A thin, uniform, metal bar, 3.00 m long and weighing 90.0 N , is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 1.60 m below the ceiling by a small 3.00-kg ball, initially traveling horizontally at 10.0 m/s . The ball rebounds in the opposite direction with a speed of 5.00 m/s. Find the angular speed of the bar just after the collision? Why linear momentum not conserved?

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Respuesta :

lcmendozaf lcmendozaf · Answer 1 · 2019-12-03T22:21:33+01:00

Answer:

[tex]\omega_f=-2.6rad/s[/tex]

Since the bar cannot translate, linear momentum is not conserved

Explanation:

By conservation of the angular momentum:

Lo = Lf

[tex]I_B*\omega_o+I_b*(V_o/d)=I_B*\omega_f+I_b*(V_f/d)[/tex]

where

[tex]I_B =1/3*M_B*L^2[/tex]

[tex]I_b=m_b*d^2[/tex]

[tex]M_B=90/g=9kg[/tex]; [tex]m_b=3kg[/tex]; d=1.6m; L=3m; [tex]V_o=-10m/s[/tex]; [tex]V_f=5m/s[/tex]; [tex]\omega_o=0 rad/s[/tex]

Solving for [tex]\omega_f[/tex]:

[tex]\omega_f=m_b*d/I_B*(V_o-V_f)[/tex]

Replacing the values we get:

[tex]\omega_f=-2.6rad/s[/tex]

Since the bar can only rotate (it canno translate), only angular momentum is conserved.