Question 32: Work, Energy and Power & Circular Motion

Question

In case of vertical circular motion of a particle by a thread of length $r$ if the tension in the thread is zero at an angle $30^{\circ}$ shown in figure, the velocity at the bottom point $(A)$ of the circular path is (g = gravitational acceleration)

Accepted Answer

Let the point where the tension becomes zero be P .
At any point in vertical circular motion, the net radial force provides the necessary centripetal force. The forces acting on the particle at point P are :
1. Tension (T) is acting towards the centre.
2. Weight (mg) is acting vertically downwards.
The component of weight acting away along the radius is $\mathrm{mg} \sin \left(30^{\circ}\right)$. Thus, the equation of motion is :
$$ \mathrm{T}+\mathrm{mg} \sin \left(30^{\circ}\right)=\frac{\mathrm{m} \mathrm{v}_{\mathrm{P}}^2}{\mathrm{r}} $$
The problem states that at this point, the tension $\mathrm{T}=0$. Substituting this :
$$ 0+m g \sin \left(30^{\circ}\right)=\frac{m v_p^2}{r} $$
$$\Rightarrow $$ $\mathrm{mg}\left(\frac{1}{2}\right)=\frac{\mathrm{mv}_{\mathrm{p}}^2}{\mathrm{r}}$
$$\Rightarrow $$ $$ \mathrm{v}_{\mathrm{p}}^2=\frac{\mathrm{gr}}{2} \ldots (i)$$
We need to find the vertical height of point P relative to the bottom point A .
From bottom A to the centre, the height is r . And from the centre to point P , the vertical displacement is $\mathrm{r} \sin \left(30^{\circ}\right)$.
Total height h from A to $\mathrm{P}=\mathrm{h}=\mathrm{r}+\mathrm{r} \sin \left(30^{\circ}\right)=\mathrm{r}+\frac{\mathrm{r}}{2}=\frac{3 \mathrm{r}}{2}$
Applying law of conservation of energy between points A and P ,
Total Mechanical Energy at $\mathrm{A}=$ Total Mechanical Energy at P
Let $\mathrm{v}_{\mathrm{A}}$ be the velocity at the bottom point. Taking point A as the reference for zero potential energy :
$$ \frac{1}{2} m v_{\mathrm{A}}^2=\frac{1}{2} m v_{\mathrm{P}}^2+m g h $$
$$\Rightarrow $$ $\frac{v_A^2}{2}=\frac{1}{2}\left(\frac{g r}{2}\right)+g\left(\frac{3 r}{2}\right)$
$$\Rightarrow $$ $\frac{\mathrm{v}_{\mathrm{A}}^2}{2}=\frac{\mathrm{gr}}{4}+\frac{3 \mathrm{gr}}{2}$
$$\Rightarrow $$ $\mathrm{v}_{\mathrm{A}}^2=\frac{\mathrm{gr}}{2}+3 \mathrm{gr}=\frac{7}{2} \mathrm{gr}$
$$\Rightarrow $$ $$ \mathrm{v}_{\mathrm{A}}=\sqrt{\frac{7}{2} \mathrm{gr}} $$
The velocity at the bottom point A is $\sqrt{\frac{7}{2} \mathrm{gr}}$.
Hence, the correct option is (A).

Answer

$\sqrt{\frac{7}{2} g r}$

Answer

$\sqrt{4 g r}$

Answer

$\sqrt{\frac{5}{2} g r}$

Answer

$\sqrt{5 g r}$

Work, Energy and Power & Circular Motion

In case of vertical circular motion of a particle by a thread of length r if the tension in the thread is zero at an angle 30^{\circ} shown in figure, the velocity at the bottom point (A) of the circular path is (g = gravitational acceleration)

Choose the correct answer: