A string of length 1 m is fixed at one end with a bob of mass 200 g and the string makes π 5 rev/s around a vertical axis through the fixed point. The angle of inclination of the string with vertical is

cos θ = ω 2 l g = ( 2 π × π 5 ) 2 × 1 10 = 20 10 = 2 1 , so θ = 6 0 ∘ .

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Physics - Circular Motion Question with Solution | TestHub

PhysicsCircular MotionDynamicsMedium2 minQB

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A string of length 1 m is fixed at one end with a bob of mass 200 g and the string makes $\frac{5}{π}$ rev/s around a vertical axis through the fixed point. The angle of inclination of the string with vertical is

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Solution:

$cos θ = \frac{g}{ω ^{2} l} = \frac{10}{( 2 π \times \frac{5}{π} ) ^{2} \times 1} = \frac{10}{20} = \frac{1}{2}, so θ = 6 0^{\circ} .$

Stream:JEESubject:PhysicsTopic:Circular MotionSubtopic:Dynamics

⏱ 2mℹ️ Source: QB

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