Question

An elastic string carrying a baby of mass 'm' extends by 'e' . The body rotates in a vertical circle with critical velocity. The externsion in the string at the lowest position is

Answer 1

Answer:

When the body is at rest, the extension is 1.5\ cm1.5 cm while force acting on it is mgmg

Force when it reaches bottom is F_b = \dfrac{mv^2}{r} + mg F

b

=

r

mv

2

+mg

Now using energy balance between topmost point and bottom most point ,

\dfrac{1}{2}mv^2=\dfrac{1}{2}m(\sqrt{rg})^2+mgh

2

1

mv

2

=

2

1

m(

rg

)

2

+mgh (h is the height difference between the top and bottom point ; \sqrt{gr}

gr

is critical velocity at topmost point )

v^2 = rg + 2ghv

2

=rg+2gh

or, v^2 = rg + 4gr = 5 grv

2

=rg+4gr=5gr

So, F_b = 6mg F

b

=6mg

Using this in: \dfrac{e_2}{e_1} = \dfrac{F_2}{F_1}

e

1

e

2

=

F

1

F

2

, we get:

e_2 = 1.5 \times 6 = 9.0\: cm e

2

=1.5×6=9.0cm