Ncert Exercises & Solutions

A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?

Mass of the stone, m = 0.25 kg

Radius of the circle, r = 1.5 m

Number of revolution per second, $n = \frac{40}{60} = 2 / 3 rps$

$A n g u l a r v e l o c i t y = ω = \frac{v}{r} = 2 π n . . . . . . . . (1)$

The centripetal force for the stone is provided by the tension T, in the string, i.e.,

$T = F_{centripetal}$
$= \frac{{mv}^{2}}{r} = {mrω}^{2} = mr {(2 πn)}^{2} [F r o m (1)]$
$= 0.25 \times 1.5 \times {(2 \times 3.14 \times \frac{2}{3})}^{2}$
$= 6.57 N$

Maximum tension in the string, $T_{m a x}$ = 200 N

$T_{\max} = \frac{{mv}^{2}_{\max}}{r}$
$\Rightarrow v_{\max} = \sqrt{\frac{t_{\max} \times r}{m}}$
$= \sqrt{\frac{200 \times 1.5}{0.25}}$
$= \sqrt{1200} = 34.64 m / s$

Therefore, the maximum speed of the stone is 34.64 m/s.

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