Find final angular velocity.

  • #1

Homework Statement


A solid disk of radius 5m and mass 8 kg rotates clockwise at 1.5 rad/s. Above this disk is a hoop of radius 2.5 m and mass 8 kg , rotating counterclockwise at 3 rad/s. The hoop drops down onto the disk, and friction causes them to rotate together. Find the final velocity. Find the kinetic energy lost in this collision.


Homework Equations



ωf=ωi+αt and θf=θi+ωit+(1/2)αt2 I=1/2 Mr^2 I=Mr^2

The Attempt at a Solution



I think that by starting to find the acceleration of each of the disk i will be able to find the velocity and times of each one of them.
 

Answers and Replies

  • #2
gneill
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Homework Statement


A solid disk of radius 5m and mass 8 kg rotates clockwise at 1.5 rad/s. Above this disk is a hoop of radius 2.5 m and mass 8 kg , rotating counterclockwise at 3 rad/s. The hoop drops down onto the disk, and friction causes them to rotate together. Find the final velocity. Find the kinetic energy lost in this collision.


Homework Equations



ωf=ωi+αt and θf=θi+ωit+(1/2)αt2 I=1/2 Mr^2 I=Mr^2

The Attempt at a Solution



I think that by starting to find the acceleration of each of the disk i will be able to find the velocity and times of each one of them.

Since you don't have a coefficient of friction to work with, you'll have a difficult time trying to find the forces and accelerations. Is there another approach?
 
  • #3
I was thinking in angular momentum Li=Lf
 
  • #4
gneill
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  • #5
(mr^2w^2)initialhoop+(1/2mr^2w^2)finaldisk=(mr^2w^2)finalhoop+(1/2mr^2w^2)initaldisk but than i find that i have two different angular velocities so i cannot factor.
 
  • #6
gneill
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(mr^2w^2)initialhoop+(1/2mr^2w^2)finaldisk=(mr^2w^2)finalhoop+(1/2mr^2w^2)initaldisk but than i find that i have two different angular velocities so i cannot factor.
You have numerical values for all the parameters, so why not begin by calculating the important quantities separately?

I1 = ? (Moment of inertia for the disk)
I2 = ? (Moment of inertia for the hoop)
L1 = ? (Angular momentum of the disk)
L2 = ? (Angular momentum of the hoop)

and you'll need

KE1 = ? (Initial rotational KE of the disk)
KE2 = ? (Initial rotational KE of the hoop)

Keep in mind that angular momentum has a "direction", it can be positive or negative depending upon the sense of rotation.
 
  • #8
I1 = 1/2mr^2=0.5(8kg)(5m)^2=50kgm^2 (Moment of inertia for the disk)
I2 = Mr^2=(8kg)(2.5)^2=100kgm^2 (Moment of inertia for the hoop)
L1 = Iw=(100kgm^2)(1.5rad/s)=-150Nm/s (Angular momentum of the disk)
L2 = Iw=(50kgm^2)(3rad/s)=150Nm/s (Angular momentum of the hoop)

KE1 = 1/2Iw^2=1/2(1/2mr^2)(v/r)^2=1/4mv^2=1/4(8kg)(1.5 rads)^2=4.5 J (Initial rotational KE of the disk)
KE2 =1/2Iw^2= 1/2(mr^2)(v/r)^2=1/2mv^2=1/2(8kg)(3rad/s)^2=36 J (Initial rotational KE of the hoop)

Is that right?
 
  • #9
gneill
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I1 = 1/2mr^2=0.5(8kg)(5m)^2=50kgm^2 (Moment of inertia for the disk)
I2 = Mr^2=(8kg)(2.5)^2=100kgm^2 (Moment of inertia for the hoop)
L1 = Iw=(100kgm^2)(1.5rad/s)=-150Nm/s (Angular momentum of the disk)
L2 = Iw=(50kgm^2)(3rad/s)=150Nm/s (Angular momentum of the hoop)

KE1 = 1/2Iw^2=1/2(1/2mr^2)(v/r)^2=1/4mv^2=1/4(8kg)(1.5 rads)^2=4.5 J (Initial rotational KE of the disk)
KE2 =1/2Iw^2= 1/2(mr^2)(v/r)^2=1/2mv^2=1/2(8kg)(3rad/s)^2=36 J (Initial rotational KE of the hoop)

Is that right?

