Question

A stone of mass m tied to a light inextensible string of length L =10/3 m is whirling in a circular path of radius L in a vertical plane if the ratio of maximum to the minimum tension is four find speed of stone at highest point of circle

Answer 1

Answer:

10m/s

Given:

Mass of stone=m

Weight=mg

Radius=L = 10/3

Length= 2x Radius= 2L

Let:

Tension at max height= Tmin

Tension at lowest point= Tmax

$\frac{Tmax}{Tmin}=4$

Velocity at Tmax = u

velocity at Tmin= v

Solution:

By the third equation of motion,

$v^2=u^2+2as \\\\ v^2=u^2+2(-g)(2L) \\\\ v^2= u^2-4gL \\\\ u^2=v^2+4gL------------(i)$

Centripetal force at Tmax(lowest point) :

$Tmax-mg=\frac{mu^2}{L} \\\\ Tmax=mg+\frac{mu^2}{L}$

Centripetal force at Tmin(highest point) :

$Tmin+mg=\frac{mv^2}{L} \\\\ Tmin=\frac{mv^2}{L}-mg$

A/Q

$\frac{Tmax}{Tmin}=\frac{mg+\frac{mu^2}{L}}{\frac{mu^2}{L}-mg}=\frac{\frac{mu^2+mgL}{L}}{\frac{mv^2-mgL}{L}} \\\\ \frac{mu^2+mgL}{mv^2-mgL} =\frac{m(u^2+gL)}{m(v^2-gL)} \\\\ =\frac{\cancel{m}(u^2+gL)}{\cancel{m}(v^2-gL)}\\\\ =\frac{u^2+gL}{v^2-gL}\\\\ From~(i) \\\\=\frac{v^2+4gL+gL}{v^2-gL} \\\\ \frac{v^2+5gL}{v^2-gL}$

But, the given ratio is 4.

So, $\frac{v^2+5gL}{v^2-gL}=4 \\\\ v^2+5gL=4v^2-4gL \: =3v^2=9gL \\\\ =>v=\sqrt{3gL} \:\: => \sqrt{\cancel{3} \times 10 \times \frac{10}{\cancel{3}}}=10ms^{-1}$