Question

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 $35cm$.

Short answer

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

Step-by-step solution

Given information

Student's arm length$=65cm$

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

Angular velocity 

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 

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

$=29.9\mathrm{rpm}$

$\approx 30rpm$

Final answer

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