Question

The position vector of two particles of mass 4.0 kg and 2.0 kg are, respectively, r 3 ˆi ˆj 2 kˆ 2 1 = t + t + t r and r 3ˆi ( 1)ˆj 4 kˆ 2 2 = + t − + t r 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
r1 = 3 t i + t j + 2 t² k  meters,    r2 = 3 i + (t² - 1) j + 4 t k    meters

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

Velocity v_com = d r/dt = 2 i + (2t+2)/3 j + 4(2t+1)/3  k    m/s
                        = 2  i + 2(t+1)/3 j  + 4(2t+1)/3 k    m/s

Acceleration of the system = a = 0 i + 2/3 j + 8 /3 k    m/sec²
Force = m a = 6 * a = 4 j + 16 k    Newtons
         Force magnitude = 4 √17 Newtons