JEE Challenger
More from System of Particles and Rotational Motion
PhyMedAdv 2009 • Paper 1

Center of Mass of Drawing Composed of Circles and Line Segments

Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness.

Diagram 0

The mass of ink used to draw each of the two inner circles, and each of the two line segments is mm. The mass of the ink used to draw the outer circle is 6m6m. The coordinates of the centres of the different parts are: outer circle (0,0)(0, 0), left inner circle (−a,a)(-a, a), right inner circle (a,a)(a, a), vertical line (0,0)(0, 0) and horizontal line (0,−a)(0, -a). The yy-coordinate of the centre of mass of the ink in this drawing is

Options

A

a10\frac{a}{10}

Correct
B

a8\frac{a}{8}

C

a12\frac{a}{12}

D

a3\frac{a}{3}

Topics & Concepts

System of Particles and Rotational MotionCenter of Mass

Step-by-Step Solution

To determine the yy-coordinate of the centre of mass of the ink used in the drawing, we can treat each distinct symmetric part as a point mass located at its respective centre of mass (given as the coordinates of their centres).

The drawing consists of five parts:

  1. Outer circle:

    • Mass, m1=6mm_1 = 6m
    • yy-coordinate of the centre, y1=0y_1 = 0

  2. Left inner circle:

    • Mass, m2=mm_2 = m
    • yy-coordinate of the centre, y2=ay_2 = a

  3. Right inner circle:

    • Mass, m3=mm_3 = m
    • yy-coordinate of the centre, y3=ay_3 = a

  4. Vertical line segment:

    • Mass, m4=mm_4 = m
    • yy-coordinate of the centre, y4=0y_4 = 0

  5. Horizontal line segment:

    • Mass, m5=mm_5 = m
    • yy-coordinate of the centre, y5=−ay_5 = -a

The total mass of the system of ink is:
Mtotal=m1+m2+m3+m4+m5=6m+m+m+m+m=10mM_{\text{total}} = m_1 + m_2 + m_3 + m_4 + m_5 = 6m + m + m + m + m = 10m

The yy-coordinate of the centre of mass, ycmy_{\text{cm}}, is given by:
ycm=∑i=15miyiMtotaly_{\text{cm}} = \frac{\sum_{i=1}^{5} m_i y_i}{M_{\text{total}}}

Substituting the given values into the equation:
ycm=(6m)(0)+(m)(a)+(m)(a)+(m)(0)+(m)(−a)10my_{\text{cm}} = \frac{(6m)(0) + (m)(a) + (m)(a) + (m)(0) + (m)(-a)}{10m}

ycm=0+ma+ma+0−ma10my_{\text{cm}} = \frac{0 + ma + ma + 0 - ma}{10m}

ycm=ma10m=a10y_{\text{cm}} = \frac{ma}{10m} = \frac{a}{10}

Therefore, the yy-coordinate of the centre of mass of the ink is a10\frac{a}{10}.

Correct Option: A

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