(hrw8c11p55) A uniform thin rod of length 0.50 m and mass 4.0 kg can rotate in a

Question

(hrw8c11p55) A uniform thin rod of length 0.50 m and mass 4.0 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is a rest when a 3.0-g bullet traveling in the horizontal plane of the rod is fired into one end of the rod. As viewed from above, the direction of the bullet velocity makes an angle of 60° with the rod (see the figure). If the bullet lodges in the rod and the angular velocity of the rod is 12.0 rad/s immediately after the collision, what is the magnitude of the bullet's velocity just before impact?

Explanation / Answer

length of the , l=0.5 m

mass of the rod, m_rod=4kg

mass of the bullet, mb=3g

angle made by the bullet velocity with the length of rod, theta=60 degrees

after collission,angular velocity, w=12 rad/sec

before impact,

angular momentum, Li=mb*v*r*sin(theta)

after impact,

angular momentum, Lf=I*w

here,

moment of inertia, I=I_rod+I_bullet

=(1/12)*m_rod*l^2 + mb*(l/2)^2

now,

Li=Lf

mb*v*r*sin(theta)=((1/12)*m_rod*l^2 + mb*(l/2)^2)*w

mb*v*(l/2)*sin(theta)=((1/12)*m_rod*l^2 + mb*(l/2)^2)*12

0.003*v*(0.25)*sin(60)=(1/12*(4*0.5^2 + 0.003*(0.25)^2)*12

===> v=1.54*10^3 m/sec

velocity of the bullet before impact, v=1.54*10^3 m/sec

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