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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

A sphere of radius \(R\) and mass \(M\) sits on a horizontal tabletop. A horizontally directed impulse with magnitude \(J\) is delivered to a spot on the ball a vertical distance \(h\) above the tabletop. a) Determine the angular and translational velocity of the sphere just after the impulse is delivered. b) Determine the distance \(h_{0}\) at which the delivered impulse causes the ball to immediately roll without slipping.

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 \(\left.4 \cdot 10^{26} \mathrm{~W} .\right)\)

A student of mass \(52 \mathrm{~kg}\) wants to measure the mass of a playground merry-go-round, which consists of a solid metal disk of radius \(R=1.5 \mathrm{~m}\) that is mounted in a horizontal position on a low-friction axle. She tries an experiment: She runs with speed \(v=6.8 \mathrm{~m} / \mathrm{s}\) toward the outer rim of the merry-go-round and jumps on to the outer rim, as shown in the figure. The merry-go-round is initially at rest before the student jumps on and rotates at \(1.3 \mathrm{rad} / \mathrm{s}\) immediately after she jumps on. You may assume that the student's mass is concentrated at a point. a) What is the mass of the merry-go-round? b) If it takes 35 s for the merry-go-round to come to a stop after the student has jumped on, what is the average torque due to friction in the axle? c) How many times does the merry-go-round rotate before it stops, assuming that the torque due to friction is constant?

A professor doing a lecture demonstration stands at the center of a frictionless turntable, holding 5.00 -kg masses in each hand with arms extended so that each mass is \(1.20 \mathrm{~m}\) from his centerline. A (carefully selected!) student spins the professor up to a rotational speed of \(1.00 \mathrm{rpm} .\) If he then pulls his arms in by his sides so that each mass is \(0.300 \mathrm{~m}\) from his centerline, what is his new rotation rate? Assume that his rotational inertia without the masses is \(2.80 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\), and neglect the effect on the rotational inertia of the position of his arms, since their mass is small compared to the mass of the body.

A uniform solid cylinder of mass \(M=5.00 \mathrm{~kg}\) is rolling without slipping along a horizontal surface. The velocity of its center of mass is \(30.0 \mathrm{~m} / \mathrm{s}\). Calculate its energy.
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