Answer:
a) [tex]v \approx 1.233\,\frac{m}{s}[/tex], b) [tex]v \approx 1.426\,\frac{m}{s}[/tex], c) [tex]\theta_{2} \approx 22.61^{\textdegree}[/tex]
Explanation:
a) The speed of the ball is determined by applying the Principle of Energy Conservation:
[tex]U_{g,1} + K_{1} = U_{g,2} + K_{2}[/tex]
The speed of the ball when the string makes an angle of 12° with the vertical is:
[tex]K_{2} = U_{g,1} - U_{g,2} + K_{1}[/tex]
[tex]\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot L \cdot [(1-\cos \theta_{1})-(1 -\cos \theta_{2})][/tex]
[tex]v = \sqrt{2\cdot g \cdot L \cdot (\cos \theta_{2} - \cos \theta_{1})}[/tex]
[tex]v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (1.2\,m)\cdot (\cos 12^{\textdegree}-\cos 24^{\textdegree})}[/tex]
[tex]v \approx 1.233\,\frac{m}{s}[/tex]
b) The maximum speed of the ball is:
[tex]v = \sqrt{2\cdot g \cdot L \cdot (\cos \theta_{2} - \cos \theta_{1})}[/tex]
[tex]v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (1.2\,m)\cdot (\cos 0^{\textdegree}-\cos 24^{\textdegree})}[/tex]
[tex]v \approx 1.426\,\frac{m}{s}[/tex]
c) The angle between the string and the vertical when the speed of the ball is one-third its maximum value is obtained by proving different values of [tex]\theta_{2}[/tex]. The solution is approximately:
[tex]\theta_{2} \approx 22.61^{\textdegree}[/tex]
