Question 209: Centre of Mass and Rotational Motion

Question

In the figure shown ABC is a uniform wire. If centre of mass of wire lies vertically below point A, then $${{BC} \over {AB}}$$ is close to :

Accepted Answer

Here   AB = x
and    BC = y
and   $$\lambda $$ = linear mass density.
As centre of mass is below point A, so horizontal distance  of the centre of mass from B is = xcos60
o
= $${x \over 2}$$
$$ 	herefore $$   X
CM
= $${x \over 2}$$ = $${{{m_1}{x_1} + {m_2}{x_2}} \over {{m_1} + {m_2}}}$$
$$ \Rightarrow $$   $${x \over 2}$$ = $${{\left( {\lambda x} ight)\left( {{x \over 2}} ight)\cos {{60}^o} + \left( {\lambda y} ight)\left( {{y \over 2}} ight)} \over {\lambda \left( {x + y} ight)}}$$
$$ \Rightarrow $$   $${x \over 2}$$ = $${{{{{x^2}} \over 4} + {{{y^2}} \over 2}} \over {x + y}}$$
$$ \Rightarrow $$   x
2
+ xy = $${{{x^2}} \over 2} + {y^2}$$
$$ \Rightarrow $$   x
2
+ 2xy $$-$$ 2y
2
= 0
$$ 	herefore $$   x = $${{ - 2y \pm \sqrt {{{\left( {2y} ight)}^2} - 4.1\left( { - 2{y^2}} ight)} } \over {2.1}}$$
=    $${{ - 2y \pm \sqrt {12{y^2}} } \over 2}$$
=   $$-$$ y $$ \pm $$ $$\sqrt 3 $$y
$${x \over y} 
e  - \sqrt 3  - 1$$  as  $${x \over y}$$ = positive.
$$ 	herefore $$   $${x \over y}$$ = $$\sqrt 3  - 1$$
$$ \Rightarrow $$   $${y \over x}$$ = $${1 \over {\sqrt 3  - 1}} 	imes {{\sqrt 3  + 1} \over {\sqrt 3  + 1}}$$
=   $${{\sqrt 3  + 1} \over 2}$$
=   $${{2.732} \over 2}$$
=   1.366 $$ \simeq $$ 1.37

Answer

1.85

Answer

1.37

Answer

1.5

Answer

3

Centre of Mass and Rotational Motion

In the figure shown ABC is a uniform wire. If centre of mass of wire lies vertically below point A, then ${{BC} \over {AB}}$ is close to :

Choose the correct answer: