Question

A particle of mass 1.30 kg is moving with velocity v⃗ =(7.4i^+5.6j^)m/s.

1) Find the angular momentum L⃗ relative to the origin when the particle is at r⃗ =(2.9j^+4.3k^)m. Enter your answers in indicated order separated by commas. Express your answer using two significant figures.

Lx,Ly,Lz =

2) At position r⃗ a force of F⃗ =5.0Ni^ is applied to the particle. Find the torque relative to the origin. Enter your answers in indicated order separated by commas. Express your answer using two significant figures.

τx,τy,τz =

Answer 1

Given

   vector V = (7.4i^+5.6j^)m/s.
   position r = (2.9j^+4.3k^)m
mass of particle m = 1.30 kg

we know that the angular momentum L = m(r X V)

the cross product of r and V

   r X V = -(4.3*5.6) i+(4.3*7.4)j-(7.4*2.9) k

angular momentum is L = 1.3(-(4.3*5.6) i+(4.3*7.4)j-(7.4*2.9) k)

       Lx = - 31.304 kg m2/s

       Ly = 41.366 kg m2/s
  
       Lz = - 27.898 kg m2/s

2. given r = (2.9j^+4.3k^)m

   FORCE VECTOR F = 5.0Ni^

we knwo that the torque T = r X F
           = (2.9j^+4.3k^) X (5.0Ni)

           = 0 i +21.5 j-14.5 k

           T x = 0 Nm
           Ty = 21.5 Nm
           Tz = 14.5 Nm