Moment of inertia from a cylinder.

  • Thread starter notsam
  • Start date
  • #1
50
0

Homework Statement


A solid, horizontal cylinder of mass 16.3 kg
and radius 1.44 m rotates with an angular
speed of 2.24 rad/s about a fixed vertical axis
through its center. A 0.239 kg piece of putty
is dropped vertically onto the cylinder at a
point 0.624 m from the center of rotation, and
sticks to the cylinder.
What is the final angular speed of the sys-
tem?
Answer in units of rad/s.


Homework Equations

I=.5mv^2



The Attempt at a Solution

Ok so I'm pretty sure that you must find the objects moment of inertia before and after the putty hits the cylinder and set them equal to each other using I=.5mv^2.
 

Answers and Replies

  • #2
137
5
Yes. Find the moment of inertias, and use the conservation of angular momentum.
 
  • #3
1,137
0
Find moment of inertia of cylinder and cylinder + putty (whatever that is)

now you have initial w

just use conservation of angular momentum about COM of cylinder
 
  • #4
50
0
Right my problem with this one is that I really don't know how to find the moment of inertia with the putty. Would it be (16.3kg+.239kg)(1.44m-.624m)^2?
 
  • #5
1,137
0
no ... use I = mr^2 for putty

and add it to I of cylinder
 
  • #6
50
0
Oh so my final equation will be (mr^2cylinder)*(angular speed initial)=((mr^2cylinder)+(mr^2putty))*(final angular speed)?
 
  • #7
1,137
0
I of cylinder is : [tex]I = \frac{mr^2}{2}[/tex]

what you've written is for cylinder shell
 

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