Question

A 4.92 kg particle-like object moves in a plane with velocity components v_x = 43.2 m/s and v_y = 35.7 m/s as it passes through the point with (x, y) coordinates of (5.46, -8.28) m. Just then, in unit-vector notation, what is its angular momentum relative to (a) the origin and (b) the point (-8.30, -8.30) m?

Answer 1

Given

   mass of the particle like object, m = 4.92 kg

   velocity components vx = 43.2 m/s , vy = 35.7 m/s

   passing through the point (x,y) = (5.46,-8.28) m

  
Angular momentum relative to the origin L = ?

we know that the angular momentum L = r X P

r is position vector and p is momentum vector

first we should calculate the position vector and momentum vectors

for momentum vector in XY plane is
P = m*vx i + m*vy j

P = 4.92*43.2 kg.m/s i + 4.92*35.7 kg m/s j

P = 212.54 kg.m/s i + 175.644 kg.m/s j

(a)
and position vector, relative to origin is

r = 5.46 m i - 8.28 m j

now the angular momentum L = r X P

   L = i(0-0)-j(0-0)+k((5.46*175.64)-((-8.28)(212.54)))

   L = 0 i + 0 j + 2718.826 k

the magnitude is L = sqrt(Lx^2+Ly^2+Lz^2) kg.m^2/s

   L = sqrt(0^2+0^2+2718.826^2) = 2718.826 kg.m^2/s
------------

(b)

relative to the point (-8.30,-8.30 )m is

the position vector of the particle is let r1 = ((-8.30)-(5.46) m i + ((-8.30)-(-8.28))m j

   r1 = -13.76 m i - 0.02 m j

now the angular momentum L = r X P

   L = i(0-0)-j(0-0)+k((-13.76)(175.64))-((-0.02)(212.54)))

   L = 0 i + 0 j -2412.56 k

the magnitude is L = sqrt(L x^2+L y^2+L z^2)

   L = sqrt (0^2+0^2+(-2412.56)^2) = 2412.56 kg.m^2/s