91Ó°ÊÓ

Let \(R\) be the radius of the snowball.

We take \(h = 0\) at the starting point and at the other of losing contact point.

Then we have,

\(\begin{aligned}{}{h_1} = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{v_1} = 0\\{h_2} = - R\left( {1 - \cos \alpha } \right),\,\,{v_2} = v\end{aligned}\)

\(\)

Since there are other forces acting other than gravity, work-energy theorem is given by,

\({K_1} + {U_1} + W = {K_2} + {U_2}\,\,\, \cdots \cdots \left( 1 \right)\) ,

where the kinetic energy is given by,

\(K = \frac{1}{2}m{v^2}\,\, \cdots \cdots \left( 2 \right)\)

And the gravitational potential energy is given by,

\(U = mgh\,\,\, \cdots \cdots \left( 3 \right)\).

From chapter 5, we know that, in circular motion, the acceleration vector is directed towards the center of the circle and its magnitude is given by,

\(a = \frac{{{v^2}}}{R}\,\,\, \cdots \cdots \left( 4 \right)\)

Let us calculate energy quantities.

Putting \({v_1}\) in \(\left( 2 \right)\), we get,

\({K_1} = 0\).

Also, substituting \(\,{h_1}\) in \(\left( 3 \right)\), we get,

\({U_1} = 0\).

Now, we substitute \({v_2}\) in \(\left( 2 \right)\), we get,

\({K_2} = \frac{1}{2}mv_2^2\,\,\).

and, substituting \(\,{h_2}\) in \(\left( 3 \right)\), we get,

\({U_2} = - mgR\left( {1 - \cos \alpha } \right)\).

We know that the normal force always acts perpendicular to the direction of motion.

Therefore, the work done is,

\(W = 0\).

Substituting all these values of energy and work quantities in \(\left( 1 \right)\), we get,

\(\begin{aligned}{}0 + 0 + 0 = \frac{1}{2}m{v^2} - mgR\left( {1 - \cos \alpha } \right)\\ \Rightarrow {v^2} = 2mgR\left( {1 - \cos \alpha } \right)\,\,\,\,\, \cdots \cdots \left( 5 \right)\end{aligned}\)

Applying Newton’s Second Law, to the skier along the radial direction, we get,

\(\sum {F = mg\cos \alpha } - n = m\,a\)

Clearly, she will lose contact with the snowball when there is no normal force i.e. \(n = 0\).

Thus, from \(\left( 4 \right)\), substituting the value of \(a\) in above equation, we get,

\(\begin{aligned}{}mg\cos \alpha = m\,\frac{{{v^2}}}{R}\\ \Rightarrow g\cos \alpha = \frac{{{v^2}}}{R}\end{aligned}\)

Now, using the value from \(\left( 5 \right)\) in above equation, we get,

\(\begin{aligned}{}g\cos \alpha = \frac{{2gR\left( {1 - \cos \alpha } \right)}}{R}\\ \Rightarrow \cos \alpha = 2\left( {1 - \cos \alpha } \right)\\ \Rightarrow 3\cos \alpha = 2\\ \Rightarrow \cos \alpha = \frac{2}{3}\end{aligned}\)

\(\begin{aligned}{} \Rightarrow \alpha = {\cos ^{ - 1}}\frac{2}{3}\\ \Rightarrow \alpha = 48.19^\circ \end{aligned}\)

Hence, A radial line from the center of the snowball to the skier makes an angle of \(\alpha = 48.19^\circ \) with the vertical.