What is the angular momentum of a particle about the origin which is travelling along the line y = x + 5 with a velocity of 5√2m/s and having mass 2 kg? Please provide the detailed solution explaining how to take the angle between v and r. The coordinate system used here is the standard coordinate system.
Hello Ishan Gupta dear, suppose the particle is at a distance r from the origin.
Moment of momentum is angular mometum. Momentum of the particle of mass 2 kg is 2 * 5√2 kg m/s
Moment = momentum x perpendicular distance = 2 * 5√2 * p
p is the perpendicular distance from the origin to the line x -y + 5 = 0.
We know the perpendicular distance of x1, y1 to any line ax + by + c = 0 as (ax1 + by1 + c) / √(a2 + b2)
Perpendicular distance p from origin 0,0 to x - y + 5 = 0 is 5 /√2
Angular momentum about origin at every instant of time = 2 * 5√2 * 5/√2 = 50 kg m2 s-1
Moment of momentum is angular mometum. Momentum of the particle of mass 2 kg is 2 * 5√2 kg m/s
Moment = momentum x perpendicular distance = 2 * 5√2 * p
p is the perpendicular distance from the origin to the line x -y + 5 = 0.
We know the perpendicular distance of x1, y1 to any line ax + by + c = 0 as (ax1 + by1 + c) / √(a2 + b2)
Perpendicular distance p from origin 0,0 to x - y + 5 = 0 is 5 /√2
Angular momentum about origin at every instant of time = 2 * 5√2 * 5/√2 = 50 kg m2 s-1