A body of mass 100 g is moving in circular path of radius 2 m on vertical plane as shown in figure. The velocity of the body at point A is 10 m/s. The ratio of its kinetic energies at point B and C is :

(Take acceleration due to gravity as 10 m/s²)

To solve this, we can use the conservation of mechanical energy principle, which states:

$$\text{Total Energy at any point} = \text{Kinetic Energy (K)} + \text{Potential Energy (U)} = \text{Constant}.$$

$$E_{\text{total}} = K + U = \frac{1}{2} m v^2 + m g h,$$

where:

• $v$ = velocity at the point,
• $h$ = height above reference point A.

We calculate kinetic energy by finding the velocity at points B and C using energy conservation.

• At A: $h_A = 0 \, \text{m}$ (reference point),
• At B: ${\mathrm{h}}_{\mathrm{B}}=\mathrm{r}(1-\cos {30}^{\circ })=2-\sqrt{3} \mathrm{m}$,
• At C: $h_C=r=2+2\cos 60°=3\text{m}$.

At point A:

$$E_A = K_A + U_A = \frac{1}{2} m v_A^2 + m g h_A.$$

Substituting the values:

$$E_A = \frac{1}{2} (0.1) (10)^2 + (0.1)(10)(0),$$

 $$E_A = 5 \, \text{J}.$$ 

Using conservation of mechanical energy:

$$E_A = E_B = E_C.$$

At B:

$$E_B = \frac{1}{2} m v_B^2 + m g h_B.$$

Substitute $E_B=5$:

$5=K_B+\left(0.1\right)\left(10\right)(2-\sqrt{3})\mathrm{.}$

 ${\mathrm{K}}_{\mathrm{B}}=3+\sqrt{3} \mathrm{J}$

At C:

$$E_C = \frac{1}{2} m v_C^2 + m g h_C.$$

Substitute $E_C=5$:

$5=K_C+\left(0.1\right)\left(10\right)\left(3\right)\mathrm{.}$

 ${\mathrm{K}}_{\mathrm{C}}=2 \mathrm{J}$ 

Thus,  $\frac{K_B}{K_C}=\frac{3+\sqrt{3}}{2}$