Physics Rotating Disk, Moment of Inertia, Angular Momentum, Angular Velocity

In summary, Homework Equations are L=r*p, L=R*m*v, and I think that's the answer, b/c they don't give us any constants to work with, but it seems too easy, so I don't trust myself on that. I think that the final angular velocity, this is where I'm totally tripped up. I thought it would be v/(answer in iii), because it's like taking v/m in straight kinematics, but translating it to rotational kinematics. Does that make sense again?iii) Here's where I get really confused. I thought the final moment of inertia would be I+(L/2*pi) because that would be like
  • #1
Darkalyan
34
0

Homework Statement



Here's the problem statement, uploaded onto google docs:
http://docs.google.com/View?docid=d277r7r_48f97jqsgs [Broken]

Homework Equations





The Attempt at a Solution



i)
L=r*p
L=R*m*v
I think that's the answer, b/c they don't give us any constants to work with, but it seems too easy, so I don't trust myself on that.

ii)
I=[M(R^2)]/2

iii) Here's where I get really confused. I thought the final moment of inertia would be I+(L/2*pi) because that would be like taking the effect of the ball, and dividing it by the entire circumference of the circle. Does that make sense?

iv) The final angular velocity, this is where I'm totally tripped up. I thought it would be v/(answer in iii), because it's like taking v/m in straight kinematics, but translating it to rotational kinematics. Does that make sense again?
 
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  • #2
Darkalyan said:
iii) Here's where I get really confused. I thought the final moment of inertia would be I+(L/2*pi) because that would be like taking the effect of the ball, and dividing it by the entire circumference of the circle. Does that make sense?
No. What's the moment of inertia of a point mass?

If you like, you can imagine the mass spread out over the circumference of the circle. What shape would that be and what is its moment of inertia?
 
  • #3
Doc Al said:
No. What's the moment of inertia of a point mass?

If you like, you can imagine the mass spread out over the circumference of the circle. What shape would that be and what is its moment of inertia?


Aaah, I see. I guess the moment of inertia would be that of a hoop, which is mr^2. But, since the mass is now spread out over the entire circumference, would it be like (m/2*pi*R)*r^2, which simplifies to (mr)/(2pi) Is that right now?
 
  • #4
Darkalyan said:
Aaah, I see. I guess the moment of inertia would be that of a hoop, which is mr^2.
Exactly. (Now stop while your ahead. :wink:)
But, since the mass is now spread out over the entire circumference, would it be like (m/2*pi*R)*r^2, which simplifies to (mr)/(2pi) Is that right now?
Nope. Looks like you're trying to apply the linear mass density of the hoop for some reason. But you already know the moment of inertia of a hoop, which is in terms of its total mass.
 
  • #5
Oh, wow! It's a lot simpler than I thought. I don't know why I was complicating it so much. Okay, so we know the moment of Inertia for the ball is mR^2, and the moment of Inertia for the disk is .5*MR^2, so is the moment of inertia for the final assembly just the other moments of inertia added up? That is to say,

I (final) = R^2(m+.5M)
 
  • #6
You got it.
 
  • #7
Sweet! You rock, Doc Al. Does that mean my answer for #1 is right too? And how do I solve for iv?
 
  • #8
Darkalyan said:
Does that mean my answer for #1 is right too?
Yes.
And how do I solve for iv?
What's conserved during the collision?
 
  • #9
Doc Al said:
What's conserved during the collision?

Energy is conserved, I guess in this case there would be no translational kinetic energy, so all the energy from the ball, the 1/2mv^2, would have to be changed into energy of rotation. So, I could set up this equation:

.5*m*v^2 = .5*I*w^2, and solve for w.

My answer is that w=the square root of (mv^2)/I
 
  • #10
Darkalyan said:
Energy is conserved...
Kinetic energy is not conserved. This is a inelastic collision--the mass sticks to the disk.

But something else is conserved. Hint: It was the first thing they asked about.
 
  • #11
Aaah, I see. The ball sticks, soo kinetic energy isn't conserved, okay. But, just like in 'normal' kinematics, momentum is conserved, so in this case Angular momentum is conserved. So, would that mean that I set up an equation that:

R*m*v=I*w
Thus, w=(Rmv)/I
 
  • #12
Darkalyan said:
So, would that mean that I set up an equation that:

R*m*v=I*w
Thus, w=(Rmv)/I
Exactly. (Where I is the total moment of inertia, of course.)
 
  • #13
All riiight. Sweeeet. Many thanks to you, Doc Al. Marvellous way of teaching without giving me the answer and without being too hard either. Really magnificent.
 

1. What is a rotating disk in physics?

A rotating disk in physics refers to a flat circular object that is spinning around an axis. It is often used as a model for understanding rotational motion and the principles of angular momentum.

2. What is the moment of inertia in physics?

The moment of inertia in physics is a measure of an object's resistance to changes in its rotational motion. It is calculated by taking into account the mass and distribution of mass around the axis of rotation.

3. How is angular momentum defined in physics?

Angular momentum in physics is a measure of an object's rotational motion. It is defined as the product of an object's moment of inertia and its angular velocity.

4. What is the relationship between angular velocity and angular momentum?

Angular velocity and angular momentum are directly proportional to each other. This means that an increase in one will result in a corresponding increase in the other, assuming the moment of inertia remains constant.

5. How does the moment of inertia affect the behavior of a rotating disk?

The moment of inertia determines how easily a rotating disk can be accelerated or decelerated. A higher moment of inertia means that more force is required to change the angular velocity of the disk, while a lower moment of inertia allows for easier changes in angular velocity.

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