Question

Ryan (mass 70 kg) is running east (along the x-axis) in a straight line at a speed of 2.5 m/s. If Ryan is at the origin, what is the magnitude and direction of the angular momentum of Ryan as measured by an observer located at a position A = (3.0 m, 3.0 m)? Draw the following vectors v and rA to show your work.

Answer 1

The magnitude of the angular momentum of Ryan, as observed from position A = (3.0 m, 3.0 m), is 262.5 kg⋅m²/s, directed perpendicular to both the position vector from Ryan to observer A and the velocity vector of Ryan.

Step-by-step explanation:

To calculate the angular momentum of Ryan as observed from position A, we first find the position vector r_A from Ryan to observer A. Given that A = (3.0 m, 3.0 m) and Ryan is at the origin, r_A is √(3.0² + 3.0²) = 3√2 m in magnitude. The velocity vector v of Ryan is 2.5 m/s along the x-axis.

The angular momentum L_A observed from A is given by L_A = r_A × p, where p is the linear momentum. As the motion is along the x-axis, the linear momentum p = mv points along the x-axis with a magnitude of 70 kg × 2.5 m/s = 175 kg⋅m/s. The direction of L_A is perpendicular to both r_A and p, following the right-hand rule, it is directed out of the plane and perpendicular to the x-axis and position vector r_A.

Therefore, the magnitude of L_A is r_A × p = (3√2 m) × (175 kg⋅m/s) = 262.5 kg⋅m²/s. The direction of the angular momentum is perpendicular to the x-axis and the position vector from Ryan to observer A, indicating a rotation-like effect around that axis when observed from A. Drawing the position vector r_A from Ryan to position A and the velocity vector v along the x-axis facilitates visualizing the directions and relationships involved in calculating the angular momentum as observed from point A.