5.40 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 g/R . What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω = 2g/R ? Neglect friction.

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8 months ago

Let θ be the angle made by the radius vector joining the bead and the centre of the wire with the downward direction. Let, N be the normal reaction.

mg = N cos?θ …….(1)

mr ω2 = N sin?θ ……(2)

m(R sin?θ ) ω2 = N sin?θ

Hence N = m(R) ω2

Substituting the value on N in eqn (1)

mg = mR ω2cos?θ

or cos?θ = g/ R ω2 ………(3)

As cos?θ  1, the bead will remain at th

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physics ncert solutions class 11th 2023

physics ncert solutions class 11th 2023

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