Conservation of momentum and a uniform rod

AI Thread Summary
The discussion centers on a physics problem involving the conservation of angular momentum with a uniform rod and a putty wad. The rod, with a rotational inertia of 0.12 kg·m², rotates at an angular speed of 2.4 rad/s before colliding with a 0.20 kg putty wad. The initial equation used for conservation of momentum was correct, but a mistake was identified in squaring the radius in the calculation. After correcting the equation, the final angular speed of the rod-putty system was determined to be 1.5 rad/s. The conversation highlights the importance of careful application of formulas in physics problems.
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Homework Statement



The uniform rod (length 0.60 m, mass 1.0 kg) in Fig. 11-54 rotates in the plane of the figure about an axis through one end, with a rotational inertia of 0.12 kg·m2. As the rod swings through its lowest position, it collides with a 0.20 kg putty wad that sticks to the end of the rod. If the rod's angular speed just before collision is 2.4 rad/s, what is the angular speed of the rod–putty system immediately after collision?

http://edugen.wiley.com/edugen/courses/crs4957/halliday9118/halliday9118c11/image_n/nt0058-y.gif

Homework Equations


IW=IW

I=mr^2

The Attempt at a Solution



IW=IW
.12(2.4) = (.12 + mr^2) Wf
.12(2.4) = (.12 + .20(.60)Wf
Wf= 1.2 rad/s

Homework Statement


Homework Equations


The Attempt at a Solution


Homework Statement


Homework Equations


The Attempt at a Solution

 
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i know .12 = 1/3ML^2
 
.12 is in the problem because its the inertia of the rod both before and after the collision
 
Sorry, it's been a while since I've done one of these problems.

Everything is correct, you just forgot to square the radius on the right hand side.

.12(2.4) = (.12 + .20(.60)Wf should be .12(2.4) = (.12 + .20(.60)^2)Wf

Wf is then 1.5 rad/s.
 
o wow forgot to square r, heh sorry
 
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