Question

A woman with mass 50 kg is standing on the rim of a large disk that is rotating at 0.80 rev/s about an axis through its center. The disk has mass 110 kg and radius 4.0 m. Calculate the magnitude of the total angular momentum of the woman\(-\)disk system. (Assume that you can treat the woman as a point.)

Short answer

The total angular momentum of the woman-disk system is approximately 8424 kg m²/s.

Step-by-step solution

Understand the System

The system consists of a woman and a disk, both rotating about a common axis. The woman can be treated as a point mass, and the angular momentum of both the woman and the disk must be calculated and then summed to find the total angular momentum.

Calculate Disk's Angular Momentum

The moment of inertia (\(I_d\)) for a disk is given by \(I_d = \frac{1}{2} m_d r^2\), where \(m_d = 110\, \text{kg}\) and \(r = 4.0\, \text{m}\). The angular velocity (\(\omega\)) is \(0.80 \times 2\pi\, \text{rad/s}\). So, \(I_d = \frac{1}{2} \times 110 \times 4^2 = 880 \; \text{kg} \, \text{m}^2\). Then, the angular momentum of the disk, \(L_d = I_d \omega = 880 \times 0.80 \times 2\pi = 4402.89 \; \text{kg} \, \text{m}^2/\text{s}\).

Calculate Woman's Angular Momentum

Treat the woman as a point mass, so her moment of inertia is \(I_w = m_w r^2\), where \(m_w = 50\, \text{kg}\) and \(r = 4.0\, \text{m}\). Therefore, \(I_w = 50 \times 4^2 = 800 \; \text{kg} \, \text{m}^2\). Her angular momentum, \(L_w = I_w \omega = 800 \times 0.80 \times 2\pi = 4021.24 \; \text{kg} \, \text{m}^2/\text{s}\).

Find Total Angular Momentum

The total angular momentum \(L_{total}\) is the sum of the woman's and the disk's angular momentum. So, \( L_{total} = L_d + L_w = 4402.89 + 4021.24 = 8424.13 \; \text{kg} \, \text{m}^2/\text{s}\).

Key concepts

Moment of Inertia

The moment of inertia is like the rotational equivalent of mass in linear motion. It tells us how hard it is for an object to change its rotational speed. Simply put, it depends on the mass of the object and how that mass is distributed relative to the axis of rotation.
To calculate the moment of inertia (I), you often use specific formulas based on the object's shape and mass distribution. For a solid disk, the formula is:
\[ I_d = \frac{1}{2} m_d r^2 \]
where \( m_d \) is the mass of the disk and \( r \) is its radius. In our step-by-step solution, the disk's moment of inertia is 880 \( \text{kg} \, \text{m}^2 \).
Similarly, when treating the woman as a point mass, her moment of inertia about the axis is:
\[ I_w = m_w r^2 \]
understanding these calculations helps when you need the rotational characteristics in tasks involving spinning objects.

Rotational Dynamics

Rotational dynamics studies how forces can affect the rotational movement. It involves concepts like torque and angular momentum, which are essential for predicting motion in systems that rotate.
Angular momentum, a central concept in rotational dynamics, tells us about the rotation of an object. It's determined by both the rotational inertia and its angular velocity. For any rotating object, like the disk in our problem, the formula is:
\[ L = I \omega \]
Here, \( L \) is the angular momentum, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity.
Practical understanding of rotational dynamics is useful in areas like engineering and physics; it explains how athletes spin faster by pulling their arms in or when an object, like our disc, conserves angular momentum when no external torques are applied.

Point Mass

In physics, we often approximate objects by treating them as a point mass to simplify problems. A point mass assumption means you consider all of an object's mass concentrated at a single point.
In the given exercise, the woman is treated as a point mass while calculating the moment of inertia. This allows for easier math because the distribution of mass is simplified into a single spot, making moment calculations straightforward:
\[ I_w = m_w r^2 \]
where \( m_w \) is the mass of the woman and \( r \) is the distance to the axis.
This simplification is particularly useful for objects at a distance from the axis or when their shape doesn't significantly affect the mass distribution about the axis. However, for extended bodies where mass distribution is significant, more complex methods might be needed for precision.