A bob of mass $200 \; gm$ is whirled in a horizontal circle of radius $50 \; cm$ by a string inclined at $30^{\circ}$ to the vertical. Calculate the tension in the string and the speed of the bob in the horizontal circle.

Given, (need figure)

Mass of the bob $(m) = 200 \; gm = 0.2\;kg$

Radius of horizontal circle $(r) = 50 \; cm = 0.5\;m$

Angle of inclination with vertical $(\theta) = 30^{\circ}$

Tension in the string $(T) =\;?$

Speed of the mass $(v) = \;?$

In case of the horizontal circle, we have

(i) To balance the weight of the body,           $T\;Cos\;\theta = mg$

$T =$ $\frac{m\;g}{Cos\; \theta} = \frac{0.2 \;*\;10}{Cos\;30}$ $= 2.3\;N$

Tension in the string $T = 2.3\;N$

(ii) To provide the necessary centripetal force,        $T\;Sin\;\theta =$ $\frac{M\;v^2}{r}$

$v^2 =$ $\frac{T\;Sin\;\theta \;*\;r}{m}$

$v =$ $\sqrt{\frac{2.3 \; * \; Sin\;30\;*\;0.5}{0.2}}$ $= 1.69\;m/s$

Speed of mass $(v) = 1.69\;m/s$

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