A particle of mass $2m$ kg hangs from a light inextensible string of length $1.3$ m attached to a fixed point A. It moves at constant speed in a horizontal circle with the string at $35^\circ$ to the vertical. Taking $g = 9.81\,\mathrm{ms^{-2}}$ , calculate its angular speed in $\mathrm{rad\,s^{-1}}$ , to 2 significant figures.

Model answer

3.0

Also accepted: 3, 3.0 rad/s, 3.04, 3.035

Explanation

Vertically

T\cos 35^\circ = 2mg

and horizontally

T\sin 35^\circ = 2m\omega^2(1.3\sin 35^\circ)

. Dividing removes both

T

and the mass and gives

\omega^2 = \dfrac{g}{1.3\cos 35^\circ} = \dfrac{9.81}{1.3 \times 0.81915} = 9.212

, so

\omega = 3.035

, that is

3.0\,\mathrm{rad\,s^{-1}}

to 2 significant figures. Taking

g = 9.8

instead gives

3.033

and

g = 10

gives

3.063

, so the answer to 2 significant figures is

3.0

on any of the usual values of

g

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

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