Answer:
Pluck velocity = 83.34 m/s
Explanation:
The initial angular speed of the rod before it was hit, [tex]w_{r} = 0[/tex]
The angular speed of the rod and pluck rotating together, [tex]w_{2} = \frac{2\pi }{T}[/tex]
The period of rotation, T = 0.736 s
[tex]w_{2} = \frac{2\pi }{0.736}\\w_{2} = 8.53 rad/s[/tex]
Using the principle of conservation of angular momentum:
[tex]I_{r} w_{r} + I_{p} w_{p} = (I_{r}+ I_{p}) w_{2}[/tex]
[tex]I_{p} = ml^{2}[/tex]
[tex]w_{p} = v_{p} /l[/tex]
[tex]I_{p} w_{p} = ml^{2} * \frac{v_{p} }{l} \\I_{p} w_{p} = mlv_{p}[/tex]
[tex]I_{r} w_{r} +mlv_{p} = (I_{r}+ I_{p}) w_{2}\\mlv_{p} =( (I_{r}+ I_{p}) w_{2} ) - I_{r} w_{r}\\v_{p} = \frac{( (I_{r}+ I_{p}) w_{2} ) - I_{r} w_{r}}{ml}[/tex]
[tex]I_{r} = 1/3 ML^{2} \\I_{r} = 1/3 * 1.9 * 2^{2} \\I_{r} = 2.533 kg m^{2}[/tex]
[tex]I_{p} = ml^{2} \\I_{p} = 0.163 * 2^{2} \\I_{p} = 0.652 kgm^{2}[/tex]
[tex]v_{p} = \frac{( (2.533+0.652) 8.53 ) - (2.533* 0)}{0.163*2}\\v_{p} = 83.34 m/s[/tex]
