A bucket containing water is revolved in a ve | vedclass

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(B) For water not to fall from a bucket revolved in a vertical circle, the minimum velocity $v$ at the highest point must be at least $\sqrt{rg}$.Give

Vedclass · Accepted Answer

$2\pi \sqrt{\frac{r}{g}}$

Vedclass · Answer

(B) For water not to fall from a bucket revolved in a vertical circle, the minimum velocity $v$ at the highest point must be at least $\sqrt{rg}$.Given the radius is $r$ and gravitational acceleration is $g$.The angular velocity $\omega$ is related to linear velocity $v$ by the formula $v = r\omega$, which implies $\omega = v/r$.Substituting the minimum velocity $v = \sqrt{rg}$, we get $\omega = \frac{\sqrt{rg}}{r} = \sqrt{\frac{g}{r}}$.The period of revolution $T$ is given by $T = \frac{2\pi}{\omega}$.Substituting the value of $\omega$, we get $T = 2\pi \sqrt{\frac{r}{g}}$.

$A$ bucket containing water is revolved in a vertical circle of radius $r$. To prevent the water from falling down, the period of revolution required is ($g =$ gravitational acceleration)

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