Question

2. Centre of mass is at point P
(1,1,1)when system of consist
of particles of masses 2, 3, 4,5 kg
if the centre of mass shifts to
point Q(2,2,2) on removing the
mass 5 kg what was its position?​

Answer 1

The position of 5 kg was (-0.8, -0.8, -0.8).

Explanation:

Here Given data

$M_{1} =2kg$ ;$M_{2} = 3kg$ ;$M_{3} = 4kg$ ;$M_{4} = 5kg$

From formula of centre of mass of discrete mass

For initial position  

1.  $(M_{1}+M_{2}+M_{3}+M_{4})\bar{X} = M_{1}X_{1}+M_{2}X_{2}+M_{3}X_{3}+M_{4}X_{4}$      

Here $\bar{X} =1$

So $M_{1}X_{1}+M_{2}X_{2}+M_{3}X_{3}+M_{4}X_{4} = 14$   ...a)

2. $(M_{1}+M_{2}+M_{3}+M_{4})\bar{Y} = M_{1}Y_{1}+M_{2}Y_{2}+M_{3}Y_{3}+M_{4}Y_{4}$      

 Here $\bar{Y} = 1$

$M_{1}Y_{1}+M_{2}Y_{2}+M_{3}Y_{3}+M_{4}Y_{4}=14$      ...b)

3. $(M_{1}+M_{2}+M_{3}+M_{4})\bar{Z} = M_{1}Z_{1}+M_{2}Z_{2}+M_{3}Z_{3}+M_{4}Z_{4}$    

Here      $\bar{Z} = 1$

$M_{1}Z_{1}+M_{2}Z_{2}+M_{3}Z_{3}+M_{4}Z_{4} = 14$      ...c)

For new position on removing 5 kg

4. $(M_{1}+M_{2}+M_{3})\bar{X} = M_{1}X_{1}+M_{2}X_{2}+M_{3}X_{3}$

Here $\bar{X} = 2$

  So $M_{1}X_{1}+M_{2}X_{2}+M_{3}X_{3} =18$               ...d)

5. $(M_{1}+M_{2}+M_{3})\bar{Y} = M_{1}Y_{1}+M_{2}Y_{2}+M_{3}Y_{3}$

 Here $\bar{Y} = 2$

So $M_{1}Y_{1}+M_{2}Y_{2}+M_{3}Y_{3} = 18$              ...e)

6. $(M_{1}+M_{2}+M_{3})\bar{Z} = M_{1}Z_{1}+M_{2}Z_{2}+M_{3}Z_{3}$

Here $\bar{Z} = 2$

So $M_{1}Z_{1}+M_{2}Z_{2}+M_{3}Z_{3}= 18$            ...f)

Now

7.  On Subtracting equation d from equation a

we get $M_{4}X_{4} = -4$

$X_{4} = -0.8$

8. On Subtracting equation e from equation b

we get $M_{4}Y_{4} = -4$

$Y_{4} = -0.8$

9. On Subtracting equation f from equation c

we get $M_{4}Z_{4} = -4$

$Z_{4} = - 0.8$

10. Old position of 5 kg was (-0.8, -0.8, -0.8)