Question 39: Work, Energy and Power & Circular Motion

Question

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 $10 \mathrm{~m} / \mathrm{s}$. The ratio of its kinetic energies at point B and C is :

(Take acceleration due to gravity as $10 \mathrm{~m} / \mathrm{s}^2$)

Accepted Answer

Step 1: Use Energy Conservation at Points A and B
At point A, the kinetic energy is from velocity and the potential energy is set at reference (usually zero).

At point B, both kinetic and potential energy are present.
The conservation of energy formula between A and B is:

$$ \frac{1}{2} m 	imes 100 + 0 = \frac{1}{2} m V_B^2 + mg \left(R - \frac{R \sqrt{3}}{2}ight) $$
Step 2: Use Given Values
Given: $ R = 2 $ m, $ g = 10\,\mathrm{m/s^2} $, and initial velocity at A is $ 10\,\mathrm{m/s} $, so $ (10)^2 = 100 $.
Substitute these into the equation:

$$ 100 = V_B^2 + 2gR\left(1 - \frac{\sqrt{3}}{2}ight) $$

Here, $ m $ cancels out because it is on both sides.
Step 3: Simplify for $ V_B^2 $
Calculate $ 2gR = 2 	imes 10 	imes 2 = 40 $.

So,

$$ 100 = V_B^2 + 40 \left(1 - \frac{\sqrt{3}}{2}ight) $$
$ 40(1 - \frac{\sqrt{3}}{2}) = 40 - 20\sqrt{3} $
Rearrange to solve for $ V_B^2 $:

$$ V_B^2 = 100 - 40 + 20\sqrt{3} $$

$$ V_B^2 = 60 + 20\sqrt{3} $$
So, kinetic energy at point B:

$$ K.E._B = \frac{1}{2} m V_B^2 = \frac{m}{2}(60 + 20\sqrt{3}) $$
Step 4: Use Energy Conservation at Points A and C
At point C, use conservation of energy again:

$$ \frac{1}{2} m (10)^2 = \frac{1}{2} m V_C^2 + mg \left(\frac{3R}{2}ight) $$
The potential energy term is $ mg 	imes \frac{3R}{2} $.

Substitute values:

$$ 100 = V_C^2 + 2gR 	imes \frac{3}{2} $$

$ 2gR = 40 $, so $ 40 	imes \frac{3}{2} = 60 $
So:

$$ 100 = V_C^2 + 60 $$

$$ V_C^2 = 100 - 60 = 40 $$
So, kinetic energy at point C:

$$ K.E._C = \frac{1}{2} m V_C^2 = \frac{1}{2} m (40) $$
Step 5: Find the Ratio
The ratio of kinetic energies at B and C is:

$$ \frac{K.E._B}{K.E._C} = \frac{\frac{1}{2} m (60 + 20\sqrt{3})}{\frac{1}{2} m (40)} = \frac{60 + 20\sqrt{3}}{40} = \frac{3}{2} + \frac{\sqrt{3}}{2} = \frac{3 + \sqrt{3}}{2} $$

Answer

$\frac{3-\sqrt{2}}{2}$

Answer

$\frac{2+\sqrt{3}}{3}$

Answer

$\frac{2+\sqrt{2}}{3}$

Answer

$\frac{3+\sqrt{3}}{2}$

Work, Energy and Power & Circular Motion

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 \mathrm{~m} / \mathrm{s}. The ratio of its kinetic energies at point B and C is : (Take acceleration due to gravity as 10 \mathrm{~m} / \mathrm{s}^2)

Choose the correct answer: