Question

A small container of water is placed on a turntable inside a microwave oven, at a radius of 12.0 cm from the center. The turntable rotates steadily, turning one revolution in each 7.25 s. What angle does the water surface make with the horizontal?

Short answer

The water surface's angle with the horizontal is $0.{527}^{°}$.

Step-by-step solution

Calculation of the centripetal acceleration

Calculate the acceleration caused by the turntable on the water's surface using the notion of centrifugal force.

Convert the following units of radius from cm to m:

$\begin{array}{c}r=12cm\\ =(12cm)\left(\frac{1m}{0.01m}\right)\\ =0.12m\end{array}$

The following is the centripetal acceleration of the water surface caused by the rotating turntable.

$\begin{array}{c}a={\omega }^{2}r\\ ={\left(\frac{2\pi }{T}\right)}^{2}r\\ =\frac{4{\pi }^{2}r}{T^2}\end{array}$

Here, r is the radius and T is the time period of revolution.

Substitute$0.12mforrand7.25sforT$

$\begin{array}{c}a=\frac{4{\pi }^2(0.12m)}{{(7.25s)}^2}\\ =9.013\times {10}^{-2}m/s^2\end{array}$

Determining the required water surface angle

The following is the angle formed by the water surface with the horizontal:

$\begin{array}{c}\theta =ta{n}^{-1}\left(\frac{ma}{mg}\right)\\ =ta{n}^{-1}\left(\frac{a}{g}\right)\end{array}$

Here, mg is the gravitational force and ma is the centrifugal force acting on it.

Substitute $9.013\times {10}^{-2}m/s^{2}foraand9.8m/s^{2}forg.$

$\begin{array}{c}\theta =ta{n}^{-1}\left(\frac{9.013\times {10}^{-2}m/s^2}{9.8m/s^2}\right)\\ =0.{527}^{°}\end{array}$

As a result, the water surface's angle with the horizontal is $0.{527}^{°}$.