Question

As shown Figure, the stones of 30, 60, 90 and 120 g are placed at 3, 6, 9 and 12 hour symbols respectively of a weightless dial of clock having radius of 10 cm. Find the co-ordinates of centre of mass of this system.​

Answer 1

Given:

The stones of 30, 60, 90 and 120 g are placed at 3, 6, 9 and 12 hour symbols respectively of a weightless dial of clock having radius of 10 cm.

To find:

Coordinate of Centre of Mass.

Calculation:

Since the radius of the watch is 10 cm, and the objects being placed along the circumference , we can take the coordinates of the objects as follows :

• 30g : (10,0)

• 60g : (0,-10)

• 90g : (-10,0)

• 120g : (0,10)

So, x coordinate of C-O-M:

$\therefore \: \bar{x} = \dfrac{ \sum(mx)}{ \sum(m)}$

$= > \: \bar{x} = \dfrac{ (30 \times 10) + (60 \times 0) + \{90 \times ( - 10) \} + (120 \times 0)}{30 + 60 + 90 + 120}$

$= > \: \bar{x} = \dfrac{ 300 + 0 + - 900+ 0}{300}$

$= > \: \bar{x} = \dfrac{ 300 - 900}{300}$

$= > \: \bar{x} = \dfrac{ - 600}{300}$

$= > \: \bar{x} = - 2$

Similarly , y coordinates of C-O-M:

$\therefore \: \bar{y} = \dfrac{ \sum(my)}{ \sum(m)}$

$= > \: \bar{y} = \dfrac{(30 \times 0) + \{60 \times ( - 10) + (90 \times 0) + (120 \times 10)}{300}$

$= > \: \bar{y} = \dfrac{0 - 600+ 0+1200}{300}$

$= > \: \bar{y} = \dfrac{600}{300}$

$= > \: \bar{y} = 2$

So, the required coordinates are:

$\boxed{ \bold{ \large{( \bar{x}, \bar{y}) = ( - 2,2)}}}$

HOPE IT HELPS.

Answer 2

Answer:

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