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Q. A rigid body consists of a 3 kg mass connected to a 2 kg mass by a massless rod. The 3 kg mass is located at →r1 = (2ˆi + 5ˆj ) m and the 2 kg mass at →r2 = ( 4ˆi + 2ˆj )m. Find the length of rod and the coordinates of the centre of mass.

Answer 1

We know that distance between the two point is equal to the magnitude of the position vector of the line joining the two point.

Given r₁ = 2i + 5j

r₂ = 4i + 2j

hence,

r ₁r₂ = (4-2)i + (2-5)j = 2i - 3 j

Hence

|r₁r₂| = √(2²+3²) = √13

Hence length of the rod is √13 m .

Centre of mass ,

r = (m₁r₁ + m₂r₂)/m₁+m₂

= 3(2i + 5 j) + 2(4i + 2j )/5

= (14i + 19j)/5

= 14/5 i + 19/5 j

Hence the centre of mass is at 14/5 i + 19/5 j  m

Answer 2

Appropriate question :-

A rigid body consists of a 3 kg mass is located at $\sf r_{1} = (2 \bar i + 5 \bar j)$m and a 2 kg mass located at $\sf r_{2} = (4 \bar i + 2\bar j)$m. The position of centre of mass is

Solution :-

$\displaystyle \bf \qquad \red{\: \implies \frac{ 14\bar I }{5} +\frac{ 19 \bar J \:}{5} }$

Step-by-step explanation :

Given,

A rigid body consists of a 3 kg mass is located at $\sf r_{1} = (2 \bar i + 5 \bar j)$m and a 2 kg mass located at $\sf r_{2} = (4 \bar i + 2\bar j)$m.

According to the question by using formula we get,

$\displaystyle \bf \implies \: r = \frac{ m_{1}r_{1} + m_{2}r_{2} }{m_{1} + m_{2} }$

$\displaystyle \sf \implies \: r = \frac{ 3( 2\bar I + 5 \bar J) +2( 4\bar I + 2 \bar J) \:}{3 + 2}$

$\displaystyle \sf \implies \: r = \frac{ 6\bar I + 15 \bar J +8\bar I + 4 \bar J \:}{5}$

$\displaystyle \sf \implies \: r = \frac{ 14\bar I + 19 \bar J \:}{5}$

$\displaystyle \bf \implies \: r = \frac{ 14\bar I }{5} +\frac{ 19 \bar J \:}{5}$

Hence,

$\displaystyle \sf ∴ The \: position \: of \: centre \: of \: mass \: is \: \frac{ 14\bar I }{5} +\frac{ 19 \bar J \:}{5}$

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