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Circular Motion question

2024 · 29 Jan · Shift 2 · Q64

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Circular Motion
2024 · 29 Jan · Shift 2 · Q64

Circular Motion question

2024 · 29 Jan · Shift 2 · Q64

JEE MainPhysicsCircular MotionMCQ+4 / −1

A stone of mass

900 \mathrm{~g}

is tied to a string and moved in a vertical circle of radius

1 \mathrm{~m}

making

10 \mathrm{~rpm}

. The tension in the string, when the stone is at the lowest point is (if

\pi^2=9.8

and

g=9.8 \mathrm{~m} / \mathrm{s}^2

) :

A
17.8 N
B
97 N
C
9.8 N
D
8.82 N

View written solutionFree

Correct answer: C

Given data

Mass of stone: $m = 900\,\text{g} = 0.9\,\text{kg}$
Radius of vertical circle: $r = 1\,\text{m}$
Speed of rotation: $10\,\text{rpm}$
$g = 9.8\,\text{m/s}^2$
Given: $\pi^2 = 9.8$

Convert rpm to angular speed

Since the stone makes $10$ revolutions per minute,

f = \frac{10}{60} = \frac{1}{6}\,\text{rev/s}

Angular speed:

\omega = 2\pi f = 2\pi \cdot \frac{1}{6} = \frac{\pi}{3}\,\text{rad/s}

Find linear speed

v = \omega r = \frac{\pi}{3} \cdot 1 = \frac{\pi}{3}\,\text{m/s}

So,

v^2 = \frac{\pi^2}{9} = \frac{9.8}{9}

Centripetal force at the lowest point

At the lowest point, the centripetal force is directed upward toward the center. Forces on the stone:

Tension $T$ upward
Weight $mg$ downward

Therefore,

T - mg = \frac{mv^2}{r}

So,

T = mg + \frac{mv^2}{r}

Substitute values

First,

mg = 0.9 \times 9.8 = 8.82\,\text{N}

Next,

\frac{mv^2}{r} = 0.9 \cdot \frac{9.8}{9} = 0.98\,\text{N}

Hence,

T = 8.82 + 0.98 = 9.80\,\text{N}

Match with options

T = 9.8\,\text{N}

So the correct option is C.

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