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ioda
3 years ago
8

A thin, uniform, metal bar, 3.00 m long and weighing 90.0 N , is hanging vertically from the ceiling by a frictionless pivot. Su

ddenly 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?
Physics
1 answer:
liraira [26]3 years ago
6 0

Answer:

\omega_f=-2.6rad/s

Since the bar cannot translate, linear momentum is not conserved

Explanation:

By conservation of the angular momentum:

Lo = Lf

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

where

I_B =1/3*M_B*L^2

I_b=m_b*d^2

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

Solving for \omega_f:

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

Replacing the values we get:

\omega_f=-2.6rad/s

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

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