Question
Download Solution PDFA bob of heavy mass m is suspended by a light string of length l. The bob is given a horizontal velocity v₀ as shown in figure. If the string gets slack at some point P making an angle θ from the horizontal, the ratio of the speed v of the bob at point P to its initial speed v₀ is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCorrect option is: (4) \(\left( \frac{\sin \theta}{2 + 3 \sin \theta} \right)^{\frac{1}{2}} \)
At Point P, mg sin θ = m v2 / l ... (1)
By conservation of mechanical energy at point P ∈ Q
(1/2) mv02 = (1/2) mv2 + mg (I + I sin θ)
⇒ v02 / 2 = v2 / 2 + gl(1 + sin θ)
Put gl = v2 / l sin θ using (1)
v02 / 2 = v2 / 2 + (v2 / sin θ)(1 + sin θ)
v02 / 2 = (3v2/2) + (v2 / sin θ)
⇒v / v0 = (sin θ /(2 + 3 sin θ))1/2
Last updated on Jun 16, 2025
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