Solved Problem on Dynamics

A small mass m is placed into a container whose inner surface is a hemisphere of radius R. The container rotates around the vertical axis with angular velocity ω. Determine:
a) The magnitude of the force that the sphere makes against the wall;
b) The radius of the circumference described by the sphere when in equilibrium to the container.

Problem data:

Mass of sphere: m;
Radius of the hemisphere: R;
Angular velocity of the hemisphere: ω.

Problem diagram:

The container is spinning, and the sphere inside is in equilibrium at a certain height on the internal surface. This system is equivalent to a container at rest with a sphere spinning inside it with speed ω.
The forces that act on the sphere are the gravitational force \( {\vec F}_{g} \) vertically down, the normal reaction force \( \vec{N} \) is perpendicular to the hemisphere wall, pointed to the center of the hemisphere (Figure 1-A).

Figure 1

The hemisphere has a radius R, the distance d is measured between the center of the hemisphere and the center of the circumference with radius r described by the sphere (Figure 1-B).

Solution

Drawing the forces in a system of coordinate xy, we can apply Newton's Second Law for the circular motion.

\[ \begin{gather} \bbox[#99CCFF,10px] {{\vec F}_{cp}=m{\vec a}_{cp}} \tag{I} \end{gather} \]

a) The force that the sphere does against the wall of the hemisphere is equal, in magnitude, to the force that the hemisphere does on the sphere. These forces are a pair of forces of action and reaction, according to Newton's Third Law.

Figure 2

The resultant centripetal force \( {\vec F}_{cp} \) is given by the component of the normal force in the x direction, \( {\vec N}_{x} \) (Figure 2)

\[ \begin{gather} F_{cp}=N_{x} \tag{II} \end{gather} \]

the N_x component is given by

\[ \begin{gather} N_{x}=N\cos \theta \tag{III} \end{gather} \]

the cosine is given by (Figure 1-B)

\[ \begin{gather} \cos \theta =\frac{r}{R} \tag{IV} \end{gather} \]

substituting the expression (IV) into expression (III)

\[ \begin{gather} N_{x}=N\frac{r}{R} \tag{V} \end{gather} \]

The centripetal acceleration is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {a_{cp}=\frac{v^{2}}{r}} \tag{VI} \end{gather} \]

the velocity as a function of angular velocity is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {v=\omega r} \tag{VII} \end{gather} \]

substituting the expression (VII) into expression (VI)

\[ \begin{gather} a_{cp}=\frac{(\omega r)^{2}}{r}\\[5pt] a_{cp}=\frac{\omega^{2}r^{\cancel{2}}}{\cancel{r}}\\[5pt] a_{cp}=\omega^{2}r \tag{VIII} \end{gather} \]

substituting the expressions (V) and (VIII) into expression (I)

\[ \begin{gather} N\frac{\cancel{r}}{R}=m\omega^{2}\cancel{r} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {N=Rm\omega ^{2}} \end{gather} \]

b) In the y direction, there is no movement, the gravitational force \( {\vec F}_{g} \) and the component of the normal reaction force \( {\vec N}_{y} \) cancel out.

\[ \begin{gather} F_{g}=N_{y} \tag{IX} \end{gather} \]

the gravitational force is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{g}=mg} \tag{X} \end{gather} \]

the N_y component is given by

\[ \begin{gather} N_{y}=N\sin \theta \tag{XI} \end{gather} \]

the sine is given by (Figure 1-B)

\[ \begin{gather} \sin \theta =\frac{d}{R} \tag{XII} \end{gather} \]

substituting the expression (XII) and the result of the normal force found in the previous item into expression (XIII)

\[ \begin{gather} N_{y}=Rm\omega ^{2}\frac{d}{R} \tag{XIII} \end{gather} \]

substituting expressions (X) and (XIII) into expression (IX)

\[ \begin{gather} \cancel{m}g=\cancel{R}\cancel{m}\omega^{2}\frac{d}{\cancel{R}}\\[5pt] g=\omega ^{2}d\\[5pt] d=\frac{g}{\omega ^{2}} \tag{XIV} \end{gather} \]

Applying the Pythagorean Theorem to the triangle of Figure 1-B

\[ \begin{gather} R^{2}=d^{2}+r^{2} \end{gather} \]

substituting the value of d found in (XIV)

\[ \begin{gather} R^{2}=\left(\frac{g}{\omega^{2}}\right)^{2}+r^{2}\\[5pt] R^{2}=\frac{g^{2}}{\omega^{4}}+r^{2}\\[5pt] r^{2}=R^{2}-\frac{g^{2}}{\omega ^{4}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {r=\sqrt{R^{2}-\frac{g^{2}}{\omega ^{4}}\;}} \end{gather} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .