# Rotating disk

1. Feb 4, 2004

### jaweibuch

if a disk which is initially spinning with an angular velocity,w, radians per second is pushed with an initial linear velocity of v, what is the resultant motion? assume a kinetic friction of mu normal to the disk

i would love to see how this problem is solved. it seems real interersting..anyone that can help would be awesome!

2. Feb 5, 2004

### HallsofIvy

What do you mean by "pushed with an initial linear velocity of v"? You push with a force, not a velocity.

3. Feb 5, 2004

### jaweibuch

so yes i think that it is a force but the force has a linear velocity? i'm not sure but just make any assumptions to solve the problem because this was the only information given to me.
thanks

4. Feb 6, 2004

### krysith

Halls of Ivy - I believe the questioner meant "pushed so that it has an initial linear velocity of v". That would make sense.

Jaweibuch - I think the best way of treating this problem is to look at the radial motion and the linear motion separately. Somebody out there correct me if these can't be treated orthogonally! So for your linear motion, you start out with an initial velocity v and you have a force = mu * N acting in the direction opposite v. N is your normal force = mass * g. You will need to know your mass and gravity, or perhaps mu is a force in your case. Basically its the same problem as a block stopping due to friction.

For the radial motion case, you also have a force acting opposite your motion. In this case your friction force is acting at different radii from the center, so we might have to integrate. hmm, no luckily we don't because the moment of inertia will cancel. See below:

The mass has some distribution with respect to the radius. For example, if the disk had uniform density and thickness, then obviously there would be more mass at higher radius. Luckily there is a parameter called the moment of inertia (I) which is the integral of the mass * radius. We use I in the following formulas:

torque = I * dw/dt
torque= intergral of friction force * radius
integral of friction force * radius = I * g * mu
so...
dw/dt = g * mu

dw/dt is just the change in w per unit time (sorry, don't know how much calculus you've had).

Cheers,
krysith

5. Feb 6, 2004

### jaweibuch

thanks for getting back about the problem

i guess i'm really having problems how we find the "resultant motion" of the disk. i would just love to know how to approach the problem. i did quite understand what you said about an integral(i've been all the way through calculus, by the way. just so you know) but i took physics quite a long time ago so i'm just trying to figure out what formulas i will use when trying to solve this problem.

back to the resultant motion. does that mean i should find the final velocity as part of the answer or does it have nothing to do with the final velocity?
thanks

6. Feb 19, 2004

### krysith

Sorry for taking such a long time to get back to this thread.

The "resultant motion" is simply the motion which results from the combination of the spinning plus the linear motion.

So Vr=Vs + Vl (for any point on the disk)

Remember that velocities are vectors, so use vector addition. The final velocity should be zero, if you go to large t, as the friction will bring the disk to a stop. To find the final displacement, integrate the velocity with respect to time. I hope this helps you, and that I haven't taken too long to get back to you.