Question

Calculate the velocity of the center of mass of the system of particles shown in figure (9-E3).

Answer 1

Answer:

m₁ = 1 kg        

$v_{1} = -1.5 \ cos 37^{\circ} \hat{i} - 1.5 \ sin 37^{\circ} \hat{j} = -1.5* \frac{4}{5} \hat{i} - 1.5*\frac{3}{5} \hat{j} = - 1.2 \hat{i} - 0.9 \hat{j}$

m₂ = 1.2 kg 

$v_{2} = 0.4 \hat{j}$        [There is no movement on x-axis]

m₃ = 1.5 kg     

$v_{3} = -1 \ cos 37^{\circ} \hat{i} + 1 \ sin 37^{\circ} \hat{j} = -1* \frac{4}{5} \hat{i} + 1*\frac{3}{5} \hat{j} = - 0.8 \hat{i} + 0.6 \hat{j}$

m₄ = 0.5 kg     

$v_{4} = 3 \hat{i}$           [There is no movement on y-axis]

m₅ = 1 kg       

$v_{5} = 2 \ cos 37^{\circ} \hat{i} - 2 \ sin 37^{\circ} \hat{j} = 2* \frac{4}{5} \hat{i} - 2*\frac{3}{5} \hat{j} = 1.6 \hat{i} - 1.2 \hat{j}$

Velocity of centre of mass of the system =>

$\frac{m_{1}v_{1} + m_{2}v_{2} + m_{3}v_{3} + m_{4}v_{4} + m_{5}v_{5} }{m_{1} + m_{2} + m_{3} + m_{4} + m_{5}}$

$\frac{1(-1.2\hat{i} - 0.9 \hat{j}) + (1.2)(0.4\hat{j}) + (1.5)(-0.8\hat{i} + 0.6 \hat{j}) + (0.5)(3\hat{i}) + (1)(1.6\hat{i}-1.2\hat{j}) }{1 + 1.2 + 1.5 + 0.5 + 1}$

$\frac{(-1.2-1.2+1.5+1.6)\hat{i}}{5.2} + \frac{(-0.9+0.48+0.9-1.2)\hat{j}}{5.2}$

$\frac{(0.7)\hat{i}}{5.2} - \frac{(0.72)\hat{j}}{5.2}$

Answer 2

Explanation:

Step 1:

$m_{1}=1 \mathrm{kg}$

$\mathrm{v}_{1}=-1.5 \cos 37^{\circ} \mathrm{i}-1.55 \sin 37^{\circ} \mathrm{j}$

$=-1.2 \mathrm{i}-0.9 \mathrm{j}$

$\mathrm{m}_{2}=1.2 \mathrm{kg}$

$v_{z}=0.4 \mathrm{J}$

$\mathrm{m}_{3}=1.5 \mathrm{kg}$

$\mathrm{v}_{3}=-0.8 \mathrm{j}+0.6 \mathrm{j}$

$m_{4}=0.5 \mathrm{kg}$

$m_{5}=1 \mathrm{kg}$

$v_{5}=1.6 \mathrm{i}-1.2 \mathrm{j}$

Step 2:

The velocity of the system's mass center

$=\frac{\left(\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}+\mathrm{m}_{3} \mathrm{v}_{3}+\mathrm{m}_{4} \mathrm{v}_{4}+\mathrm{m}_{5} \mathrm{v}_{5}\right)}{\left(\mathrm{m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}+\mathrm{m}_{4}+\mathrm{m}_{5}\right)}$

$=\frac{[1(-1.2 \mathrm{i}-0.9 \mathrm{j})+1.2(0.4 \mathrm{j})+1.5(-0.8 \mathrm{j}+0.6 \mathrm{j})+0.5(3 \mathrm{i})+1(1.6 \mathrm{i}-1.2 \mathrm{j})}{(1+1.2+1.5+0.5+1)}$

$=\frac{(0.7 i-0.72 j)}{5.2}$

$=\left(\frac{0.7 \mathrm{i}}{5.2}-\frac{0.72 \mathrm{j}}{5.2}\right)$