Question

Consider a rigid body rotating with angular velocity \(\omega\) about a fixed axis. (You could think of a door rotating about the axis defined by its hinges.) Take the axis of rotation to be the \(z\) axis and use cylindrical polar coordinates \(\rho_{\alpha}, \phi_{\alpha}, z_{\alpha}\) to specify the positions of the particles \(\alpha=1, \cdots, N\) that make up the body. (a) Show that the velocity of the particle \(\alpha\) is \(\rho_{\alpha} \omega\) in the \(\phi\) direction. (b) Hence show that the \(z\) component of the angular momentum \(\ell_{\alpha}\) of particle \(\alpha\) is \(m_{\alpha} \rho_{\alpha}^{2} \omega .\) (c) Show that the \(z\) component \(L_{z}\) of the total angular momentum can be written as \(L_{z}=I \omega\) where \(I\) is the moment of inertia (for the axis in question),$$I=\sum_{\alpha=1}^{N} m_{\alpha} \rho_{\alpha}^{2}$$

Short answer

The velocity is \( \rho_{\alpha} \omega \); \( \ell_{\alpha} = m_{\alpha} \rho_{\alpha}^{2} \omega \); \( L_{z} = I \omega \).

Step-by-step solution

Understanding Particle Velocity

In cylindrical coordinates, the position of a particle \( \alpha \) is given by \( (\rho_{\alpha}, \phi_{\alpha}, z_{\alpha}) \). The body is rotating about the z-axis with angular velocity \( \omega \). In a rotational system, the tangential velocity \( v_{\alpha} \) of a particle at a distance \( \rho_{\alpha} \) from the axis is given by \( v_{\alpha} = \rho_{\alpha} \omega \). This velocity is oriented in the \( \phi \) direction, perpendicular to the radius vector.

Angular Momentum of a Particle

The angular momentum \( \vec{\ell}_{\alpha} \) of a particle \( \alpha \) in a rotational system about the z-axis is given by \( \vec{\ell}_{\alpha} = \vec{r}_{\alpha} \times m_{\alpha} \vec{v}_{\alpha} \). The position vector \( \vec{r}_{\alpha} \) in cylindrical coordinates is \( \vec{r}_{\alpha} = (\rho_{\alpha} \cos \phi_{\alpha}, \rho_{\alpha} \sin \phi_{\alpha}, z_{\alpha}) \) and the velocity \( \vec{v}_{\alpha} = (0, \rho_{\alpha}\omega, 0) \). The cross product \( \vec{r}_{\alpha} \times m_{\alpha} \vec{v}_{\alpha} \) results in \( \ell_{z\alpha} = m_{\alpha} \rho_{\alpha}^{2} \omega \) since only the z-component of angular momentum is considered.

Total Angular Momentum

The total angular momentum about the z-axis \( L_{z} \) is the sum of the angular momenta of all particles: \( L_{z} = \sum_{\alpha=1}^{N} \ell_{z\alpha} = \sum_{\alpha=1}^{N} m_{\alpha} \rho_{\alpha}^{2} \omega \). This can be factored as \( L_{z} = \left( \sum_{\alpha=1}^{N} m_{\alpha} \rho_{\alpha}^{2} \right) \omega \). The term \( I = \sum_{\alpha=1}^{N} m_{\alpha} \rho_{\alpha}^{2} \) is the moment of inertia for the axis in question. Thus, \( L_{z} = I \omega \), demonstrating the relationship between total angular momentum, moment of inertia, and angular velocity.

Key concepts

Rigid Body Dynamics

When we talk about rigid body dynamics, we are focused on how solid objects move and rotate without changing their shape. Think of an object like a spinning top or a rotating door; these are examples of rigid bodies. In these cases, the body's particles move as if they are fixed relative to each other, and the body rotates around an axis.

The behavior of objects under rotational motion can be studied by analyzing angular velocity and angular momentum. In our exercise, the object is rotating about the z-axis with a constant angular velocity, denoted as \( \omega \). Angular momentum is a key concept here and is a measure of the extent to which an object will continue to rotate around its axis. It depends on both the distribution of the mass of the body and the angular velocity.

In the case of multiple particles making up the rigid body, each contributes to the body's total angular momentum, and we add these contributions together to find the overall effect on the motion.

Moment of Inertia

The moment of inertia can be thought of as the rotational equivalent of mass in linear motion. It determines how much torque is required for the object to achieve a certain angular acceleration. In our problem, each particle of the rigid body has a "resistance" to change in rotational motion, which is defined by its moment of inertia.

Mathematically, we calculate the moment of inertia \( I \) about an axis by summing up the product of each particle's mass \( m_{\alpha} \) and the square of its distance from the axis \( \rho_{\alpha}^2 \). This is given by the formula:

\[ I = \sum_{\alpha=1}^{N} m_{\alpha} \rho_{\alpha}^{2} \]

This equation tells us that the further a particle is from the axis, the more it contributes to the moment of inertia. Therefore, larger \( \rho \) values significantly increase the moment of inertia. This value is crucial in the equation \( L_{z} = I \omega \) where it connects the rotational inertia with angular momentum and velocity.

Cylindrical Coordinates

Cylindrical coordinates are very useful for problems involving rotations around a single axis, particularly the z-axis. This system uses three parameters to define the position of a particle: radial distance from the axis \( \rho \), angular position \( \phi \), and height \( z \) along the axis.

In our exercise, we use cylindrical coordinates because they make it simple to describe the motion of particles around an axis. The angular velocity \( \omega \) only affects the radial component \( \rho \). The velocity \( v \) then becomes \( \rho \omega \), directed along the \( \phi \) direction, highlighting the circular motion around the z-axis.

Understanding the cylindrical coordinate system is crucial to analyze the forces and motions involved in rotational dynamics. By describing positions this way, we can more easily compute the necessary elements like velocity, angular momentum, and moment of inertia when dealing with rotating systems.