the position vector of two particles of mass 4 kg and 2 kg are , respectively, r1 = 3t i^ + t j^ +2t2 k^ and r2 = 3 i^ +(t2-1)j^ +4t k^ 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.
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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
magnitude = √(36+4t²+4+8t+64t²+16+16t) /3 m
= √(17 t²+ 6t + 10) * 2/3 m
Acceleration of cm = a = dv/dt = 2/3 j + 8/3 k
Net force acting on the system = (m1+m2) * a
= 4 j + 16 k
Magnitude = 4√17 Newtons
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
magnitude = √(36+4t²+4+8t+64t²+16+16t) /3 m
= √(17 t²+ 6t + 10) * 2/3 m
Acceleration of cm = a = dv/dt = 2/3 j + 8/3 k
Net force acting on the system = (m1+m2) * a
= 4 j + 16 k
Magnitude = 4√17 Newtons
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