Question

Three point masses of 3 kg each have the following position vectors:
r i k r j k r i j mˆ 3 mˆ
; ( ) 3( )1 m ˆ 3 mˆ
( ) 4 mˆ m ˆ
( ) 2( 3 )
2 2 2
1 2 3
t = t + t + t ; t = t + t = t − + t
r r r
Determine the velocity and acceleration of the centre of mass of the system.

Answer 1

Answer on Question #54940-Physics-Mechanics-Kinematics-Dynamics

Three point masses of $3\,\mathrm{kg}$ each have the following position vectors:

$$\widetilde{r_1} = (2t + 3t^2)m\tilde{i} + tm\tilde{k}; \widetilde{r_2} = 4t^2m\tilde{j} + 3m\tilde{k}; \widetilde{r_3} = (3t - 1)m\tilde{i} + 3t^2m\tilde{j}$$

Determine the velocity and acceleration of the centre of mass of the system.

Solution

The position of the centre of mass of the system is

$$\tilde{x} = \frac{\sum m_i \widetilde{r_i}}{\sum m_i}.$$

Thus,

$$\tilde{x} = \frac{3\widetilde{r_1} + 3\widetilde{r_2} + 3\widetilde{r_3}}{3 + 3 + 3} = \frac{\widetilde{r_1} + \widetilde{r_2} + \widetilde{r_3}}{3} = \frac{1}{3}(5t - 1 + 3t^2; 7t^2; t + 3)m.$$

The velocity of the centre of mass of the system is

$$\tilde{v} = \frac{d\tilde{x}}{dt} = \frac{1}{3}(5 + 6t; 14t; 1) \frac{m}{s} = \left(\frac{5 + 6t}{3}; \frac{14t}{3}; \frac{1}{3}\right) \frac{m}{s}.$$

The acceleration of the centre of mass of the system is

$$\ddot{a} = \frac{d\tilde{v}}{dt} = \left(2; \frac{14}{3}; 0\right) \frac{m}{s}.$$

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