Physics - W.E.P. Question with Solution | TestHub

The situation is shown in the diagram. The angle θ made by the string with the vertical is given by

$\sin \left(\theta \right)=\frac{r}{L}$...(i)



The forces on the particle are

(1) the tension T along the string and

(2) the weight mg vertically downward.

The particle is moving in a circle with constant speed$v$. Thus, the radial acceleration towards the centre has magnitude$\frac{v^2}{r}$. Resolving the forces along the radial direction and applying Newton's second law,

$T \sin \theta = \frac{m{v}^2}{r}$...(ii)

As there is no acceleration in the vertical direction, net force in the vertical direction is zero,

$T \cos \theta = m\mathrm{g}$...(iii)

Dividing (ii) by (iii),

$\text{tan}\text{θ}=\frac{{\text{v}}^2}{\text{rg}}$

$\mathrm{or}\text{v}=\sqrt{\text{rg}\text{tan}\text{θ}}$

And from (iii),

$\text{T}=\frac{\text{mg}}{\text{cos θ}}$

Using (i),

$\text{v}=\frac{\text{r}\sqrt{\text{g}}}{{\left({\text{L}}^2-{\text{r}}^2\right)}^{\text{1/4}}}\text{and}\text{T}=\frac{\text{mgL}}{{\left({\text{L}}^2-{\text{r}}^2\right)}^{\text{1/2}}}$​