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Chapter 10: Problem 26

Does a particle traveling in a straight line have an angular momentum? Explain.

Short Answer

Expert verified
Answer: Yes, a particle traveling in a straight line does have angular momentum. Its magnitude depends on the perpendicular distance from the reference point and the linear momentum of the particle.

Step by step solution

01

Define Angular Momentum

Angular momentum is a property associated with the rotation of any object, or system of objects, about a specific reference point. We use the letter 'L' to represent it, and it is defined as the cross product of the distance vector 'r' (the vector from the reference point to the position of the particle) and the linear momentum 'p' of the particle. Mathematically, L = r x p.
02

Establish the situation

We have a particle traveling in a straight line. Let's assume this line is parallel to the x-axis. We will consider a reference point on the y-axis, at a perpendicular distance d from the particle's path.
03

Determine the linear momentum and position vector

Since the particle is moving in a straight line along the x-axis, its linear momentum, p, will be in the x-direction. Let's denote its magnitude as m*v, where m represents the mass of the particle, and 'v' is its constant velocity. The position vector, r, will be the vector from the reference point to the particle's position. As the particle moves in a straight line parallel to the x-axis, its position vector will have components in both the x and y directions. The y-component will be equal to the perpendicular distance d from the reference point to the particle's path.
04

Calculate the cross product of r and p

To find the angular momentum, we need to calculate the cross product of the position vector r and the linear momentum p. The cross product of two vectors in three-dimensional space can be given by: L = r x p = (r_y * p_z - r_z * p_y) i + (r_z * p_x - r_x * p_z) j + (r_x * p_y - r_y * p_x) k Since the particle is moving in a straight line parallel to the x-axis, the linear momentum has no component in the y or z directions, so p_y = p_z = 0. The position vector r is confined to the xy-plane, so the z-component of r will also be zero i.e., r_z = 0. Given this information, we can now simplify the cross product L = (r_y * 0 - 0 * 0) i + (0 * p_x - r_x * 0) j + (r_x * 0 - r_y * p_x) k
05

Evaluate the cross product and determine angular momentum

After simplifying the expression for the cross product, we obtain: L = - r_y * p_x k Since r_y = d (the perpendicular distance from the reference point to the particle's path), we can write the angular momentum as: L = - d * p_x k Since d and p_x are both non-zero, we can conclude that the particle traveling in a straight line does have an angular momentum. The negative sign indicates that the angular momentum is in the opposite direction of the z-axis, and its magnitude depends on the perpendicular distance d from the reference point and the linear momentum of the particle.

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Most popular questions from this chapter

The Crab pulsar \(\left(m \approx 2 \cdot 10^{30} \mathrm{~kg}, R=5 \mathrm{~km}\right)\) is a neutron star located in the Crab Nebula. The rotation rate of the Crab pulsar is currently about 30 rotations per second, or \(60 \pi \mathrm{rad} / \mathrm{s} .\) The rotation rate of the pulsar, however, is decreasing; each year, the rotation period increases by \(10^{-5}\) s. Justify the following statement: The loss in rotational energy of the pulsar is equivalent to 100,000 times the power output of the Sun. (The total power radiated by the Sun is about \(\left.4 \cdot 10^{26} \mathrm{~W} .\right)\)

In another race, a solid sphere and a thin ring roll without slipping from rest down a ramp that makes angle \(\theta\) with the horizontal. Find the ratio of their accelerations, \(a_{\text {ring }} / a_{\text {sphere }}\)

A uniform solid sphere of mass \(M\) and radius \(R\) is rolling without sliding along a level plane with a speed \(v=3.00 \mathrm{~m} / \mathrm{s}\) when it encounters a ramp that is at an angle \(\theta=23.0^{\circ}\) above the horizontal. Find the maximum distance that the sphere travels up the ramp in each case: a) The ramp is frictionless, so the sphere continues to rotate with its initial angular speed until it reaches its maximum height. b) The ramp provides enough friction to prevent the sphere from sliding, so both the linear and rotational motion stop (instantaneously).

A projectile of mass \(m\) is launched from the origin at speed \(v_{0}\) at angle \(\theta_{0}\) above the horizontal. Air resistance is negligible. a) Calculate the angular momentum of the projectile about the origin. b) Calculate the rate of change of this angular momentum. c) Calculate the torque acting on the projectile, about the origin, during its flight.

A 25.0 -kg boy stands \(2.00 \mathrm{~m}\) from the center of a frictionless playground merry-go-round, which has a moment of inertia of \(200 . \mathrm{kg} \mathrm{m}^{2} .\) The boy begins to run in a circular path with a speed of \(0.600 \mathrm{~m} / \mathrm{s}\) relative to the ground. a) Calculate the angular velocity of the merry-go-round. b) Calculate the speed of the boy relative to the surface of the merry-go- round.
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