- #1
Kushal
- 438
- 1
Homework Statement
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.
Homework Equations
E = (FL)/(Ae)
F = mrw^2
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.