Question

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?

Answer 1

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.