Question

A small ball is tied to a light inextensible thread of length \(64\,cm\) and is whirled in a vertical circle. If the thread is just taut at the highest point of the circle, then the minimum speed of the ball at the lowest point is: (Take \(g=10\,m/s^{2}\))

• A. \(2.53\,m/s\)
• B. \(5.66\,m/s\)
• C. \(8.0\,m/s\)
• D. \(10.0\,m/s\)

Hint:

For vertical circular motion, remember: \[ \boxed{ v_{\text{top(min)}}=\sqrt{gr} } \] and \[ \boxed{ v_{\text{bottom(min)}}=\sqrt{5gr} } \] These two formulas are among the most frequently used results in vertical circle problems.

Answer 1

The Correct Option is B

Concept: For a body moving in a vertical circle, the minimum speed at the highest point occurs when the tension in the string just becomes zero. Hence, \[ T=0. \] Applying Newton's second law at the highest point, \[ mg=\frac{mv_t^2}{r}. \] Therefore, \[ \boxed{v_t=\sqrt{gr}} \] where \[ v_t=\text{Speed at the highest point}, \] \[ r=\text{Radius of the circular path}. \] Using the law of conservation of mechanical energy between the lowest and the highest points, \[ \frac12 mv_b^2 = \frac12 mv_t^2 + mg(2r), \] where \[ v_b=\text{Speed at the lowest point}. \] This gives \[ \boxed{v_b=\sqrt{5gr}}. \]

Step 1: Write the given data.
Given, \[ r=64\,cm=0.64\,m, \] \[ g=10\,m/s^2. \]

Step 2: Determine the minimum speed at the highest point.
Since the string is just taut, \[ T=0. \] Hence, \[ v_t=\sqrt{gr}. \] Substituting, \[ v_t = \sqrt{10\times0.64} = \sqrt{6.4} = 2.53\,m/s. \]

Step 3: Apply conservation of mechanical energy.
Using, \[ v_b=\sqrt{5gr}, \] we get \[ v_b = \sqrt{5\times10\times0.64}. \] \[ v_b = \sqrt{32}. \] Therefore, \[ v_b = 5.66\,m/s. \] Thus, \[ \boxed{v_b=5.66\,m/s.} \] Hence, the correct answer is \[ \boxed{\textbf{Option (B)}}. \]

Important Observation: Using the standard theory of vertical circular motion, \[ \boxed{v_{\text{bottom}}=\sqrt{5gr}} \] For \[ r=0.64\,m, \] the minimum speed at the lowest point is \[ \boxed{5.66\,m/s.} \] Thus, Option (B) is mathematically correct.