Mass per unit area of the disc in first quadrant is σ , in the second quadrant is 4σ , in the third quadrant is 5σ and in the fourth quadrant is 8σ . The coordinates of centre of mass of this system are :

Question

<img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-eem7szgcacvtfvsu.jpg" /> Mass per unit area of the disc in first quadrant is σ , in the second quadrant is 4σ , in the third quadrant is 5σ and in the fourth quadrant is 8σ . The coordinates of centre of mass of this system are : (A) (0 , (- 16 R/27 π )) (B) (0 , (- 2 R/3 π )) (C) (0 , (- 16 R/65)) (D) (0 , (- 4 R/3 π ))

Tardigrade Admin · Accepted Answer

xcm=(m (x) + 4 m (- x) + 5 m (- x) - 8 m (x)/m + 4 m + 5 m + 8 m)=0 ycm=(m (x) + 4 m (x) + 5 m (- x) + 8 m (- x)/18 m) =-(8 m x/18 m)=(- 4/9)x=(- 4/9)× ((4 R/3 π ))=-(16 R/27 π ) Solution

Q.
Mass per unit area of the disc in first quadrant is $σ,$ in the second quadrant is $4 σ,$ in the third quadrant is $5 σ$ and in the fourth quadrant is $8 σ$ . The coordinates of centre of mass of this system are :

$(0, \frac{- 16 R}{27 π})$

$(0, \frac{- 2 R}{3 π})$

$(0, \frac{- 16 R}{65})$

$(0, \frac{- 4 R}{3 π})$

$x_{c m} = \frac{m ( x ) + 4 m ( - x ) + 5 m ( - x ) - 8 m ( x )}{m + 4 m + 5 m + 8 m} = 0$
$y_{c m} = \frac{m ( x ) + 4 m ( x ) + 5 m ( - x ) + 8 m ( - x )}{18 m}$
$= - \frac{8 m x}{18 m} = \frac{- 4}{9} x = \frac{- 4}{9} \times (\frac{4 R}{3 π}) = - \frac{16 R}{27 π}$

Q. Mass per unit area of the disc in first quadrant is σ, in the second quadrant is 4σ, in the third quadrant is 5σ and in the fourth quadrant is 8σ . The coordinates of centre of mass of this system are :

(0,27π−16R​)

(0,3π−2R​)

(0,65−16R​)

(0,3π−4R​)