Question

A mass $m$ is revolving in a vertical circle at the end of a string of length $20cm$.By how much does the tension of the string at the lowest point exceed the tension of the top most point?

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

Answer 1

The correct option is C $6mg$

Tension at Top $=T_1$, velocity at top $=\overline{v_1}$

Tension at Bottom $=T_2$, velocity at Bottom $=\overline{v_2}$

Now

$T_1+Mg=\frac{M{v}_1^2}{l}$

$T_2-M_y=\frac{M{v}_2^2}{l}$

$T_2-T_1=\frac{M{v}_2^2}{l}-\frac{m{v}_1^2}{l}+2mg$

$T_2-T_1=\frac{m}{l}(v_2^2-v_1^2)+2mg$

Acc. to newton's $3^{rd}e{q}^n.$

$v_1^2=v_2^2-2g\left(2l\right)$

$v_2^2-v_1^2=4gl$

So

$T_2-T_1=4m_g+2mg=6mg\text{Answer:-}6mg\text{.}$