Three point masses of 3 kg each have the following position vectors:
r1 (t)= (2t+3t^2)mî+tmk
r2 (t)=4t^2mj+3mk
r3 (t)=(3t-1) mî+3t^2mj
Determine the velocity and acceleration of the centre of mass of the system.
Answers
Answered by
1
r1 = (2 t + 3 t²) m i + t m k
r2 = (4 t² m j + 3 m k
r3 = (3 t -1) m i + 3 t² m j
Position vector of center of mass = [r1 + r2 + r3]/ 3
r = (3 t² + 5 t - 1) m i + 7 t² m j + (3 + t) m k
velocity is = dr /dt
= (6 t + 5) m i + 14 t m j + m k
acceleration = dv/dt = 6 m i + 14 m j
r2 = (4 t² m j + 3 m k
r3 = (3 t -1) m i + 3 t² m j
Position vector of center of mass = [r1 + r2 + r3]/ 3
r = (3 t² + 5 t - 1) m i + 7 t² m j + (3 + t) m k
velocity is = dr /dt
= (6 t + 5) m i + 14 t m j + m k
acceleration = dv/dt = 6 m i + 14 m j
Similar questions