A thin circular loop of radius R rotates about its vertical diameter with an angular frequency ω . Show that a small bead on the wire loop remains at its lowermost point for ω ≤ g / R . What is the angle made by the radius vector joining the centre to the bead with the vertically downward direction for ω = 2 g / R ? Neglect friction.

Assume that the radius vector joining the bead with the centre make an angle θ, with the vertically downward direction. O P = R = Radius of the circle N = Normal reaction Then from the above diagram, vertical and horizontal equations of forces can be written as: m g = N c o s θ ……….. (i) m l ω 2 = N s i n θ ……... (ii) In Δ O P Q , we have: s i n θ = l R l = R s i n θ ……… (iii) Substituting equation (iii) in equation (ii), we get: m ( R s i n θ ) ω 2 = N s i n θ m R ω 2 = N ……. (iv) Substituting equation (iv) in equation (i), we get: m g = m R ω 2 c o s θ c o s θ = g R ω 2 …….. (v) Since we know that c o s θ ≤ 1 ⇒ g R ω 2 ≤ 1 ∴ ω > g R for bead to rotate in circular motion. if ω ≤ g R then bead will remain at lowest point of the circular loop. For ω = 2 g R or ω 2 = 2 g R …….. (vi) On equating equations (v) and (vi), we get: 2 g R = g R c o s θ c o s θ = 1 2 ∴ θ = c o s - 1 ( 0 .5 ) = 60 °

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Any net force causing uniform circular motion is called a tangential force.

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Important Questions on Circular Motion

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Science>Physics>Circular Motion>Tangential and Radial Acceleration

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Science>Physics>Circular Motion>Tangential and Radial Acceleration

A thin circular loop of radius

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ω = \sqrt{2 g / R}

? Neglect friction.

Any net force causing uniform circular motion is called a tangential force.

Important Questions on Circular Motion

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