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Standard X

Physics

Centripetal Force and Circular Motion

A stone of ma...

Question

A stone of mass $1 k g$ is whirled in horizontal circle attached at the end of a 1 m long string. If the string makes an angle of $30^{o}$ with vertical, calculate the centripetal force acting on the stone. $(g = 9.8 m / s^{2})$ . (in $N$ )

6.321

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5.658

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5.369

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5.123

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Solution

The correct option is B $5.658$
as the diagram shows
$T cos 30^{\circ} = m g . . . (1)$
$T sin 30^{\circ} = F_{C} . . . (2)$
dividing eqn(2) by eqn (1)
$\Rightarrow F_{C} = m g tan 30^{\circ}$
$\Rightarrow F_{C} = 1 \times 9.8 \times \frac{1}{\sqrt{3}} = 5.658 N$

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A stone of mass 1kg is whirled in horizontal circle attached at the end of a 1 m long string. If the string makes an angle of 30o with vertical, calculate the centripetal force acting on the stone.(g=9.8m/s2). (in N)

The correct option is B 5.658as the diagram showsTcos30∘=mg...(1)Tsin30∘=FC...(2)dividing eqn(2) by eqn (1)⇒FC=mgtan30∘⇒FC=1×9.8×1√3=5.658N

A stone of mass $1 k g$ is whirled in horizontal circle attached at the end of a 1 m long string. If the string makes an angle of $30^{o}$ with vertical, calculate the centripetal force acting on the stone. $(g = 9.8 m / s^{2})$ . (in $N$ )

The correct option is B $5.658$
as the diagram shows
$T cos 30^{\circ} = m g . . . (1)$
$T sin 30^{\circ} = F_{C} . . . (2)$
dividing eqn(2) by eqn (1)
$\Rightarrow F_{C} = m g tan 30^{\circ}$
$\Rightarrow F_{C} = 1 \times 9.8 \times \frac{1}{\sqrt{3}} = 5.658 N$