Angular momentum and perfectly inelastic collision

  • Thread starter cozzbp
  • Start date
  • #1
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Hi, I just don't really understand rotational motion very well, and I don't know how to proceed with this problem.

Homework Statement


You throw a 235 g ball 26 m/s at the apparatus shown in the figure.
http://volta.byu.edu/ph121/homework/hw18f3.png [Broken]
The apparatus catches the ball, causing it to rotate. There is a kinetic frictional force on the bearing of the ball of 17 Newtons. The bearing has a radius of a=2.5 cm and the apparatus has a distance of r=1.2 m between the catcher and the axis of rotation. The catcher has a mass of 408 g and the rest of the aparatus has negligible mass. How many revolutions does the apparatus rotate before it stops?

Homework Equations


L = mvr
[tex]\Sigma[/tex][tex]\tau[/tex] = I[tex]\alpha[/tex]
[tex]\omega[/tex]=v/r


The Attempt at a Solution


Here is what I have tried so far...
First, I need to find the velocity after the ball and the apparatus collide. So,
mball*vball = (mball + mcatcher)*vf
.235 * 26 = (.235 + .408)*vf
solving for vf yields 9.5 m/s
then I can find the angular velocity using [tex]\omega[/tex] = vf/r
9.5/2.5 = 7.92 rad/s
I can also get the angular momentum from this.
My problem is, is that I have no idea how to factor gravity into the whole equation. I figure if I could find [tex]\alpha[/tex], then I could use a rotational kinematic equation to find the change in [tex]\vartheta[/tex]. My other thought was that I could somehow use the friction and rotational kinetic energy to find when it stops. Any help would be greatly appreciated.
 
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Answers and Replies

  • #2
consider the energy of the system.

the ball has an initial K.E. The catcher and the ball are associated with some gravitational potential energy (depending on where you take your reference)

for each revolution, the system lose energy as heat W=2(pi)a*friction.
 
  • #3
6
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When the ball collides with the catcher, does the system lose some kinetic energy as well?
 
  • #4
LowlyPion
Homework Helper
3,097
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First of all is the catcher rotating about a horizontal axis or a vertical one?

I would read it as not necessarily horizontal axis and hence you need only find the resisting torque of the friction applied to slowing the system after determining initial ω. In which case aren't you just using simple rotational kinematic means to solve for θ?

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#rlin
 
  • #5
6
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Ok! I never even thought of the possibility that it wasn't vertical. So now my thinking is to use torque = 17 N * .025 m and divide that by the moment of inertia MR^2 to get angular acceleration. Then I can just use simple kinematics. Does that sound correct?
 
  • #6
LowlyPion
Homework Helper
3,097
5
Ok! I never even thought of the possibility that it wasn't vertical. So now my thinking is to use torque = 17 N * .025 m and divide that by the moment of inertia MR^2 to get angular acceleration. Then I can just use simple kinematics. Does that sound correct?

Looks like a plan to me.

If it's vertically rotating it will more likely at some point either fall forward or backward to settle at the bottom after some number of revolutions. I would think that case doesn't teach the concepts as readily.
 

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