The rotational inertia of the putty rod system after

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, the rotational inertia of the putty-rod system (after the collision) is I = I 0 + (0.20) r 2 = 0.19 kg ∙ m 2 . Invoking angular momentum conservation 0 f L L or 0 0 I I , we have 2 0 0 2 0.12 kg m 2.4rad/s 1.5rad/s. 0.19 kg m I I 10. We make the unconventional choice of clockwise sense as positive, so that the angular velocities (and angles) in this problem are positive. Mechanical energy conservation applied to the particle (before impact) leads to mgh mv v gh 1 2 2 2 for its speed right before undergoing the completely inelastic collision with the rod. The collision is described by angular momentum conservation: 2 md I mvd rod where I rod is found using Table 10-2(e) and the parallel axis theorem: . 3 1 2 12 1 2 2 2 Md d M Md I rod Thus, we obtain the angular velocity of the system immediately after the collision: 2 2 2 ( /3) md gh Md md
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which means the system has kinetic energy 2 2 rod / 2, I md which will turn into potential energy in the final position, where the block has reached a height H (relative to the lowest point) and the center of mass of the stick has increased its height by H /2. From trigonometric considerations, we note that H = d (1 cos ), so we have 2 2 2 2 rod 2 2 2 1 1 1 cos 2 2 2 ( /3) 2 m d gh H M I md mgH Mg m gd Md md from which we obtain     2 1 1 1 1 / cos 1 cos 1 / 2 /3 1 / 2 1 /3 (20 cm/ 40 cm) cos 1 cos (0.85) (1 1)(1 2/3) 32 . m h h d m M m M M m M m
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