HRW7 11-56 In Fig. 12-45, a 1.10 g bullet is fired into a 0.535 kg block that is
ID: 1298535 • Letter: H
Question
HRW7 11-56 In Fig. 12-45, a 1.10 g bullet is fired into a 0.535 kg block that is mounted on the end of a 0.565 m nonuniform rod of mass 0.479 kg. The block-rod-bullet system then rotates about a fixed axis at point A. The rotational inertia of the rod alone about A is 0.0575 kg*m2.
Assume the block is small enough to treat as a particle on the end of the rod.
(a) What is the rotational inertia of the block-rod-bullet system about point A?
kg*m2
(b) If the angluar speed of the system about A just after the bullet's impact is 5.85 rad/s, what is the speed of the bullet just before the impact?
m/s
Explanation / Answer
A) I = 0.0575 + M*L^2 + m*L^2
= 0.0575 + (M+m)*L^2
= 0.0575 + (0.535 + 0.0011)*0.565^2
= 0.228 kg.m^2
B) Apply momentum conseravtion
m*v*L = I*w
v = I*w/(m*L)
= 0.228*5.85/(0.0011*0.565)
= 2152 m/s
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