(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 11.0 rad/s immediately after the collision, what is the magnitude of the bullet's velocity just before impact?

Explanation / Answer

analyzing of the moment inertia,

The rod has a moment of inertia of I=mL2/12. The mass of the rod is m=4.0 kg, so the moment of inertia is

I_rod = (1/12)ML²

The bullet can be treated as a point mass, with moment of inertia mR2

I_total = I_rod + mR²

I_total = (1/12) ML² + (m_bullet) R²

I_total =  (1/12) * 4.0kg*0.50² + 3*10-3 kg*0.25² [since the radius is half the length of the rod R = 0.25m]

I_total = 0.0833 + 1.875*10-4

I_total = 0.08352

To solve this problem, we need to convert the speed of the bullet to the angular velocity, w.
wr=v(tangent)=vsin(60).
Therefore, Wbullet=v/r sin(60)

Using the law of conservation of angular momentum, we can then calculate the velocity of the bullet.

I*Wbullet=I_total*wrod
( 1.875*10-4 ) (v/0.25) (sin60) = ( 0.08352) (11)
6.495*10-4 v = 0.91872

v=1414.50 m/s

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