A particle of mass $1$ kg threaded on a smooth circular wire of radius $0.5$ m reaches the lowest point of the wire with $v^2 = 2g$ . Which expression gives the reaction $R$ between the wire and the particle there?

R = g - \dfrac{2g}{0.5} = -3g

, taking the centripetal force as acting away from the centre O.

R = \dfrac{2g}{0.5} = 4g

, since at the lowest point the reaction supplies the centripetal force by itself.

R = g + \dfrac{2g}{0.5} = 5g

, since

R

must both support the weight and supply the centripetal force.

R = g

, since the particle is momentarily moving horizontally and so is in vertical equilibrium.

Explanation

At the lowest point the centre O is directly above the particle, so the acceleration

\dfrac{v^2}{r}

is directed upwards. Newton's second law upwards gives

R - mg = \dfrac{mv^2}{r}

. With

m = 1

v^2 = 2g

and

r = 0.5

R = g + \dfrac{2g}{0.5} = g + 4g = 5g\,\mathrm{N}

, which is

49.05

N when

g = 9.81\,\mathrm{ms^{-2}}

Derived from ZIMSEC Additional Mathematics Paper 2, Specimen Paper, Q3

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