The position vector of two particles of mass 4.0 kg and 2.0 kg are, respectively, r i j kˆ 2 ˆ ˆ 3 2 1 = t + t + t r and r i ( )j kˆ 4 ˆ 1 ˆ 3 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. (10)
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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
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