91Ó°ÊÓ

Here we have, the mass of rock is\(m = 0.12{\rm{ }}kg\).

The length of the spring is \(l = 0.8{\rm{ }}m\).

The maximum angle with vertical is \(\theta = 45^\circ \).

Here there are other forces than gravity is acting on the rock.

According to the work-energy theorem, the net work done by forces acting on an item equals the change in its kinetic energy.

So, the work-energy theorem is given by:

\({K_1} + {U_1} + {W_{other}} = {K_2} + {U_2}\)

Where kinetic energy is given by:

\(K = \frac{1}{2}m{v^2}\) ….. (2)

And the gravitational potential energy is given by:

\({U_{grav}} = mgy\) ….. (3)

In a uniform circular motion, the acceleration is directed toward the center of the circle.

The acceleration is given by:

\({a_{rad}} = \frac{{{v^2}}}{R}\) ….. (4)

Let\(y = 0\)at the vertical position.

By taking point 1 at the vertical point and the point 2 at the maximum angle point, then you have,

\({y_1} = 0\)

\(\begin{aligned}{}{y_2} &= l\left( {1 - \cos \theta } \right)\\ &= 0.8\left( {1 - \cos 45} \right)\\ &= 0.234{\rm{ }}m\end{aligned}\)

Since the rock starts from the zero potential energy level, then you have

\({U_1} = 0\)

Since at the maximum height, the rock comes to rest, so you get,

\({K_2} = 0\)

Now, put value of\({y_2}\)and \({\rm{m}}\) in equation (3),

\(\begin{aligned}{}{U_2} &= \left( {0.12} \right) \times 9.8 \times 0.234\\ &= 0.29{\rm{ }}J\end{aligned}\)

The tension in the string is always perpendicular to the motion direction,

So you obtain,

\({W_{other}} = 0\)

Now, put all these values in equation (1).

\(\begin{aligned}{}{K_1} + 0 + 0 = 0 + 0.29\\{K_1} = 0.29{\rm{ }}J\end{aligned}\)

Now, put value of\({K_1}\) and \(m\) in equation (2).

\(\begin{aligned}{}0.29 &= \frac{1}{2}\left( {0.12} \right){v_1}^2\\{v_1} &= 2.14{\rm{ }}{m \mathord{\left/ {\vphantom {m s}} \right.\\} s}\end{aligned}\)

Hence the speed of the rock when the string passes through the vertical position\(2.14{\rm{ }}{m \mathord{\left/ {\vphantom {m s}} \right.\\} s}\).

Now, apply Newton’s second law to the rock at point 2 along radial direction, so you get,

\(\begin{aligned}{}\sum {{F_{red}} = {T_2} - mg\cos \theta } &= m{a_{rad}}\\{T_2} &= m{a_{rad}} + mg\cos \theta \end{aligned}\)

Now from equation (4)

\({T_2} = m\frac{{{v^2}}}{R} + mg\cos \theta \)

Here you have \({v_2} = 0\)

The tension at \(\theta = 45^\circ \) is,

\(\begin{aligned}{}{T_2} &= mg\cos \theta \\ &= 0.12 \times 9.8 \times \cos 45^\circ \\ &= 0.83{\rm{ }}N\end{aligned}\)

Hence, the tension in the string when it makes an angle of \(45^\circ \) with the vertical is\({T_2} = 0.83{\rm{ }}N\).

Now, apply Newton’s second law to the rock at point 1 along radial direction, so you obtain,

\(\begin{aligned}{}\sum {{F_{red}} = {T_1} - mg} &= m{a_{rad}}\\{T_1} = m{a_{rad}} + mg\end{aligned}\)

Now from equation (4), you can write,

\({T_1} = m\frac{{{v_1}^2}}{R} + mg\)

Let\(R = l\), then you get,

\(\begin{aligned}{}{T_1} &= m\frac{{{v_1}^2}}{l} + mg\\ &= 0.12 \times \frac{{{{\left( {2.14} \right)}^2}}}{{0.8}} + 0.12 \times 9.8\\ &= 1.87{\rm{ }}N\end{aligned}\)

Hence, the tension in the string when the string as it passes through the vertical is\(1.87{\rm{ }}N\).