Question

Derive expressions for linear velocity at lowest position, mid-way position, and the top-most position for a particle revolving in a vertical circle, if it has to just complete circular motion without string slackening at the top.

Hint:

The velocity at different positions in a vertical circle can be found by using conservation of energy and considering the forces at the top, middle, and lowest positions.

Answer 1

Step 1: Mechanical Energy in a Vertical Circle.
For a particle revolving in a vertical circle, the mechanical energy is conserved. The total energy at any position is the sum of kinetic energy and potential energy. At the top-most position, the tension in the string is zero (just before slackening). The forces acting on the particle are gravitational force and centripetal force.
Step 2: Lowest Position Velocity.
At the lowest position, the total energy is purely kinetic: \[ K_1 = \frac{1}{2} m v_{\text{lowest}}^2 \] The potential energy at the lowest position is zero, so the total energy is just the kinetic energy.
Step 3: Mid-Way Position Velocity.
At the mid-way position, the energy is split between kinetic and potential energy. We can derive the velocity by considering the work-energy theorem.
Step 4: Top Position Velocity.
At the top-most position, the velocity is derived using the balance of forces. For circular motion to continue, the centripetal force must be provided by the tension and gravity. The condition for slackening is: \[ T_{\text{top}} = 0 \quad \text{so} \quad m \frac{v_{\text{top}}^2}{r} = mg \] Thus: \[ v_{\text{top}} = \sqrt{g r} \]