Question

The velocity of a particle at highest point of the vertical circle is $\sqrt{3rg}$. The tension at the lowest point, if mass of the particle is $m$, is

• A. $2mg$
• B. $4mg$
• C. $6mg$
• D. $8mg$

Answer 1

Answer: (D)

$8mg$

Let the velocity at lowest point be $v$. then according to law of conservation of energy.
energy at lowest point= energy at highest point

$\Rightarrow \frac{1}{2}m{v}^2=mg\left(2r\right)+\frac{1}{2}m(\sqrt{3rg}{)}^2\Rightarrow v=\sqrt{7rg}$

At lowest point,

$\frac{m{v}^2}{r}=T-mg\Rightarrow T=mg+\frac{m{v}^2}{r}=mg+\frac{m(\sqrt{7rg}{)}^2}{r}$

$\Rightarrow T=mg+7mg=8mg$