Q35P - Fundamentals Of Physics

Question

A boy is initially seated on the top of a hemispherical ice mound of radius R = 13.8 m. He begins to slide down the ice, with a negligible initial speed (Figure). Approximate the ice as being frictionless. At what height does the boy lose contact with the ice?

Step-by-Step Solution

Answer

The height at which the boy lose contact with the ice is h = 9.20 m.

Step 1: Given

The radius of the hemispherical ice mound is R = 0.138 m . 

Step 2: Determining the concept

Use the concept of the energy conservation law and gravitational potential energy of the boy. According to the law of energy conservation, energy can neither be created, nor be destroyed.

 

Formulae:

$$\begin{gathered} U=mgh \\ K=\frac{1}{2}m{v}^2 \end{gathered}$$ 

 Where,

 K is kinetic energy, U is potential energy, m is mass, v is velocity, g is an acceleration due to gravity and h is height.

Step 3: Determining theheight at which the boy lose contact with the ice

Consider,$F_N$is the normal force acting on the boy in the upward direction as shown in the free body diagram. The acceleration of the boy at that point is centripetal acceleration,

$a=\frac{v^2}{2}$ 

According to the Newton’s second law,

$$\begin{gathered} F_{net}=ma \\ mg \cos \theta -F_N=m\frac{v^2}{2} \end{gathered}$$ 

 

The boy leaves the ice at the point$F_N=0$ , then,

 $$\begin{gathered} mg \cos \theta =m\frac{v^2}{2} \\ g \cos \theta =\frac{v^2}{2} \end{gathered}$$                                                 

                                                                                                                    (i)

The gravitational potential energy of the at the point where he leaves the mound is,

 $U=-mgR\left(1-\cos \theta \right)$

The body starts from the rest. The kinetic energy at that point is,

$K=\frac{1}{2}m{v}^2$ 

According to the energy conservation law,

$$\begin{gathered} 0=\frac{1}{2}m{v}^2-mgR\left(1-\cos \theta \right) \\ \frac{1}{2}m{v}^2=mgR\left(1-\cos \theta \right) \\ \frac{1}{2}{v}^2=gR\left(1-\cos \theta \right) \\ {v}^2=2gR\left(1-\cos \theta \right) \\ \mathrm{Equation} \left(\mathrm{i}\right) \mathrm{becomess} \mathrm{as}, \\ \mathrm{g} \cos \mathrm{\theta }=\frac{2\mathrm{gR}\left(1-\cos \mathrm{\theta }\right)}{\mathrm{R}} \\ \cos \mathrm{\theta }=2\left(1-\cos \mathrm{\theta }\right) \\ \cos \mathrm{\theta }=2-2 \cos \mathrm{\theta } \\ 3 \cos \mathrm{\theta }=2 \\ \cos \mathrm{\theta }=\frac{2}{3} \end{gathered}$$ 

 The height of the boy above the bottom of the mound is,

 $$\begin{gathered} h=R \cos \theta \\ h=13.8 \mathrm{mx}\frac{2}{3} \\ h=9.20 \mathrm{m} \end{gathered}$$

 Hence, the height at which the boy lose contact with the ice is h = 9.20 m

 

Therefore, the height of the boy at the point where he leaves the mound can be found by using the concept of conservation of energy and the gravitational potential energy.