Question

A stone tied to a string $1$ m long is whirled in a circular path in a verticle plane. If the difference in the tension on the string at the bottom and top is $120$ N, determine the mass of the stone. $(Takeg=10{ms}^{-2})$

• A. $5$ kg
• B. $1$ kg
• C. $2$ kg
• D. $12$ kg

Answer 1

The correct option is C $2$ kg

Tension, $T=\frac{m{V}^2}{l}+mg\cos \theta$

At the top:

$T_C=0$

$⟹\frac{m{V}_C^2}{l}+mg\cos \left({180}^o\right)=0$

$⟹V_C^2=gl$

$⟹V_C=\sqrt{gl}$

From conservation of energy:

$E_A=E_C$

$⟹\frac{1}{2}m{V}_A^2=\frac{1}{2}m{V}_C^2+2mgl$

$⟹\frac{V_A^2}{2}=\frac{gl}{2}+2mgl$

$⟹V_A=\sqrt{5gl}$

$T_A=\frac{m{V}_A^2}{l}+mg\cos \left(0^o\right)$

$⟹T_A=5$ $mg+1$ $mg=6$ $mg$

Now, $T_A-T_C=120$ $N$

So, $6mg=120$ $N$

$⟹m=\frac{20}{10}$

$⟹m=2$ $kg$