I think you swapped the values for the moments of inertia when you copied them for I1 and I2, but I see that you used the correct ones for the angular momenta, and they look good.

The KE calculations look very strange (and the results are incorrect). What are those v terms? Angular momentum is simply (1/2) I ω2. And you have values for I already calculated and ω is given to you for each object.
 
  • #10
well the v terms are w=v/r and doesnt the KE rot cancel out the radius when you turn w into (v/r)^2?? was that wrong ??
 
  • #11
gneill
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well the v terms are w=v/r and doesnt the KE rot cancel out the radius when you turn w into (v/r)^2?? was that wrong ??
Well, there's no v involved anywhere! You're dealing with angular velocities only. You really only need to deal with v's or a's (linear stuff) when there's some interface between an angular motion part of a system and a linear motion part.
 
  • #12
so you are saying that instead of doing 1/2(mr^2)(v/r)^2 i should just leave 1/2(mr^2)w^2 instead and solve for KE rot?? also leaving the r alone too??
 
  • #13
gneill
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so you are saying that instead of doing 1/2(mr^2)(v/r)^2 i should just leave 1/2(mr^2)w^2 instead and solve for KE rot?? also leaving the r alone too??

Sure, if you wish. But you've already calculated the moments of inertia so I don't know why you're repeating the effort when calculating the KE. Why not just plug in your already calculated values? ##KE = \frac{1}{2} I \omega^2##. Not a v in sight :smile:
 
  • #15
KE1 = 1/2(100kg/m)(1.5)^2=112.5 J (Initial rotational KE of the disk)
KE2 =1/2(50kg/m)(3rad/s)^2=225 J (Initial rotational KE of the hoop)

is that now correct??
 
  • #16
gneill
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KE1 = 1/2(100kg/m)(1.5)^2=112.5 J (Initial rotational KE of the disk)
KE2 =1/2(50kg/m)(3rad/s)^2=225 J (Initial rotational KE of the hoop)

is that now correct??

The results are correct.

This doesn't affect the numerical results here, but watch your units: the moment of inertia has units kg*m2.
 
  • #17
so from here how do I find the total angular velocity??
 
  • #18
gneill
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so from here how do I find the total angular velocity??

How would you do it for a linear system if you knew the total angular momentum? This is the angular version of a linear inelastic collision.
 
  • #20
gneill
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http://www.real-world-physics-problems.com/images/inelastic_collision_8.png like this, i know all my momentum. but 150 does not equal -150 un less you set the equation to 0=150-150 which equals 0. But what happened to the angular velocity??? don't i need to find like a magnitude in angular in velocity??

Zero is a perfectly respectable magnitude :smile:

If you have two equal mass objects traveling with the same speeds in a head-on collision and they stick together, what's their final velocity?

By the way, the equation shown in your link is for elastic collisions, not inelastic. In a perfectly inelastic collision the two objects fuse -- m1+m2.
 
  • #21
0? but this 2 objects had different speeds! does i mean that doesn't matter how fast the objects are moving , when they stick together they both have 0 velocity??
 
  • #22
gneill
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0? but this 2 objects had different speeds! does i mean that doesn't matter how fast the objects are moving , when they stick together they both have 0 velocity??

Conservation of mometum. They had different rotational speeds but they had equal and opposite momenta. In the linear example I just gave, suppose you doubled the speed of one object but cut its mass in half. It would have the same momentum as before. Same result of zero net momentum and zero velocity for the stuck pair.
 
  • #23
oh so it was only finding the moments of inertia to calculate angular momentum and use conservation of energy of inelastic collision??? that was it? ok now for part 2 i know the answer is 338 J
 
  • #24
If i sum up both of my KE rot they give me the answer, but here is my doubt why do i have to sum it up? don't I have to find the KE total and subtract that from KE rot f to get the kinetic energy lost??
 
  • #25
gneill
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If i sum up both of my KE rot they give me the answer, but here is my doubt why do i have to sum it up? don't I have to find the KE total and subtract that from KE rot f to get the kinetic energy lost??

Yes. So what's the initial total KE and the final total KE?
 

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