A bob is whirled in a horizontal plane by means of a string with an initial speed of rpm. The tension in the string is T. If speed becomes while keeping the same radius, the tension in the string becomes [2024]

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Solution

Q.
A bob is whirled in a horizontal plane by means of a string with an initial speed of $ω$ rpm. The tension in the string is T. If speed becomes $2 ω$ while keeping the same radius, the tension in the string becomes [2024]

1 $T$

2 $4 T$

3 $\frac{T}{4}$

4 $\sqrt{2} T$

Ans.
(2)

When the bob is moving with speed $' ω'$ , then FBD is,

Let $' r'$ be the radius of the circle.

Applying Newton’s second law on mass of the bob $' m'$ along centripetal direction, we have

$T = m r ω^{2}$ ...(i)

When speed becomes $2 ω$ , the FBD is

$T' = m r {(2 ω)}^{2} = m r (4 ω^{2})$

or $m r = \frac{T'}{4 ω^{2}}$ ...(ii)

Using value of equation (ii) in equation (i),

$T = \frac{T' ω^{2}}{4 ω^{2}} = \frac{T'}{4}$

Hence, $T' = 4 T$

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