Question

n particles of masses m ,2m, 3m ,4m .......... grams are placed on the same line at distances l, 2l, 3l ,4l , ........ c.m. from a fixed point. The distance of centre of mass of the particles from the fixed point in centimeters is.​

Answer 1

answer : (2n + 1)l/3

centre of mass is given as ,

$X_{cm}=\frac{m_1x_1+m_2x_2+m_3x_3+.......+m_nx_n}{m_1+m_2+m_3+.....+m_n}$

here, m1 = m, m2 = 2m, m3 = 3m, m4 = 4m...... m_n = nm

x1 = l, x2 = 2l, x3 = 3l, x4, = 4l.... x_n = nl

so, centre of mass of the particles from the fixed point in the centimetre is

$\frac{ml+(2m)(2l)+(3m)(3l)+(4m)(4l)+....+(nm)(nl)}{m+2m+3m+4m+5m+.....+(nm)}$

= (ml + 2²ml + 3²ml + 4²ml + .... + n²ml)/m(1 + 2 + 3 + 4 + 5 + 6 + ... + n)

= l[ (1² + 2² + 3² + 5² + ..... + n²)/(1 + 2 + 3 + 4 + 5 + .. + n)]

= l [{n(n + 1)(2n + 1)/6}/{n(n + 1)/2} ]

= l [(2n + 1)/3 ]

= (2n + 1)l/3

Answer 2

Answer:

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