Using the energy conservation, derive the expressions for the minimum speeds at different locations along a vertical circular motion controlled by gravity. Is zero speed possible a... Using the energy conservation, derive the expressions for the minimum speeds at different locations along a vertical circular motion controlled by gravity. Is zero speed possible at the uppermost point? Under what condition/s? Also prove that the difference between the extreme tensions (or normal forces) depends only upon the weight of the object. Read more

Steps to Solve

1. Understanding Energy Conservation
   The principle of conservation of energy states that the total mechanical energy (kinetic + potential) in a system remains constant if only conservative forces (like gravity) act on it.

2. Kinetic and Potential Energy at Different Points
   At any point in the vertical circle of radius ( r ):

• Kinetic Energy (KE) is given by ( KE = \frac{1}{2}mv^2 )
• Potential Energy (PE) at height ( h ) is given by ( PE = mgh )

3. At the Top of the Circle
   At the top of the circle, the height is ( 2r ). Setting up the energy equation:
   $$ KE_{top} + PE_{top} = KE_{bottom} + PE_{bottom} $$
   Substituting gives:
   $$ \frac{1}{2}mv_t^2 + mg(2r) = \frac{1}{2}mv_b^2 + mg(0) $$

4. Minimum Speed at the Top
   For minimum speed (( v_t )) at the top to maintain circular motion, the centripetal force must be provided by the gravitational force:
   $$ \frac{mv_t^2}{r} = mg \implies v_t^2 = gr
   $$
   Thus,
   $$ v_t = \sqrt{gr} $$

5. At the Bottom of the Circle
   At the bottom of the circle (height = 0):
   Using the energy conservation principle:
   $$ \frac{1}{2}mv_b^2 + mg(0) = \frac{1}{2}mv_t^2 + mg(2r) $$
   Simplifying gives: $$ \frac{1}{2}mv_b^2 = \frac{1}{2}m(gr) + mg(2r) $$
   Thus,
   $$ v_b^2 = gr + 4gr = 5gr \implies v_b = \sqrt{5gr} $$

6. Comparing Tensions at Different Points
   At the top of the circle: $$ T_{top} + mg = \frac{mv_t^2}{r} \implies T_{top} = \frac{mv_t^2}{r} - mg $$
   At the bottom of the circle: $$ T_{bottom} - mg = \frac{mv_b^2}{r} \implies T_{bottom} = \frac{mv_b^2}{r} + mg $$

7. Condition for Zero Speed at the Top
   Zero speed (( v_t = 0 )) at the top can occur if the height is such that potential energy equals total energy, which means vertical motion must stop at the top: $$ mg(2r) = \frac{1}{2}mv_t^2 = 0 $$ This occurs at the critical point of the highest height feasible without motion.