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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

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

Step by step solution

01

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.
02

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.
03

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Consider a system comprising two extended bodies, which have masses \(M_{1}\) and \(M_{2}\) and centers of mass at \(\mathbf{R}_{1}\) and \(\mathbf{R}_{2}\). Prove that the \(\mathrm{CM}\) of the whole system is at $$ \mathbf{R}=\frac{M_{1} \mathbf{R}_{1}+M_{2} \mathbf{R}_{2}}{M_{1}+M_{2}} $$ This beautiful result means that in finding the CM of a complicated system, you can treat its component parts just like point masses positioned at their separate centers of mass - even when the component parts are themselves extended bodies.

Consider a planet orbiting the fixed sun. Take the plane of the planet's orbit to be the \(x y\) plane, with the sun at the origin, and label the planet's position by polar coordinates \((r, \phi)\). (a) Show that the planet's angular momentum has magnitude \(\ell=m r^{2} \omega,\) where \(\omega=\dot{\phi}\) is the planet's angular velocity about the sun. (b) Show that the rate at which the planet "sweeps out area" (as in Kepler's second law) is \(d A / d t=\frac{1}{2} r^{2} \omega,\) and hence that \(d A / d t=\ell / 2 m .\) Deduce Kepler's second law. SECTION 3.5 Angular Momentum for Several Particles

Use spherical polar coordinates \(r, \theta, \phi\) to find the CM of a uniform solid hemisphere of radius R, whose flat face lies in the \(x y\) plane with its center at the origin. Before you do this, you will need to convince yourself that the element of volume in spherical polars is \(d V=r^{2} d r \sin \theta d \theta d \phi\). (Spherical polar coordinates are defined in Section 4.8. If you are not already familiar with these coordinates, you should probably not try this problem yet.)

A uniform spherical asteroid of radius \(R_{\mathrm{o}}\) is spinning with angular velocity \(\omega_{\mathrm{o}}\). As the aeons go by, it picks up more matter until its radius is \(R\). Assuming that its density remains the same and that the additional matter was originally at rest relative to the asteroid (anyway on average), find the asteroid's new angular velocity. (You know from elementary physics that the moment of inertia is \(\frac{2}{5} M R^{2}\).) What is the final angular velocity if the radius doubles?

[Computer] A grenade is thrown with initial velocity \(\mathbf{v}_{\mathrm{o}}\) from the origin at the top of a high cliff, subject to negligible air resistance. (a) Using a suitable plotting program, plot the orbit, with the following parameters: \(\mathbf{v}_{\mathrm{o}}=(4,4), g=1,\) and \(0 \leq t \leq 4\) (and with \(x\) measured horizontally and \(y\) vertically up). Add to your plot suitable marks (dots or crosses, for example) to show the positions of the grenade at \(t=1,2,3,4 .\) (b) At \(t=4,\) when the grenade's velocity is \(\mathbf{v},\) it explodes into two equal pieces, one of which moves off with velocity \(\mathbf{v}+\Delta \mathbf{v} .\) What is the velocity of the other piece? (c) Assuming that \(\Delta \mathbf{v}=(1,3),\) add to your original plot the paths of the two pieces for \(4 \leq t \leq 9 .\) Insert marks to show their positions at \(t=5,6,7,8,9\). Find some way to show clearly that the CM of the two pieces continues to follow the original parabolic path.

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