Problem 25 - Master Resource Book in JEE Main Physics

Question

A stone tied to a string of length \(L\) is whirled in a vertical circle, with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position, and has a speed \(u\). The magnitude of change in its velocity as it reaches a position, where the string is horizontal is (a) \(\sqrt{u^{2}-2 g L}\) (b) \(\sqrt{2 g L}\) (c) \(\sqrt{u^{2}-g L}\) (d) \(\sqrt{2\left(u^{2}-g L\right)}\)

Step-by-Step Solution

Answer

The answer is (a) \(\sqrt{u^2 - 2gL}\).

Step 1: Understand the Problem

We need to calculate the magnitude of the change in velocity of a stone tied to a string, moving in a vertical circle from the lowest point to the horizontal position. The length of the string is given as \(L\), and the initial speed of the stone at the lowest point is \(u\).

Step 2: Identify Relevant Physics Concepts

This problem involves circular motion and energy conservation. When the stone moves from the lowest point to the horizontal position, kinetic energy is partially converted to potential energy. We use these concepts to find the change in velocity.

Step 3: Apply Energy Conservation

At the lowest point, the stone has kinetic energy \(KE_1 = \frac{1}{2}mu^2\) and zero potential energy (choosing the lowest point as reference). At the horizontal position, it rises to height \(h = L\) and potential energy is \(PE_2 = mgL\).

Step 4: Plug in the Formulas

At the horizontal position, using energy conservation: \(KE_1 + PE_1 = KE_2 + PE_2\). Thus, \(\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgL\), where \(v\) is the speed at the horizontal position.

Step 5: Solve for Final Speed

Rearrange the equation \(\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgL\) to find \(v\):\[v = \sqrt{u^2 - 2gL}\]

Step 6: Calculate Change in Velocity Magnitude

The initial magnitude of velocity is \(u\) and the final magnitude is \(v = \sqrt{u^2 - 2gL}\). The magnitude of the change in velocity, \(\Delta v\), is given by the absolute difference: \[\Delta v = |u - \sqrt{u^2 - 2gL}|\].

Step 7: Compare with Given Options

The magnitude of the change in velocity was derived as \(\Delta v = \sqrt{u^2 - 2gL}\). This corresponds to option (a) \(\sqrt{u^2 - 2gL}\).

Key Concepts

Conservation of Energy

The principle of conservation of energy is key to analyzing the motion of the stone in this problem. Energy conservation states that within a closed system, the total energy remains constant. For the stone tied to a string, the system comprises two types of mechanical energy: kinetic energy and potential energy.
Kinetic energy is associated with motion, calculated using the formula \(KE = \frac{1}{2}mv^2\), where \(m\) is the mass and \(v\) is the velocity. Potential energy, on the other hand, depends on the stone's height and is given by \(PE = mgh\), where \(h\) is the height above the reference level and \(g\) is the acceleration due to gravity.
As the stone moves from the lowest point to the horizontal position, its potential energy increases while kinetic energy decreases proportionally. The transformation between these energies allows us to use conservation laws to find unknown variables in the stone's motion.

Change in Velocity

When analyzing the stone's motion, the change in velocity is crucial. Initially, the stone is at the lowest point with a known speed \(u\), giving it a specific kinetic energy. As it ascends to the horizontal position, its speed is reduced as some of this kinetic energy is converted to potential energy.
The formula for change in velocity can be derived from energy conservation. Using the equation \(\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgL\), where \(L\) is the length of the string and \(v\) is the speed at the horizontal position, we identify the final speed \(v = \sqrt{u^2 - 2gL}\). Then, the magnitude of change in velocity is calculated as \(|u - v|\). This approach gives a clear picture of how velocity changes due to varying energy forms.

Kinetic and Potential Energy

Kinetic and potential energy are fundamental concepts describing the energy states in this stone's motion. Kinetic energy arises from the stone's movement, while potential energy is related to its position in the gravitational field.

• Initially, at the lowest point, the stone has maximum kinetic energy: \(KE = \frac{1}{2}mu^2\). This is because its speed \(u\) is at its highest and the height from the reference level is zero.
• As the stone rises to the horizontal position, a portion of this kinetic energy is transformed into potential energy: \(PE = mgL\), where \(L\) is the vertical distance risen by the stone.
• This energy conversion results in decreased kinetic energy at the horizontal position, corresponding to a slower speed \( v = \sqrt{u^2 - 2gL}\).

Understanding these energy forms helps in visualizing how the stone's energy dynamically shifts between kinetic and potential as it moves along the circular path.