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For a stone whirled in a horizontal circle on a string, the components of the tension are:
A$T\sin\theta=mg$ and $T\cos\theta=\dfrac{mv^2}{r}$
B$T\cos\theta=mg$ and $T\sin\theta=\dfrac{mv^2}{r}$
C$T\cos\theta=\dfrac{mv^2}{r}$ and $T\sin\theta=mgr$
D$T\tan\theta=mg$ and $T\cos\theta=\dfrac{mv^2}{r}$
Answer & Solution
Correct answer: B. $T\cos\theta=mg$ and $T\sin\theta=\dfrac{mv^2}{r}$
1. The stone moves in a horizontal circle, so the string sweeps out a cone.
2. Vertically there is no acceleration, so the vertical component of tension balances the weight: $T\cos\theta=mg$.
3. Horizontally the net force is the centripetal force, supplied entirely by the horizontal component: $T\sin\theta=\dfrac{mv^2}{r}$.
4. Dividing the second by the first gives $\tan\theta=\dfrac{v^2}{rg}$, hence $v^2=rg\tan\theta$.
5. Swapping the sine and cosine, a recorded error, makes the vertical balance depend on the speed, which is wrong.
_Source: NECTA ACSEE 2023 Physics 131/1, Question 3: Mechanics (Uniform Circular Motion)_