Question

A uniform bar 8m long has a mass of
6kg . A 10kg mass is tied to one end . The position of centre of mass of the system is

Answer 1

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Answer 2

The position of center of mass of the system is 1.5 meters.

Explanation:

It is given that,

Mass of object 1, $m_1=6\ kg$

Mass of object 2, $m_2=10\ kg$

Length of the bar, l = 8 m

Object 2 is tied to one end. Let C is the center of mass of the system. It is given by :

l' = 0

Now according to the given data we can say that the center of mass, initially, is at 4 m position,

$C=\dfrac{m_1l+m_2l'}{m_1+m_2}$

$C=\dfrac{6\times 4+10\times 0}{6+10}$

C = 1.5 m

So, the position of center of mass of the system is 1.5 meters. Hence, this is the required solution.

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