Solved Problem on Dynamics

A cylinder has the axis in the vertical direction and radius R, spinning inside the cylinder in a horizontal plane there is a small sphere. Assuming that the coefficient of friction between the sphere and the wall of the cylinder is μ and the local acceleration due to gravity is g, calculate the minimum tangential speed of the particle spinning inside the cylinder without falling.

Problem data:

Radius of the cylinder: R;
Coefficient of friction: μ;
Acceleration due to gravity: g.

Problem diagram:

Figure 1

The forces that act on the sphere are the gravitational force \( {\vec F}_{g} \), pointing down, the force of friction \( {\vec F}_{f} \) between the sphere and the cylinder wall, that prevents the sphere fell by opposing the gravitational force, the normal reaction force \( \vec{N} \) of the cylinder on the sphere.

Solution

Applying Newton's Second Law for circular motion

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

In the horizontal direction, the centripetal force F_cp is given by the normal force of reaction N

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

the centripetal acceleration is given by

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

substituting expressions (II) and (III) into expression (I)

\[ \begin{gather} N=m\frac{v^{2}}{R} \tag{IV} \end{gather} \]

In the vertical direction, there is no motion, the gravitational force \( {\vec F}_{g} \) and the force of friction \( {\vec F}_{f} \) cancel

\[ \begin{gather} F_{g}=F_{f} \tag{V} \end{gather} \]

the gravitational force is given by

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

the force of friction is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{f}=\mu N} \tag{VII} \end{gather} \]

substituting expressions (VI) and (VII) into expression (V)

\[ \begin{gather} mg=\mu N\\[5pt] N=\frac{mg}{\mu } \tag{VIII} \end{gather} \]

substituting the expression (VIII) into expression (IV)

\[ \begin{gather} \frac{\cancel{m}g}{\mu }=\cancel{m}\frac{v^{2}}{R}\\[5pt] \frac{v^{2}}{R}=\frac{g}{\mu }\\[5pt] v^{2}=\frac{Rg}{\mu} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {v_{min}=\sqrt{\frac{Rg}{\mu }\;}} \end{gather} \]

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