A string of length 1 m is fixed at one end with a bob of mass 100 g and the string makes 2 π r e v s - 1 around a vertical axis through a fixed point. The angle of inclination of the string with vertical is

T cos ⁡ θ = m g .....(i) T sin ⁡ θ = m ω 2 r .....(ii) So tan ⁡ θ = ω 2 r g ; ⇒ sin ⁡ θ cos ⁡ θ = ω 2 l sin ⁡ θ g ⇒ cos ⁡ θ = g ω 2 l ⇒ cos ⁡ θ = 10 16 × 1 = 5 8 ⇒ θ = cos - 1 5 8

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PhysicsW.E.P.Vertical circular motionEasy2 minQB

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A string of length $1 m$ is fixed at one end with a bob of mass $100 g$ and the string makes $\frac{2}{π} r e v s^{- 1}$ around a vertical axis through a fixed point. The angle of inclination of the string with vertical is

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Answer:

Solution:

$T \cos θ = m g$ .....(i)

$T \sin θ = m ω^{2} r$ .....(ii)

So $\tan θ = \frac{ω^{2} r}{g}$ ;

$\Rightarrow \frac{\sin θ}{\cos θ} = \frac{ω^{2} l \sin θ}{g} \Rightarrow \cos θ = \frac{g}{ω^{2} l}$

$\Rightarrow \cos θ = \frac{10}{16 \times 1} = \frac{5}{8}$

$\Rightarrow θ = \cos^{- 1} (\frac{5}{8})$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Vertical circular motion

⏱ 2mℹ️ Source: QB

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