Spinning disk, friction and pure roll

  • #1

Homework Statement


[/B]
A uniform solid disk of radius R is set into rotation with an angular speed ωi about an axis through its center. While still rotating at this speed, the disk is placed into contact with a horizontal surface and immediately released as shown in figure below (a) What is the angular speed of the disk once pure rolling takes place? (b) Find the fractional change in kinetic energy from the moment the disk is set down until pure rolling occurs. (c) Assume the coefficient of friction between disk and surface is μ. What is the time interval after setting the disk down before pure rolling motion begins? (d) How far does the disk travel before pure rolling begins?

Homework Equations



τ = r x ΣF = ΣIα
L = Iω
ΔKr = 1/2Iω^2
I = 1/2mR^2

The Attempt at a Solution



The full solution is in the manual;

[1] μmg = ma = m (v-0) / t

[2] μmgt = mv

[3] -μmgR = Iα = (1/2mR^2) (ω-ωi) / t

[4] -μmgRt = 1/2mR^2ω - 1/2mR^2ωi -> [5] Rωi - 2μgt = Rω

[6] Rωi - 2v = Rω -> Rωi - 2Rω = Rω -> ω = ωi / 3

My questions:

At first I tried an energy approach with Krot and Ufriction turning into Krot and Ktrans. My solution differed quite a bit from the textbook solution. However, I do have questions about the solution itself;

In [1], why is there (v - 0) ? v is never zero during that transition, only a = 0 when v reaches μs. Correct? I think they mean v = at -> a = v / t in general? Maybe zero is the speed of the surface?

In [3] the sign turns negative. I am guessing it is because they mean τ = R x -μmg opposes the rotation of the wheel? And the rotation of the wheel's effective torque Iα is acting the other way?

In [3] α is replaced with (ω - ωi) / t in the same manner as in [1], but this time there is a ωi unlike [1].

The rest of the entire problem is straightforward.


I just really want to find a way to understand this problem properly. I have spent at least two hours running through concepts in my head, drawings and scribbles..

Thank you
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
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Hello.

In [1], why is there (v - 0) ? v is never zero during that transition, only a = 0 when v reaches μs. Correct?
v represents the velocity of the center of mass of the wheel. What is the value of v at the initial instant when the wheel is released?

In [3] the sign turns negative. I am guessing it is because they mean τ = R x -μmg opposes the rotation of the wheel? And the rotation of the wheel's effective torque Iα is acting the other way?
Yes. A free body diagram of the wheel should make this clear.

In [3] α is replaced with (ω - ωi) / t in the same manner as in [1], but this time there is a ωi unlike [1].
Hopefully, the comments regarding [1] will clear this up.
 
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  • #3
*SLAPS FOREHEAD*

I need more coffee.

Thank you for taking your time, I appreciate it! I feel better now hahaha..
 

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