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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 figure skater draws her arms in during a final spin. Since angular momentum is conserved, her angular velocity will increase. Is her rotational kinetic energy conserved during this process? If not, where does the extra energy come from or go to?

The flywheel of an old steam engine is a solid homogeneous metal disk of mass \(M=120 . \mathrm{kg}\) and radius \(R=80.0 \mathrm{~cm} .\) The engine rotates the wheel at \(500 .\) rpm. In an emergency, to bring the engine to a stop, the flywheel is disengaged from the engine and a brake pad is applied at the edge to provide a radially inward force \(F=100 .\) N. If the coefficient of kinetic friction between the pad and the flywheel is \(\mu_{\mathrm{k}}=0.200,\) how many revolutions does the flywheel make before coming to rest? How long does it take for the flywheel to come to rest? Calculate the work done by the torque during this time.

-10.62 The Earth has an angular speed of \(7.272 \cdot 10^{-5} \mathrm{rad} / \mathrm{s}\) in its rotation. Find the new angular speed if an asteroid \(\left(m=1.00 \cdot 10^{22} \mathrm{~kg}\right)\) hits the Earth while traveling at a speed of \(1.40 \cdot 10^{3} \mathrm{~m} / \mathrm{s}\) (assume the asteroid is a point mass compared to the radius of the Earth) in each of the following cases: a) The asteroid hits the Earth dead center. b) The asteroid hits the Earth nearly tangentially in the direction of Earth's rotation. c) The asteroid hits the Earth nearly tangentially in the direction opposite to Earth's rotation.

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.

If the iron core of a collapsing star initially spins with a rotational frequency of \(f_{0}=3.2 \mathrm{~s}^{-1},\) and if the core's radius decreases during the collapse by a factor of \(22.7,\) what is the rotational frequency of the iron core at the end of the collapse? a) \(10.4 \mathrm{kHz}\) b) \(1.66 \mathrm{kHz}\) c) \(65.3 \mathrm{kHz}\) d) \(0.46 \mathrm{kHz}\) e) \(5.2 \mathrm{kHz}\)
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