Solved Problem on Energy, Work and Power

A boy is sitting on an igloo in a hemispherical shape, as shown in the figure. If it starts to slide from rest, neglecting friction, at what height h relative to the horizontal is the point O at which it will lose contact with the hemispherical surface of radius R?

Problem data:

Igloo radius: R.

Problem diagram:

When the boy is on top of the igloo the forces acting on him are the gravitational force \( {\vec F}_{T} \) and the normal force \( \vec{N} \). The condition for the boy to lose contact with the ice surface must be when the normal force is equal to zero, he takes off from the igloo, and there is no more reaction from the igloo on the boy. At that moment, the only force acting on him will be the gravitational force which can be decomposed into two components, a normal component, \( {\vec F}_{gN} \), in the radial direction and another tangential component, \( {\vec F}_{gT} \), in the direction of the velocity (Figure 1).

Figure 1

Solution

Using the Principle of Conservation of Mechanical Energy, the energy of the boy on the top of the igloo must equal the energy at point O where he loses contact with the hemisphere. Taking the ground as the Reference Level (R.L.), from which the Potential Energy will be measured, we have that, at the top of the igloo, the boy is at rest, his speed is equal to zero, so he only has Potential Energy proportional to R. At point O it has Kinetic Energy due to velocity v, and Potential Energy proportional to height h

\[ \begin{gather} E_{M}^{top}=E_{M}^{O}\\[5pt] U^{top}=K^{O} \end{gather} \]

the Potential Energy is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {U=mgH} \end{gather} \]

the Kinetic Energy is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {K=\frac{mv^{2}}{2}} \end{gather} \]

\[ \begin{gather} \cancel{m}gR=\frac{\cancel{m}v^{2}}{2}+\cancel{m}gh\\[5pt] gR=\frac{v^{2}}{2}+gh \tag{I} \end{gather} \]

For the calculation of v, we see from Figure 2 that the angle between the radius R of the igloo and the height h is θ

\[ \begin{gather} \cos \theta =\frac{h}{R} \tag{II} \end{gather} \]

Using Newton's Second Law for circular motion

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

Figure 2

The normal component of the gravitational force in the radial direction is given by

\[ \begin{gather} F_{gN}=F_{g}\cos \theta \tag{IV} \end{gather} \]

is the centripetal force that acts on the boy, and the centripetal acceleration is given by

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

substituting expressions (IV) and (V) into expression (III)

\[ \begin{gather} F_{g}\cos \theta =m\frac{v^{2}}{R} \end{gather} \]

the gravitational force is given by

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

\[ \begin{gather} \cancel{m}g\cos \theta =\cancel{m}\frac{v^{2}}{R}\\[5pt] g\cos \theta =\frac{v^{2}}{R} \tag{VI} \end{gather} \]

substituting expression (II) into expression (VI)

\[ \begin{gather} g\frac{h}{\cancel{R}}=\frac{v^{2}}{\cancel{R}}\\[5pt] v^{2}=gh \tag{VII} \end{gather} \]

substituting expression (VII) into expression (I)

\[ \begin{gather} \cancel{g}R=\frac{\cancel{g}h}{2}+\cancel{g}h\\[5pt] R=\frac{h}{2}+h\\[5pt] R=\frac{h+2h}{2}\\[5pt] R=\frac{3h}{2} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {h=\frac{2}{3}R} \end{gather} \]

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