A particle of mass $2m$ kg on a string of length $L$ moves in a horizontal circle with the string at an angle $\theta$ to the vertical, at angular speed $\omega$ . Resolving vertically gives $T\cos\theta = 2mg$ and horizontally gives $T\sin\theta = 2m\omega^2 L\sin\theta$ . Which expression for $\omega^2$ follows from these two equations?

\omega^2 = \dfrac{g}{L\sin\theta}

, taking the radius as

L

and the vertical component as

g\sin\theta

\omega^2 = \dfrac{g\tan\theta}{L}

, using the full string length

L

as the radius of the circle.

\omega^2 = \dfrac{gL\cos\theta}{2m}

, keeping the mass

2m

in the result rather than cancelling it.

\omega^2 = \dfrac{g}{L\cos\theta}

, since dividing the horizontal equation by the vertical one leaves

\tan\theta

on the left.

Explanation

Dividing the horizontal equation by the vertical one removes the tension and the mass:

\tan\theta = \dfrac{\omega^2 L\sin\theta}{g}

. Writing

\tan\theta = \dfrac{\sin\theta}{\cos\theta}

and cancelling

\sin\theta

leaves

\omega^2 = \dfrac{g}{L\cos\theta}

. The mass cancels, so no numerical value of

m

is ever needed.

Derived from ZIMSEC Additional Mathematics Paper 2, Specimen Paper, Q1

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