Respuesta :
Answer:
Final angular speed equals 3 revolutions per second
Explanation:
We shall use conservation of angular momentum principle to solve this problem since the angular momentum of the system is conserved
[tex]L_{disk}=I_{disk}\omega \\\\L_{disk}=\frac{1}{2}mr^{2}\\\therefore L_{disk}=\frac{1}{2}mr^{2}\times10rad/sec[/tex]
After the disc and the dropped rod form a single assembly we have the final angular momentum of the system as follows
[tex]L_{final}=I_{disk+rod}\times \omega_{f} \\\\I_{disk+rod}=\frac{1}{2}mr_{disc}^{2}+\frac{1}{12}mL_{rod}^{2}\\I_{disk+rod}=\frac{1}{2}mr_{disc}^{2}+\frac{1}{12}m\times (2r_{disc})^{2}\\\\I_{disk+rod}=\frac{1}{2}mr_{disc}^{2}+\frac{1}{3}mr_{disc}^{2}\\\\L_{final}=\frac{5mr_{disc}^{2}}{6}\times \omega _{f}\\\\[/tex]
Equating initial and final angular momentum we have
[tex]\frac{5mr_{disc}^{2}}{6}\times \omega _{f}=\frac{1}{2}m_{disc}\times r_{disc}^{2}\times 10\pi rad/sec[/tex]
Solving for [tex]\omega_{f}[/tex] we get
[tex]\omega_{f}=6\pi rad/sec[/tex]
Thus no of revolutions in 1 second are 6π/2π
No of revolutions are 3 revolutions per second
