Question

two uniform rods A and B of length 5m and 3m are placed end to end. if their linear densities are 3kg/m and 2kg/m the position of thier centre of mass from the interface is

Answer 1

mass of each rod is given as

$m = \lambda * L$

$m_1 = 3*5 = 15 kg$

$m_2 = 2* 3 = 6 kg$

now let say the interface is our reference which is assumed to be (0,0)

now the coordinates of center of two rods is (-2.5, 0) and (1.5, 0) respectively

now by the formula of center of mass

$r = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}$

$r = \frac{15* (-2.5) + 6* 1.5}{15 + 6}$

$r = -1.36$

so the position of center of mass of the system of two rods will be at distance 1.36 m from the interface towards the center of 5 m rod

Answer 2

Explanation:

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