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Chapter 10: Problem 3

Show that the angular momentum about any point of a single particle moving with constant velocity remains constant throughout the motion.

Short Answer

Expert verified
The angular momentum does remain constant for a particle moving with constant velocity because the mass, velocity, and the direction of the cross product between the radius and velocity vectors remain constant.

Step by step solution

01

Express Angular Momentum in Vector Form

Angular momentum, \( L \), can be expressed in vector form as \( L = m(\vec{r} \times \vec{v}) \), where \( m \) is the mass of the particle, \( \vec{r} \) is the position vector of the particle from the point about which we are calculating angular momentum and \( \vec{v} \) is the velocity vector of the particle.
02

Express Velocity as Constant

Given that the particle is moving with a constant velocity, this implies that \( \vec{v} \) is a constant vector, \( \vec{v} = constant \). Additionally, as \( m \) is a quantity inherent to the particle, \( m \) is also a constant.
03

Determine the Cross Product

The cross product \( \vec{r} \times \vec{v} \) obtains a vector that is perpendicular to the plane containing \( \vec{r} \) and \( \vec{v} \), following the right-hand rule. As the particle moves, \( \vec{r} \) changes its direction, but the plane defined by \( \vec{r} \) and \( \vec{v} \) remains the same. Therefore, the direction of \( \vec{r} \times \vec{v} \) remains constant even as the particle moves.
04

Conclude that Angular Momentum Remains Constant

Substituting the constant values of \( m \) and \( \vec{v} \), and the fact that the direction of \( \vec{r} \times \vec{v} \) remains constant into the equation for angular momentum, \( L = m(\vec{r} \times \vec{v}) \), it can be deduced that, \( L \) also remains constant. Hence, the angular momentum of a single particle moving with constant velocity remains constant throughout the motion.

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Most popular questions from this chapter

With center and spokes of negligible mass, a certain bicycle wheel has a thin rim of radius \(36.3 \mathrm{~cm}\) and mass \(3.66 \mathrm{~kg}\); it can turn on its axle with negligible friction. A man holds the wheel above his head with the axis vertical while he stands on a turntable free to rotate without friction; the wheel rotates clockwise, as seen from above, with an angular speed of \(57.7 \mathrm{rad} / \mathrm{s}\), and the turntable is initially at rest. The rotational inertia of wheel-plus-man-plus-turntable about the common axis of rotation is \(2.88 \mathrm{~kg} \cdot \mathrm{m}^{2} .(a)\) The man's hand suddenly stops the rotation of the wheel (relative to the turntable). Determine the resulting angular velocity (magnitude and direction) of the system. (b) The experiment is repeated with noticeable friction introduced into the axle of the wheel, which, starting from the same initial angular speed \((57.7 \mathrm{rad} / \mathrm{s})\), gradually comes to rest (relative to the turntable) while the man holds the wheel as described above. (The turntable is still free to rotate without friction.) Describe what happens to the system, giving as much quantitative information as the data permit.

A uniform stick has a mass of \(4.42 \mathrm{~kg}\) and a length of \(1.23 \mathrm{~m}\). It is initially lying flat at rest on a frictionless horizontal surface and is struck perpendicularly by a puck imparting a horizontal impulsive force of impulse \(12.8 \mathrm{~N} \cdot \mathrm{s}\) at a distance of \(46.4 \mathrm{~cm}\) from the center. Determine the subsequent motion of the stick.

A wheel of radius \(24.7 \mathrm{~cm}\), moving initially at \(43.3 \mathrm{~m} / \mathrm{s}\), rolls to a stop in \(225 \mathrm{~m} .\) Calculate \((a)\) its linear acceleration and \((b)\) its angular acceleration. ( \(c\) ) The wheel's rotational inertia is \(0.155 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) Calculate the torque exerted by rolling friction on the wheel.

A wheel with rotational inertia \(1.27 \mathrm{~kg} \cdot \mathrm{m}^{2}\) is rotating with an angular speed of 824 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with rotational inertia \(4.85 \mathrm{~kg} \cdot \mathrm{m}^{2}\), is suddenly coupled to the same shaft. What is the angular speed of the resultant combination of the shaft and two wheels?

Let \(\overrightarrow{\mathbf{r}}_{\mathrm{cm}}\) be the position vector of the center of mass \(C\) of a system of particles with respect to the origin \(O\) of an inertial reference frame, and let \(\overrightarrow{\mathbf{r}}_{i}^{\prime}\) be the position vector of the \(i\) th particle, of mass \(m_{i}\), with respect to the center of mass \(C\). Hence \(\overrightarrow{\mathbf{r}}_{i}=\overrightarrow{\mathbf{r}}_{\mathrm{cm}}+\overrightarrow{\mathbf{r}}_{i}^{\prime}\) (see Fig. \(\left.10-22\right)\). Now define the total angular momentum of the system of particles relative to the center of mass \(C\) to be \(\overrightarrow{\mathbf{L}}^{\prime}=\Sigma\left(\overrightarrow{\mathbf{r}}_{i}^{\prime} \times \overrightarrow{\mathbf{p}}_{i}^{\prime}\right), \quad\) where \(\overrightarrow{\mathbf{p}}_{i}^{\prime}=m_{i} d \overrightarrow{\mathbf{r}}_{i}^{\prime} / d t\) (a) Show that \(\overrightarrow{\mathbf{p}}_{i}^{\prime}=m_{i} d \overrightarrow{\mathbf{r}}_{i} / d t-\) \(m_{i} d \overrightarrow{\mathbf{r}}_{\mathrm{cm}} / d t=\overrightarrow{\mathbf{p}}_{i}-m_{i} \overrightarrow{\mathbf{v}}_{\mathrm{cm}}\) (b) Show next that \(d \overrightarrow{\mathbf{L}}^{\prime} / d t=\) \(\Sigma\left(\overrightarrow{\mathbf{r}}_{i}^{\prime} \times d \overrightarrow{\mathbf{p}}_{i}^{\prime} / d t\right) .(c)\) Combine the results of \((a)\) and \((b)\) and, using the definition of center of mass and Newton's third law, show that \(\vec{\tau}_{\mathrm{ext}}^{\prime}=d \overrightarrow{\mathbf{L}}^{\prime} / d t\), where \(\vec{\tau}_{\mathrm{ext}}^{\prime}\) is the sum of all the \(\mathrm{ex}\) - ternal torques acting on the system about its center of mass.
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