Wheels and rotational motion

In summary: F so TB would still be double Ta and for (b) it T= mw/F, and for (d) I'm a little confused...is it wi^2/(2Fr/mf) = (wi^2)/(2F/m) = (mwi^2)/(2F)? Right? Thanks for looking over all this.In summary, the conversation is about two grinding wheels. The first wheel, A, has an initial angular speed of w, while the second wheel, B, has an initial angular speed of 2w. Both wheels are being used to sharpen tools. The problem is that the coefficient of kinetic friction between the wheel
  • #1
candyq27
71
0
Hi. I did a 5 part problem and I just want to know if I did it right or if I'm on the right track. Please help. Thank you. The problem is as follows:

Two identical grinding wheels of mass m and radius r are initially spinning about their centers. Wheel A has an intial angular speed of w, while wheel B has an initial angular speed of 2w. Both wheels are being used to sharpen tools. For both wheels the tool is being pressed against the wheel with a force F directed toward the center of the wheel, and the coefficient of kinetic friction between the wheel and the tool is uk. You are holding the tool firmly so it does not move tangentially to the wheel.

(a) If it takes wheel A a time T to come to a stop, how long does it take for wheel B to come to a stop?

My work: p=Ft, so m(wf-wi)=uFt, so t=(m(wf-wi)/umg)= (mwi)/(umg)=TA, so since B=2w then the time for B would be 2TA.

(b) Find an expression for T in terms of teh variables specified above.

My work: T=mwi/umg, so T=wi/ug

(c) If wheel A rotates through an angle @ before coming to rest, through what angle does wheel B rotate before coming to rest?

My work: wf^2-wi^2=2a@, so @=(wf^2-wi^2)/(2a) so @=wi^2/2a...So the @B would be 2wi^2/2a, so @B is 4@A. (@ is angle and a is alpha)

(d) Find an expression for @ in terms of teh variables specified above.

My work: @= (wi^2/2a) = (wi^2/2(Ffric/m)) = (wi^2/2(uF/m)) = (wi^2/2(umg/m)) = (wi^2/2ug). So @= (wi^2)/(2ug)

(e) If you doubled the value of uk, how would that affect the time required to stop wheel A?

My work: Doubling the value of uk doubles the friction, which would reduce the time to stop wheel A by 1/2. Double uk would cause 1/2TA.

Thank you for reading through this!
 
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  • #2
You have uF = umg. But F is simply the force of the tool against the wheel and has no relation to the wheel's mass. Just leave F is it is, F.
So, check your answers for parts a), b), and d).
Also, in d), a (alpha) should be a = (2Fr)/(mr). {I_disc = (1/2)mr²}
All your other working looks OK.

Edit: corrected F = umg to uF = umg
 
Last edited:
  • #3
Ok so if F can't be simplified then for (a) it is mwi/F so TB would still be double Ta...for (b) it T= mwi/F, and for (d) I'm a little confused...
is it wi^2/(2Fr/mf) = (wi^2)/(2F/m) = (mwi^2)/(2F)? Right? Thanks for looking over all this.
 
  • #4
candyq27 said:
... for (a) it is mwi/F so TB would still be double Ta...

correct.

candyq27 said:
...for (b) it T= mwi/F, ...

correct, but use w , rather than wi, since you are asked to use the variables specified above.

candyq27 said:
... and for (d) I'm a little confused...
is it wi^2/(2Fr/mf) = (wi^2)/(2F/m) = (mwi^2)/(2F)? Right? Thanks for looking over all this.
I'm not sure how you got alpha = Fr/mf, but ...

F = normal reaction of disc against tool face.
μ is coefft of (kinetic) friction
Fr = frictional force on rim of wheel
and
Fr = μF

Decelerating torque on wheel is T = Fr x r
and
T = Iα, where I = (1/2)mr²
Therefore,
(1/2)mr² x α = Fr x r
(1/2)mr x α = μF
α = 2μF/(mr)

Now, you should have

θ = ω²/(2α)
θ = ω²/(4μF/(mr))
θ = mrω²/(4μF)
 
  • #5
So for the entire problem I don't have to factor in the coefficient of kinetic friction between the wheel and the tool? That's why I used the umg in each of the steps
 
  • #6
You have Fr = uF = umg, but F = mg is wrong.

Also, you have P = Ft, which is also wrong. I assumed that was a typo. It should be P = Fr x t.

So, you can't use umg in each of the steps.
 
  • #7
wait so,

(a) (mwi)/F ?
(b) (mw)/F ?

wouldnt you have to factor in uk?
 

1. How do wheels rotate?

Wheels rotate when a force, such as a push or a pull, is applied to them. This causes the wheels to turn on their axis, creating rotational motion.

2. What factors affect the rotational motion of a wheel?

The factors that affect the rotational motion of a wheel include the force applied to it, the mass of the wheel, the size of the wheel, and the surface it is rolling on.

3. What is the difference between rolling and sliding motion?

Rolling motion is when a wheel rotates on its axis while moving forward, while sliding motion is when an object moves without rotating, such as when a box is pushed across a surface.

4. How do wheels help with transportation?

Wheels allow for efficient transportation by reducing friction between the object and the surface it is moving on. This reduces the amount of force needed to move the object and increases its speed.

5. What is the principle behind how wheels work?

The principle behind how wheels work is called the law of conservation of angular momentum. This states that the total amount of angular momentum in a system remains constant, so when a force is applied to a wheel, its rotational motion increases while the speed of the object it is attached to remains constant.

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