Question

A body of mass m is rotated in a vertical circle with help of light string such that velocity of be a point is equal to critical velocity at that point. If $T_1, T_2$ be the tensions in the string when the b crossing the highest and the lowest positions then the following relation is correct.

• A. $T_2 - T_1 = 6mg$
• B. $T_2 - T_1 = 4mg$
• C. $T_2 - T_1 = 3mg$
• D. $T_2 - T_1 = 2mg$

Answer 1

Answer: (A)

$T_2 - T_1 = 6mg$

Answer 2

To solve this problem, we need to analyze the forces acting on the body at the highest and lowest points of its vertical circular motion and use the principle of conservation of mechanical energy.

Let:

• $m$ be the mass of the body.
• $R$ be the radius of the vertical circle (length of the string).
• $v_1$ be the velocity of the body at the highest point.
• $v_2$ be the velocity of the body at the lowest point.
• $T_1$ be the tension in the string at the highest point.
• $T_2$ be the tension in the string at the lowest point.

1. Forces at the Highest Point: At the highest point, both the tension $T_1$ and the gravitational force $mg$ act downwards, towards the center of the circle. The net force provides the centripetal force: $T_1 + mg = \frac{mv_1^2}{R} \quad \text{(Equation 1)}$

The problem states that the velocity at a point is equal to the critical velocity at that point. For a body rotating in a vertical circle with a string, the critical velocity at the highest point ($v_1$) is the minimum velocity required to complete the circle, which is $\sqrt{gR}$. At this critical velocity, the tension in the string at the highest point ($T_1$) is zero. Substituting $v_1 = \sqrt{gR}$ into Equation 1: $T_1 + mg = \frac{m(\sqrt{gR})^2}{R}$ $T_1 + mg = \frac{mgR}{R}$ $T_1 + mg = mg$ $T_1 = 0 \quad \text{(This confirms the critical velocity condition implies zero tension at the top)}$

2. Forces at the Lowest Point: At the lowest point, the tension $T_2$ acts upwards (towards the center), and the gravitational force $mg$ acts downwards (away from the center). The net force provides the centripetal force: $T_2 - mg = \frac{mv_2^2}{R} \quad \text{(Equation 2)}$

3. Conservation of Mechanical Energy: We can relate the velocities at the highest and lowest points using the conservation of mechanical energy. Let's take the lowest point as the reference level for potential energy (PE = 0). Energy at the lowest point: $E_{lowest} = KE_{lowest} + PE_{lowest} = \frac{1}{2}mv_2^2 + 0$ Energy at the highest point: The height of the highest point above the lowest point is $2R$. $E_{highest} = KE_{highest} + PE_{highest} = \frac{1}{2}mv_1^2 + mg(2R)$ By conservation of mechanical energy, $E_{lowest} = E_{highest}$: $\frac{1}{2}mv_2^2 = \frac{1}{2}mv_1^2 + 2mgR$ Multiplying by $\frac{2}{m}$: $v_2^2 = v_1^2 + 4gR \quad \text{(Equation 3)}$

Now, substitute $v_1^2 = (\sqrt{gR})^2 = gR$ into Equation 3: $v_2^2 = gR + 4gR$ $v_2^2 = 5gR$

4. Calculate $T_2$: Substitute the value of $v_2^2$ into Equation 2: $T_2 - mg = \frac{m(5gR)}{R}$ $T_2 - mg = 5mg$ $T_2 = 5mg + mg$ $T_2 = 6mg$

5. Calculate the Difference $T_2 - T_1$: $T_2 - T_1 = 6mg - 0$ $T_2 - T_1 = 6mg$

This relationship, $T_2 - T_1 = 6mg$, is a general result for vertical circular motion, irrespective of whether the velocity is critical or not, as long as the motion completes the circle. The "critical velocity" condition simply sets $T_1=0$ for this specific case.