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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

A particle moves under the influence of a central force directed toward a fixed origin \(O\). (a) Explain why the particle's angular momentum about \(O\) is constant. (b) Give in detail the argument that the particle's orbit must lie in a single plane containing \(O\).

A juggler is juggling a uniform rod one end of which is coated in tar and burning. He is holding the rod by the opposite end and throws it up so that, at the moment of release, it is horizontal, its \(\mathrm{CM}\) is traveling vertically up at speed \(v_{\mathrm{o}}\) and it is rotating with angular velocity \(\omega_{\mathrm{o}} .\) To catch it, he wants to arrange that when it returns to his hand it will have made an integer number of complete rotations. What should \(v_{\mathrm{o}}\) be, if the rod is to have made exactly \(n\) rotations when it returns to his hand?

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 uniform solid disk of mass \(M\) and radius \(R\), rolling without slipping down an incline which is at angle \(\gamma\) to the horizontal. The instantaneous point of contact between the disk and the incline is called \(P\). (a) Draw a free-body diagram, showing all forces on the disk. (b) Find the linear acceleration \(\dot{v}\) of the disk by applying the result \(\dot{\mathbf{L}}=\Gamma^{\text {ext }}\) for rotation about \(P .\) (Remember that \(L=I \omega\) and the moment of inertia for rotation about a point on the circumference is \(\frac{3}{2} M R^{2} .\) The condition that the disk not slip is that \(v=R \omega \text { and hence } \dot{v}=R \dot{\omega} .)\) (c) Derive the same result by applying \(\dot{\mathbf{L}}=\mathbf{\Gamma}^{\mathrm{ext}}\) to the rotation about the CM. (In this case you will find there is an extra unknown, the force of friction. You can eliminate this by applying Newton's second law to the motion of the CM. The moment of inertia for rotation about the \(\mathrm{CM}\) is \(\frac{1}{2} M R^{2}\).)

In the early stages of the Saturn V rocket's launch, mass was ejected at about \(15,000 \mathrm{kg} / \mathrm{s}\), with a speed \(v_{\mathrm{ex}} \approx 2500 \mathrm{m} / \mathrm{s}\) relative to the rocket. What was the thrust on the rocket? Convert this to tons (1 ton \(\approx 9000\) newtons) and compare with the rocket's initial weight (about 3000 tons).

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