Question

In vertical circular motion, what is the minimum velocity at the lowest point?

Hint:

Memorize these critical values for a string in vertical circle of radius $r$:
- Min velocity at bottom: $\sqrt{5gr}$
- Min velocity at top: $\sqrt{gr}$
- Min velocity at horizontal position: $\sqrt{3gr}$

Answer 1

Step 1: Understanding the Concept:
For a particle attached to a string to just complete a vertical circle of radius '\(r\)', it must not lose tension at the highest point. The critical condition is that tension \(T \ge 0\) at the highest point.
Step 2: Key Formula or Approach:
Use Newton's second law at the top of the circle to find the minimum velocity there (\(v_{top}\)). Then, apply the principle of conservation of mechanical energy between the lowest point and the highest point to find the minimum velocity at the bottom (\(v_{bottom}\)).
Step 3: Detailed Explanation:
At the highest point:
Forces acting towards the center are Tension (\(T\)) and weight (\(mg\)).
\[ T_{top} + mg = \frac{m v_{top}^2}{r} \]
For minimum velocity to complete the circle, the string just becomes slack at the top, so \(T_{top} \to 0\).
\[ 0 + mg = \frac{m v_{top}^2}{r} \implies v_{top}^2 = gr \implies v_{top} = \sqrt{gr} \]
Applying Conservation of Energy:
Total Energy at bottom = Total Energy at top
Let the lowest point be the reference level for potential energy (\(h=0\)). The height of the top point is \(2r\).
\[ (K.E. + P.E.)_{bottom} = (K.E. + P.E.)_{top} \]
\[ \frac{1}{2} m v_{bottom}^2 + 0 = \frac{1}{2} m v_{top}^2 + mg(2r) \]
Substitute \(v_{top}^2 = gr\):
\[ \frac{1}{2} m v_{bottom}^2 = \frac{1}{2} m (gr) + 2mgr \]
Divide by \(m\) and multiply by 2:
\[ v_{bottom}^2 = gr + 4gr \]
\[ v_{bottom}^2 = 5gr \]
\[ v_{bottom} = \sqrt{5gr} \]
Step 4: Final Answer:
The minimum velocity required at the lowest point to complete a vertical circle is \(\sqrt{5gr}\).