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PhysicsW.E.P.Vertical circular motionHard2 minQB

PhysicsHardsingle choice

A bob attached to a string is released from horizontal position as shown in figure. For what value of $α$ (angle between string and vertical) during its motion, the total acceleration is directly horizontally?

Options:

Answer:

Solution:

$a_{t} a_{n} m g l (cos α) tan α ∴ cos α = g sin α = \frac{v ^{2}}{ℓ} = \frac{1}{2} m v^{2} \Rightarrow \frac{v ^{2}}{ℓ} = 2 g cos α = a_{n} = \frac{a _{n}}{a _{t}} = \frac{2 g cos α}{g sin α} \Rightarrow tan α = 2 = \frac{1}{3}$

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

⏱ 2mℹ️ Source: QB

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