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A student sits at rest on a piano stool that can rotate withoutfriction. The moment of inertia of the student-stool system is 4.0kg*m^2. A second student tosses a 1.7 kg mass with a speed of 2.7m/s to the student on the stool, who catches it at a distance of0.35 m from the axis of rotation.

What is the resulting angular speed of thestudent and the stool?

= ? rad/s

Respuesta :

Answer:

[tex]\omega_{f}=0.381 \ rad/s[/tex]  

Explanation:

given,                                        

moment of inertia of the stool = 4 kg m²

mass tosses by the student = 1.7 Kg

speed = 2.7 m/s                          

distance = 0.35 m from axis of rotation

initial angular momentum of the system  

[tex]L_i = mvR[/tex]                              

final angular momentum of the system  

[tex]L_f = (I_1 +mR^2)\omega_{f}[/tex]  

from conservation of angular momentum  

[tex]L_i = L_f[/tex]                          

[tex]mvR= (I_1 +mR^2)\omega_{f}[/tex]                            

[tex]\omega_{f}= \dfrac{mvR}{I_1 +mR^2}[/tex]                          

[tex]\omega_{f}= \dfrac{1.7 \times 2.7 \times 0.35}{4 +1.7 \times 0.35^2}[/tex]   

[tex]\omega_{f}=0.381 \ rad/s[/tex]