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 #54836-Physics-Mechanics-Kinematics-Dynamics

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

$$\vec{r}_1 = (2t + 3t^2) m \vec{i} + t m \vec{k}; \vec{r}_2 = 4t^2 m \vec{j} + 3 m \vec{k}; \vec{r}_3 = (3t - 1) m \vec{i} + 3 t^2 m \vec{j}$$

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

$$\vec{x} = \frac{\sum m_i \vec{r}_i}{\sum m_i}.$$

Thus,

$$\vec{x} = \frac{3 \vec{r}_1 + 3 \vec{r}_2 + 3 \vec{r}_3}{3 + 3 + 3} = \frac{\vec{r}_1 + \vec{r}_2 + \vec{r}_3}{3} = \frac{1}{3} (5t - 1 + 3t^2; 7t^2; t + 3) m.$$

The velocity of the centre of mass of the system is

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

The acceleration of the centre of mass of the system is

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

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