}\)

Moment of Inertia of All the Pods

Since there are 32 pods evenly spaced along the edge of the disk, the total moment of inertia of all the pods can be calculated by summing the moment of inertia of each pod: \(I_{\text { total pods }}=32 I_{\mathrm{p}}=32\left(m_{\mathrm{p}} R^{2}\right)\)

Combined Moment of Inertia of the Disk and Pods

Now we will sum the moment of inertia of the disk and the total moment of inertia of the pods, to find the London Eye's moment of inertia: \(I_{\text { London Eye }}=I_{\mathrm{d}}+I_{\text { total pods }}=\frac{1}{2} m_{\mathrm{d}} R^{2}+32\left(m_{\mathrm{p}} R^{2}\right)\) \(I_{\text { London Eye }}=\left(32 m_{\mathrm{p}}+\frac{1}{2} m_{\mathrm{d}}\right) R^{2}\)

Matching the Answer with the Given Options

The expression we found for the moment of inertia of the London Eye matches with option 'c': \(\left(32 m_{\mathrm{p}}+\frac{1}{2} m_{\mathrm{d}}\right) R^{2}\) Therefore, the correct answer is option c.

Key concepts

Physics of Rotational Motion

The realm of physics that deals with bodies in rotation is not only fascinating but crucial when understanding how objects behave in our world. Rotational motion involves the movement of an object around a central point or axis. This could be as simple as a child's spinning top or as complex as celestial bodies moving through space.

Every rotating body possesses rotational inertia or moment of inertia, which is essentially its resistance to changes in its rotation. Similar to how mass influences how an object resists linear motion changes, the moment of inertia plays that role in rotational motion. It depends on both the mass of the object and the distribution of that mass relative to the axis of rotation. A key thing to note is that the farther the mass is from the axis, the greater the moment of inertia.

Rotational kinetic energy, angular momentum, and torques are other essential aspects of this physics area. These concepts also interact with the principle of conservation of angular momentum, which is pivotal in explaining phenomena such as why a figure skater can spin faster by pulling her arms in toward her body. Understanding the basics of rotational motion is crucial for tackling more complicated problems in physics and engineering.

Moment of Inertia Calculation

The moment of inertia (\(I\)) calculation is critical for understanding how difficult it is to change an object's rotational speed about an axis. It's mathematically represented as the sum of the product of the mass elements and the square of their distance from the rotation axis. The formula varies based on the geometry of the object and the axis about which it rotates.

For point masses, the calculation is straightforward: just multiply the mass (\(m\times\)) times the square of the distance (\(r^2\times\)) to the axis. However, for extended bodies, the integral calculus comes into play as mass distribution must be considered over the volume of the object. The standard moments of inertia for simple shapes like rods, discs, and spheres are often given in physics textbooks and serve as building blocks for more complex shapes.

For our example, the solid disk's inertia and the point-mass pods must be calculated separately and then summed to give a total moment of inertia for the system. In problems such as these, being meticulous with the mass distribution and geometry is crucial for arriving at the correct answer.

Solid Disk Inertia

A solid disk's moment of inertia is one of the classic calculations in physics. The disk is a common object with significant relevance in real-world applications like car wheels, machine parts, and, as seen in our example, Ferris wheels. The inertia of a solid disk is higher than that of a hollow one of the same mass since more mass is distributed further from the center.

The moment of inertia for a solid disk with mass (\(m_{\text{d}}\times\)) and radius (\(R\times\)) rotating about an axis through its center is given by the formula \times\left(I_{\text{d}}=\frac{1}{2} m_{\text{d}} R^{2}\times\right\). This equation shows that the moment of inertia increases with both mass and radius, highlighting the role of mass distribution once more.

Incorporatingthe concept of the solid disk inertia into calculations helps students understand the foundational physics behind rotating bodies. It also enhances their ability to approach problems involving composite systems, like the London Eye example, with more confidence and accuracy.