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 e→0.
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(1−e) , where e is the eccentricity.
r(φ)=a(1−e)√e(e−2)cos2φ+1≈a(1−esin2φ)
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+(4ecos2φ−2e)˙φ=(4ecos2φ−2e+1)˙φ
The angular acceleration of the ellipse ˙ω=dωdt=∂ω∂φ˙φ+∂ω∂˙φ¨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(1−esin2φ)
The vertical speed of the ellipse is
˙y=(∂∂φa(1−esin2φ))˙φ=(-2aesinφcosφ)˙φ
˙y=(-2aesinφcosφ)(4ecos2φ−2e+1)ω
¨y=(∂∂φ˙y)˙φ+(∂∂˙φ˙y)¨φ=(2ae(1−2cos2φ))˙φ2+(-2aesinφcosφ)¨φ
=-2ae(2cos2φ+2e−1)˙φ2+sinφcosφ˙ω4ecos2φ−2e+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+2e−1)˙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
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