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Chapter 5: Problem 617

A mass \(\mathrm{m}\) is moving with a constant velocity along the line parallel to the \(\mathrm{x}\) -axis, away from the origin. Its angular momentum with respect to the origin \(\\{\mathrm{A}\\}\) Zero \\{B \\} remains constant \(\\{\mathrm{C}\\}\) goes on increasing \\{D\\} goes on decreasing

Short Answer

Expert verified
The angular momentum L with respect to the origin can be calculated as \(L = r × p\). Using the given position vector \(r = (x(t), 0, 0)\), and the velocity vector \(v = (v_x, 0, 0)\), we obtain the angular momentum in the z-direction as \(L_z = r_x p_y - r_y p_x\). Plugging in the values, we get \(L_z = 0\). Therefore, the angular momentum remains constant with respect to the origin. The correct answer is (B) remains constant.

Step by step solution

01

Recall the Definition of Angular Momentum

The angular momentum L of a particle with respect to a point is given by the cross product of the position vector r and its linear momentum p, which can be defined as: \(L = r × p\) Note that linear momentum p is defined as the product of the mass (m) and velocity (v) of the particle: \(p = m · v\)
02

Analyze the Position and Velocity Vectors

In this problem, mass m is moving with a constant velocity parallel to the x-axis. We can express its position vector and velocity vector as: Position vector: \(r = (x(t), 0, 0)\), where x(t) is the distance from the origin. Velocity vector: \(v = (v_x, 0, 0)\) As the mass is moving along the x-axis, v_x is constant.
03

Calculate the Angular Momentum

Using the definition of angular momentum and the given position and velocity vectors, we can determine the angular momentum L: Since the cross product of parallel vectors is zero, we only need to obtain the cross product of the non-parallel components. Here, we have no components in the y or z direction, so we only need to calculate L in the z-direction: \(L_z = r_x p_y - r_y p_x\)
04

Analyze the Terms

From Step 3, we notice that r_x is same as x(t) and r_y = 0. Also, the mass is moving parallel to the x-axis, which means p_y = 0 and p_x = m*v_x. Thus, we can write L_z as: \(L_z = x(t) · 0 - 0 · (m · v_x)\) \(L_z = 0\)
05

Determine the Behavior of Angular Momentum

Since L_z and thus the entire angular momentum L is always equal to 0, it remains constant with respect to the origin. Hence, the correct option is: (B) remains constant

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