A bucket tied at the end of a 1.6 m long string is whirled in a vertical circle with constant speed. What shou

Question

A bucket tied at the end of a 1.6 m long string is whirled in a vertical circle with constant speed. What should be the minimum speed so that the water from the bucket does not spill, when the bucket is at the highest position? (Take $g = 10 	ext{ m/s}^2$)

Accepted Answer

1. **Concept:** For water not to spill from the bucket at the highest point of a vertical circle, the weight of the water must be just sufficient to provide the necessary centripetal force. This corresponds to the condition where the normal reaction (or tension) becomes zero [Source 85].
2. **Critical Speed Formula:** The minimum speed ($v_{min}$) at the highest point is given by equating the gravitational force to the centripetal force:
   $$ mg = \frac{mv^2}{R} $$
   $$ v_{min} = \sqrt{Rg} $$
   [Source 85]
3. **Calculation:**
   - Radius ($R$) = Length of string = $1.6 	ext{ m}$
   - Gravity ($g$) = $10 	ext{ m/s}^2$
   Substitute the values:
   $$ v_{min} = \sqrt{1.6 	imes 10} $$
   $$ v_{min} = \sqrt{16} $$
   $$ v_{min} = 4 	ext{ m/s} $$

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