Question

A bucket of water is swung in a circle in a vertical plane. The radius of the circle is 0.85 m. What is the minimum speed that will just prevent the water from falling out of the bucket when it is upside down (i.e., at the top of the circle)

Answer 1

Answer:2.88 m/s

Step-by-step explanation:

Given

radius of circle r=0.85 m

At highest Point weight will Provide the centripetal Force

thus

$weight=mg$

$Centripetal\ Force=\frac{mv^2}{r}$

where m is the mass of water in bucket

$mg=\frac{mv^2}{r}$

$v=\sqrt{gr}$

$v=\sqrt{9.8\times 0.85}$

$v=2.88 m/s$