Saturday, 7 November 2015

homework and exercises - elliptical ring rolling on a horizontal surface



Consider a ring rolling without slipping along a horizontal surface. Regardless of the speed of the ring, it is continuously in contact with the surface.
Let's deform the ring slightly so that it becomes an ellipse with small eccentricity of e0.


Now, consider the deformed ring rolling without slipping along the surface. At low speed, the ring still keeps contact with the surface continuously. However, at some critical speed the ring will jump.


I am interested in how to find this speed?



Answer



The polar shape of the ellipse is r(φ), with φ=0 at major radius rmajor=a , and φ=π2 at the minor radius rminor=a(1e) , where e is the eccentricity.


r(φ)=a(1e)e(e2)cos2φ+1a(1esin2φ)


The angle between the contact normal and the polar location of the contact point tanα=-ddφr(φ)r(φ)α=2esinφcosφ


The angular position of the ellipse as a function of the contact point location angle φ θ=φ+α(φ)=φ+2esinφcosφ


The angular velocity of the ellipse ω=˙φ+(ddφ2esinφcosφ)˙φ=˙φ+(4ecos2φ2e)˙φ=(4ecos2φ2e+1)˙φ



The angular acceleration of the ellipse ˙ω=dωdt=ωφ˙φ+ω˙φ¨φ

=(-8esinφcosφ)˙φ2+(4ecos2φ2e+1)¨φ


The above equations are used to solve for ˙φ(ω)= and ¨φ(˙ω)=


The vertical position of the ellipse as a function of the contact point location angle φ y=rcosα=a(1esin2φ)

since cosα1 .


The vertical speed of the ellipse is


˙y=(φa(1esin2φ))˙φ=(-2aesinφcosφ)˙φ


˙y=(-2aesinφcosφ)(4ecos2φ2e+1)ω

And the vertical acceleration


¨y=(φ˙y)˙φ+(˙φ˙y)¨φ=(2ae(12cos2φ))˙φ2+(-2aesinφcosφ)¨φ


=-2ae(2cos2φ+2e1)˙φ2+sinφcosφ˙ω4ecos2φ2e+1


Before I expand even more, I look at the peak acceleration is at φ=0 or where ˙φ=12e+1ω and ¨φ=12e+1˙ω


¨y=-2ae(2+2e1)˙φ2+02e+1=-2ae˙φ2=-2ae(12e+1ω)2



¨y=-2aeω2(2e+1)2


When this acceleration equals gravity with ¨y=-g then the ellipse jumps. This occurs at the critical angular velocity of


ωC=(2e+1)g2ae


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...