# Homework Help: Elasticity and young's modulus

1. Sep 26, 2007

### Kushal

1. The problem statement, all variables and given/known data

The strain in a rubber ring on a rim of a wheel of radius 0.40m is 3*10^-3 when the wheel is stationary. The normal push of the rim on the ring just becomes zero when the wheel is rotating at angular speed 'omega'. Calculate the value of 'omega' if Young's Modulus for the rubber is 0.50G Nm-2, and its density is 9.4 * 10^2 kgm-3.

2. Relevant equations

E = (FL)/(Ae)
F = mrw^2

3. The attempt at a solution

what i want to know is what is implied when it says that 'the normal push of the rim on the ring just becomes zero when the wheel is rotating at angular speed 'omega'. then i think i'll be able to do the number.

2. Sep 26, 2007

### JoAuSc

The strain on the ring is due to a stress which can be calculated, and from the stress you can calculate the acceleration the ring would have if the wheel suddenly disappeared and allowed the ring to collapse.

I hope this helps.

3. Sep 28, 2007

### Kushal

Thnks JoAuSc,....i understood the calculation part......started my calculations by equating force from stress to centripetal force instead...

what i still can't understand is what " the normal push of the rim on the ring just becomes zero when the wheel is rotating at angular speed 'omega' " means. can someone explain this to me, what actually happens??

thnks

4. Sep 28, 2007

### JoAuSc

We have three forces on the ring, all pointing in the radial direction either inwards or out. There is the stress where the ring essentially pulls itself towards the center due collectively to its tension. We'll call this F_t, for "tension". There is the normal force of the wheel on the ring; this force points outwards. We'll call this F_n, for "normal". Finally, there is the centrifugal "force", which points outwards. I use quotes because it's not really a force, unless you look at things from a rotating frame of reference. This is due to the centripetal acceleration v^2/r. (I hope I'm using "centripetal" and "centrifugal" right.) We'll call this force F_c.

When the wheel is not rotating, we have two forces which sum together to 0. We can solve for the normal force in this case. When the wheel is rotating fast enough, the normal force goes to zero, the tension force remains the same, and we have a centrifugal force which replaces the normal force.

I hope this helps.