Perfectly Inelastic Disk and Stick Collision- Angular Speed

  • Thread starter bcca
  • Start date
  • #1
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I think I'm not understanding something about this problem. Can someone help me out?

Homework Statement


A 2.0 kg disk traveling a 3.0 m/s strikes a 1.0 kg stick of length 4.0 m that is lying flat on nearly frictionless ice. The disk strikes the endpoint of the stick, at a distance r = 2.0 m from the stick's center. Suppose the collision is perfectly inelastic so that the disk adheres to the stick at the endpoint at which it strikes. The moment of inertia of the stick about its center of mass is 1.33 kg*m2. Find the angular speed of the system after the collision.

Homework Equations


(I think) Parallel axis theorem: I= Icm + mr2
L= r cross p
L= I*omega
m= mass disk

The Attempt at a Solution


L = L'
r cross p = I(system)*omega
mvr = (I(stick) + I(disk) + mvr2)*omega
mvr = (I(stick) + (1/2)mr2 + mr2)*omega
omega = mvr/(I(stick) + 2/3(mr2))
omega = 2.0kg(3.0m/s)(2.0m)/(1.33kg*m2+1.5(2.0kg)(2.0m)2)
omega= .90 rad/s

The answer is 1.0 rad/s.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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hi bcca! :smile:

(have an omega: ω :wink:)

nope, start again …

you need to decide which point you're taking moments about

the easiest point is the combined centre of mass (if you take it about anywhere else, you'll need to add mrc.o.m. x vc.o.m. to get the final angular momentum) …

what do you get? :smile:
(and you can assume the disk is a point)
 
  • #3
9
0
I got it! Thanks :)
 

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