Question

Consider a system of three equal mass particles moving in a plane; their positions are given by
ˆi ˆj
i i a + b .
For particle 1, 3 4, 1 0
2
a1 = t + b =
For particle 2, a2 = 7t + 5, b2 = 2
For particle 3, a3 = 2t, b3 = 3t + 4
Determine the position and velocity of the centre of mass as functions of time.

Answer 1

Answer on Question #40732, Physics, Mechanics | Kinematics | Dynamics

Consider a system of three equal mass particles moving in a plane; their positions are given by ai vectors and bi vectors.

For particle 1, $a_1 = 3t^2 + 4b_1 = 0$

For particle 2, $a_2 = 7t + 5b_2 = 2$

For particle 3, $a_3 = 2t b_3 = 3t + 4$

Determine the position and velocity of the center of mass as functions of time

Solution

For the 'a' component of the center of mass's position vector is $R_{a}$ where:

$$R_{a} = \frac{\sum_{i} m_{i} a_{i}}{\sum_{i} m_{i}} = \frac{m a_{1} + m a_{2} + m a_{3}}{m + m + m} = \frac{1}{3} (a_{1} + a_{2} + a_{3}) = \frac{1}{3} (3t^{2} + 4 + 7t + 5 + 2t) = t^{2} + 3t + 3.$$

The 'a' component of the center of mass's velocity, $V_{a}$ is found by differentiating with respect to time

$$V_{a} = \frac{d R_{a}}{d t} = 2t + 3.$$

For the 'b' component of the center of mass's position vector is $R_{b}$ where:

$$R_{b} = \frac{\sum_{i} m_{i} b_{i}}{\sum_{i} m_{i}} = \frac{m b_{1} + m b_{2} + m b_{3}}{m + m + m} = \frac{1}{3} (b_{1} + b_{2} + b_{3}) = \frac{1}{3} (2 + 3t + 4) = t + 2.$$

The 'b' component of the center of mass's velocity, $V_{b}$ is found by differentiating with respect to time

$$V_{b} = \frac{d R_{b}}{d t} = 1.$$

The position of the center of mass:

$$\vec{R} = (t^{2} + 3t + 3) \vec{i} + (t + 2) \vec{j}.$$

The velocity of the center of mass:

$$\vec{V} = (2t + 3) \vec{i} + \vec{j}.$$