A student has $65-cm$long arms. What is the minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any? The distance from the handle to the bottom of the bucket is $35cm$.

Student's arm length$=65cm$

The distance from the handle to the bottom of the bucket$=35cm$

Angular velocity can be characterized as the pace of progress of angular dislodging.

Calculate the minimum angular velocity of the bucket of the water. The total distance from the shoulder to the bucket is,

$R=65\mathrm{cm}+35\mathrm{cm}$

localid="1649398245082" $=\left(100\mathrm{cm}\left(\frac{{10}^{-2}\mathrm{m}}{1\mathrm{cm}}\right)\right)$

$=1\mathrm{m}$

The net force on the bucket of water is,

$F_{\text{net}}=F_{\text{cent}}$

Here,$F_{cent}$ is the centripetal force.
$mg=m{\omega }^{2}R$

$g={\omega }^{2}R$

Rearrange the above equation for$\omega$.

Rearrange the equation for $\omega$.

$\omega =\sqrt{\frac{g}{R}}$

Substitute the values

localid="1649398261321" $g=9.81\mathrm{m}/{\mathrm{s}}^2$

localid="1649398276295" $R=1\mathrm{m}$

localid="1649398291087" $\omega =\sqrt{\left(\frac{\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)}{1\mathrm{m}}\right)}$

localid="1649398301907" $=\left(3.13\mathrm{rad}/\mathrm{s}\left(\frac{1\mathrm{rev}}{2\pi }\right)\right)$

$=29.9\mathrm{rpm}$

$\approx 30rpm$

The minimum angular velocity of the bucket of water is $29.9rpm$or$30\mathrm{rpm}$.