Two objects of mass 10 kg and 20 kg respectively are connected to the two ends of a rigid rod of length 10 m with negligible mass. The distance of the center of mass of the system from the 10 kg mass is [2022]

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Solution

Q.
Two objects of mass 10 kg and 20 kg respectively are connected to the two ends of a rigid rod of length 10 m with negligible mass. The distance of the center of mass of the system from the 10 kg mass is [2022]

1 $\frac{10}{3} m$

2 $\frac{20}{3} m$

3 10 m

4 5 m

Ans.
(2)

Masses, $m_{1} = 10 kg$ and $m_{2} = 20 kg$

Let the distance between the two masses be $d = 10 m$ .

Now, centre of masses for two particle system is written as,

$X_{COM} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$ ...(i)

Considering point A as origin, so $B = (d, 0)$

$X_{COM} = (\frac{m_{1} \times 0 + m_{2} d}{m_{1} + m_{2}}) = \frac{m_{2} d}{m_{1} + m_{2}}$

$X_{COM} = \frac{20 kg \times 10 m}{10 kg + 20 kg} = \frac{20}{3} m$

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