Question

In: Physics

the position vector of a mass 2.0 kg is given as a function of time, t,...

the position vector of a mass 2.0 kg is given as a function of time, t, as

r= (3.0 m) i + (4.0 m/s)t j.

What is the angular momentum of the particle about the z axis?

  • (a) 0 kg* m^2/s
  • (b) 24 kg*m^2/s
  • (c) -24 kg*m^2/s
  • (d) 32 kg*m^2/s
  • (e) -32 kg*m^2/s
  • (f) 40 kg*m^2/s
  • (g) -40kg*m^2/s

please explain and answer!

Solutions

Expert Solution

The position vector is,

          r= (3.0 m) i + (4.0 m/s)t j

The velocity of the particle is,

          v = dr/dt

            = 4.0 m/s j

The cross product between velocity and position vector is,

     r x v = {(3.0 m) i + (4.0 m/s)t j } x {4.0 m/s j}

             = (12.0 k) m^2/s

Therefore, the angular momentum of the particle is,

      L = m(r x v)

               = (2.0 kg) [12.0 k] kg. m^2/s

               = [24 kg. m^2/s] k

Magnitude of angular momentum is,

           L = 24 kg.m^2/s

Answer: option (b) is correct

genius_generous answered 1 hour ago
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