A rod of length is 3 m and its mass acting per unit length is directly proportional to distance x from one of its end then its centre of gravity from that end will be at :

Center of Mass and Collision question

2002 · Q157

Center of Mass and Collision question

2002 · Q157

View written solutionFree

Correct answer: B

Let us consider an elementary length dx at a distance x from one end.

It’s mass = k ·x ·dx

[k = proportionality constant]

Then centre of gravity of the rod x_c is given by

${x_c} = {{\int\limits_0^3 {kxdx.x} } \over {\int\limits_0^3 {kxdx} }} = {{\int\limits_0^3 {{x^2}dx} } \over {\int\limits_0^3 {xdx} }} = {{\left. {{{{x^3}} \over 3}} \right|_0^3} \over {\left. {{{{x^2}} \over 2}} \right|_0^3}}$

$\Rightarrow {x_c} = {{{{27} \over 3}} \over {{9 \over 2}}} = 2$

$\therefore$ Centre of gravity of the rod will be at distance of 2 m from one end.

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Center of Mass and Collision question

2002 · Q157

Center of Mass and Collision question

2002 · Q157

View written solutionFree

Correct answer: B

Let us consider an elementary length dx at a distance x from one end.

It’s mass = k ·x ·dx

[k = proportionality constant]

Then centre of gravity of the rod x_c is given by

${x_c} = {{\int\limits_0^3 {kxdx.x} } \over {\int\limits_0^3 {kxdx} }} = {{\int\limits_0^3 {{x^2}dx} } \over {\int\limits_0^3 {xdx} }} = {{\left. {{{{x^3}} \over 3}} \right|_0^3} \over {\left. {{{{x^2}} \over 2}} \right|_0^3}}$

$\Rightarrow {x_c} = {{{{27} \over 3}} \over {{9 \over 2}}} = 2$

$\therefore$ Centre of gravity of the rod will be at distance of 2 m from one end.