The figure below shows the overhead view of a uniform rod of length 0.400 m and

Question

The figure below shows the overhead view of a uniform rod of length 0.400 m and mass 3.45 kg that can rotate in a horizontal plane about the vertical axis p, through its center. The rod is initially at rest when a bullet of mass 2.80 g traveling in the rotation plane is fired into one end of the rod. The angle between the rod and the path of the bullet is theta = 50.0degree. The speed of the bullet before the impact is v = 1.2 km/s. What is the angular speed of the bullet-rod system after the impact? Assume that the bullet becomes lodged in the rod.

Explanation / Answer

because net external torque about the center = 0

so anguler momentum will be conserve

applying angular momentum about p

m_b * v_b(i) * sin(theta) * L/2 = IW(rod) + m_b * v_b(f) * L/2

I = moment of inertia of rod about p = mL^2 / 12 = 3.45 * (0.4)^2 / 12 = 0.046 kgm^2

w = v_b(f) / r = v_b(f) / (L/2) = v_b(f) / 0.2 = 5*v_b(f)

so now

2.8*10^(-3) * 1.2*10^(3) * sin(50) * 0.2 = 2.8*10^(-3) * v_b(f) * 0.2 + 0.046* 5 * v_b(f)

0.515 = 0.56*v_b(f) + 0.23*v_b(f)

v_b(f) = 0.65 m/s

w = 5 * v_b(f) = 5 * 0.65 = 3.25 rad/s

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