Chapter 6: Problem 76
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 the 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) \(\sqrt{2\left(u^{2}-g L\right)}\)
Short Answer
Step by step solution
Identify the types of motion and energy involved
Write the energy conservation equation
Solve for the horizontal speed v
Determine the velocity change
Choose the correct option from the given choices
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Energy Conservation
- At the lowest point, the stone has its maximum kinetic energy because its speed is highest.
- As the stone moves up to the horizontal position, some of this kinetic energy is converted into potential energy because it is against gravity.
Vertical Circular Motion
- The object experiences a change in gravitational potential energy as it moves up or down.
- Gravity acts as a centripetal force component helping to keep the object in motion.
- The tension in the string varies based on the object's position in the circle.
Mechanical Energy
- Potential energy increases as the stone is raised against gravity.
- Kinetic energy relates directly to the stone's speed; higher speeds mean more kinetic energy.
- In a closed system like our example, potential energy transformations directly affect kinetic energy.
Change in Velocity
- When the stone is at the bottom, the velocity is directed vertically with speed \(u\).
- As the stone reaches the horizontal position, this speed reduces to \(\sqrt{u^2 - 2gL}\) due to energy conservation.
- The directions of velocity also change, from vertical to horizontal.