speed upon impact: A uniform thin rod of length 0.800 m and mass 3.80 kg can rot

Question

speed upon impact:

A uniform thin rod of length 0.800 m and mass 3.80 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3.00 g bullet traveling in the rotation plane is fired into one end of the rod. As viewed from above, the bullet's path makes angle theta = 60.0 degree with the rod. 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 bullet's speed just before impact?

Explanation / Answer

conservation of angular momentum

just before impact ,H = (m of bullet x V x sin60 x l/2 ) = 3 x 10^(-3) x V x sqrt(3)/2 x 0.4

after impact , H = moment of inertia x ang. velocity =( (Mx l^2/12)+ (m x (l/2)^2) ) x w = (3.8 x 0.8^2/12 + 0.003 x 0.4^2) x 12

3 x 10^(-3) x V x sqrt(3)/2 x 0.4 = (3.8 x 0.8^2/12 + 0.003 x 0.4^2) x 12

=> V = 2345.7 m/s

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.