Question

6. The position vector of two particles of mass 4.0 kg and 2.0 kg are, respectively, r i j kˆ 2ˆ ˆ 321 = t + t + tr and r i ( )j kˆ 4ˆ 1ˆ 322 = + t − + tr where t is in seconds and the position in metres. Determine the position vector of the centre of mass of the system, the velocity of the cm and the net force acting on the system.

Answer 1

m1 = 4.0 kg         m2 = 2.0 kg
Given   vector  r1 =  3 t i + t  j + 2 t² k   m
             vector r2 = 3 i + (t² -1) j + 4 t k   meters

Position vector of Center of mass =

 r = (m1 r1  + m2 r2 )/ (m1+m2)
 r = [ (4* 3t + 2*3) i + (4* t+ 2* (t²-1)) j + (4*2t² + 2 * 4 t) k ]/(4+2)  meters
  = [ (12t +6) i + (2t² + 4t -2) j + (8t² + 8 t) k ] /6  m

Velocity of cm = v = dr/dt = 2 i + (2t+2)/3 j + (8 t + 4)/3 k    m/s
    magnitude = √(36+4t²+4+8t+64t²+16+16t)  /3  m/s
                      = √(17 t²+ 6t + 10) * 2/3  m/s

Acceleration of cm = a = dv/dt = 2/3 j + 8/3 k   m/s^2
     
Net force acting on the system =  (m1+m2) * a
          = 4 j + 16 k
    Magnitude = 4√17 Newtons