Question

A stone is tied to a string of length 'l' and is whirled in a vertical circle with the other end of the string as the center. At a certain instant of time, the stone is at its lowest position and has a speed of 'u'. The magnitude of the change in velocity as it reaches a position where the string is horizontal (g being acceleration due to gravity) is

• A. \(\sqrt{u^2-gl}\)
• B. \(u-\sqrt{u^2-2gl}\)
• C. \(\sqrt{2gl}\)
• D. \(\sqrt{2(u^2-gl)}\)

Answer 1

The Correct Option is D

To solve this problem, we need to understand the motion of the stone tied to a string being whirled in a vertical circle. When the stone is at its lowest position, it has a speed 'u'. Our task is to find the magnitude of the change in velocity when the stone reaches a horizontal position.

Step 1: Understanding the Motion

The stone moves in a vertical circle, and we are interested in the positions when the string is at the bottom and when it is horizontal. At the lowest point, the stone's speed is 'u'. At the horizontal position, we need to find the new speed and subsequently the change in velocity.

Step 2: Conservation of Mechanical Energy

The mechanical energy of the stone is conserved as it moves from the lowest point to the horizontal point (ignoring air resistance and other non-conservative forces). At the lowest position:

• Kinetic Energy, \( KE_1 = \frac{1}{2} m u^2 \)
• Potential Energy, \( PE_1 = 0 \) (taking the lowest point as the reference level)

When the stone reaches the horizontal position:

• Kinetic Energy, \( KE_2 = \frac{1}{2} m v^2 \)
• Potential Energy, \( PE_2 = mgh = mg \left( \frac{l}{2} \right) \), where height \( h = \frac{l}{2} \)

By conservation of energy, \( KE_1 + PE_1 = KE_2 + PE_2 \), we get:

\(\frac{1}{2} m u^2 = \frac{1}{2} m v^2 + mg \frac{l}{2}\)

Simplifying, we have:

\(u^2 = v^2 + gl\)

Thus, the speed at the horizontal position, \( v = \sqrt{u^2 - gl} \).

Step 3: Calculating the Change in Velocity

The change in velocity vector is given by the difference in the velocity magnitudes from the lowest point to the horizontal point. However, since these velocities are in perpendicular directions (lowest point being vertical and horizontal at the required position), we need to calculate the magnitude of the vector change:

\(\Delta V = \sqrt{u^2 + (\sqrt{u^2-gl})^2}\)

Simplifying:

\(= \sqrt{u^2 + (u^2 - gl)} = \sqrt{2u^2 - gl}\)

Hence, the magnitude of the change in velocity as it reaches the horizontal position is:

\(\sqrt{2(u^2-gl)}\), which matches the provided correct answer.