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Question
Solve the following problem.
Four uniform solid cubes of edges 10 cm, 20 cm, 30 cm and 40 cm are kept on the ground, touching each other in order. Locate centre of mass of their system.
Solution
The given cubes are arranged as shown in the figure. Let one of the corners of the smallest cube lie at the origin.
As the cubes are uniform, let their center of masses lie at their respective centers.
∴ rA ≡ (5, 5), rB ≡ (20, 10), rC ≡ (45, 15) and rD ≡ (80, 20)
Also, masses of the cubes are,
`"m"_"A" = l_"A"^3 xx rho = 10^3rho`
`"m"_"B" = (20)^3 rho`
`"m"_"C" = (30)^3 rho`
`"m"_"D" = (40)^3 rho`
As the cubes are uniform, ρ is same for all of them.
∴ For X co-ordinate of centre of mass of the system,
`"X"_"cm" = (sum "m"_"i""x"_"i")/"M"`
`= ("m"_"A""x"_"A" + "m"_"B""x"_"B" + "m"_"C""x"_"C" + "m"_"D""x"_"D")/("m"_"A" + "m"_"B" + "m"_"c" + "m"_"D")`
`= ([10^3 xx rho xx 5] + [20^3 xx rho xx 20] + [30^3 xx rho xx 45] + [40^3 xx rho xx 80])/(10^3 xx rho + 20^3 xx rho + 30^3 xx rho + 40^3 xx rho)`
`= 6500/100`
= 65 cm
Similarly, Y - co-ordinate of centre of mass of system is,
`"Y"_"cm" = (sum "m"_"i""y"_"i")/"M"`
`= ("m"_"A""y"_"A" + "m"_"B""y"_"B" + "m"_"C""y"_"C" + "m"_"D""y"_"D")/("m"_"A" + "m"_"B" + "m"_"c" + "m"_"D")`
`= (10^3 xx (1 xx 5 + 8 xx 10 + 27 xx 15 + 64 xx 20))/(10^3 xx (1 + 8 + 27 + 64))`
`= 1770/100`
= 17.7 cm
Centre of mass of the system is located at point (65 cm, 17.7 cm).
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