Question

Water in a bucket is whirled in a vertical circle with a string to it. The water does not fallen even when the bucket is inverted at the top of its path . We conclude that in this position:

• A. $mg=\frac{m{v}^2}{r}$
• B. $mg$ is greater than $\frac{m{v}^2}{r}$
• C. $mg$ is not greater than $\frac{m{v}^2}{r}$
• D. $mg$ is not less than $\frac{m{v}^2}{r}$

Answer 1

Answer: (C)

$mg$ is not greater than $\frac{m{v}^2}{r}$

Its because a centrifugal force $\frac{m{v}^2}{r}$ acts on the water which is away from the center and opposite to force of gravity $mg$. If the $mg$ is greater then $\frac{m{v}^2}{r}$ water will spill out. but if $\frac{m{v}^2}{r}$ is greater then $mg$ water will not spill out of the bucket.
Note: for keeping the water in bucket while rotating it in a vertical circle you will have to keep the velocity above a certain value to make $\frac{m{v}^2}{r}$ greater then $mg$.