A uniform thin rod of L = 0.70 m and M = 5.0 kg can rotate in a horizontal plane

Question

A uniform thin rod of L = 0.70 m and M = 5.0 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 2.5 g bullet (m) 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's velocity makes an angle of 75degree with the rod (see the figure below). The bullet lodges in the end of the rod and the angular velocity of the rod is 2 rad s immediately after the collision. Write an equation for the initial velocity, v_0, of the bullet in terms of M, m, L, theta, and omega_f. L = 0.70 m, M = 5.0 kg, m = 2.5 g, theta = 75 degree with the rod (see the figure) and omega_f = 2 rad/s, what is initial velocity of the bullet before the collision?

Explanation / Answer

the initial angular momentum of the system is only for bullet so

L1 = m * v0 * r * sin (theta)

here r = L/2

m * v0 * L/2 * sin (theta)

Afterwards the total angular angular momentum is

L2 = I * omega

I = 1/12 * M * L^2 + ( m* r^2)

here r = L/2

I = 1/12 * M * L^2 + ( m * L^2/4)

I = L^2 ( M/12 + m/4)

from conservation of momentum

L1 = L2

m * v0 * L/2 * sin (theta) = L^2 ( M/12 + m/4)

v0 = [L^2 ( M/12 + m/4)] /   [m/2 * L * sin(theta)]

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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