(no this isn't hw, I'm just studying physics on my own)The goal is basically to find the instantaneous axis of rotation of the small discs that's on the big disc and the angular velocity around this axis.Now I suspect it's either on the edge of the small disc that's closest to the central axis or even more trivially, it' just it's own fucking axis that moves around the central axis with w1.But what's the exact theory behind finding it and calculating angular velocity around IAR?I literally went through all the textbooks I could find AND ALL OF THEM LITERALLY ONLY SHOW THAT ROLLING CYLINDER PROBLEM BUT SAY NOTHING GENERAL ABOUT THE IAR FFS.The only bit I know is that it's supposed to be in the point where the object is not moving atm.Any sources or smth? Thanks.
>>16451425Isn't it 2r*dw/dR because thats the difference in speed of the outer edge over the inner edgew(R) = w1*(R/Rf) (calling Rf the full radius)So 2r*w1*1/2*R^2/Rf = r*w1*R^2/Rf ?
>>16451425Need some clarification here - by instantaneous axis of rotation do you mean you're looking for the principal axes? Or do you mean you're looking for like a center of angular momentum frame?
>>16451503By definition it means the axis where in a given instance the point can be described as rotating only around it with some angular velocity
>>16451614How does it make sense when it rotates around two axis at the same time? If you imagine it rotated around only one axis the trajectory would be a circular arc but now it's never circular.
>>16451681>How does it make sense when it rotates around two axis at the same time?The IAR can itself move.
>>16451614Oh, okay, so you're just looking for the point on the little circle where its instantaneous velocity is zero then?Treat it as a polar coordinate problem with some distance from Disk 2's center, [math]r[/math] and some phase relative to the rotation of Disk 1, [math]\phi[/math].[math]x = R \cos \left(\omega_1 t \right) + r \cos \left(\omega_2 t + \phi \right)[/math][math]y = R \sin \left(\omega_1 t \right) + r \sin \left(\omega_2 t + \phi \right)[/math][math]\dot{x} = -\omega_1 R \sin \left(\omega_1 t \right) - \omega_2 r \sin \left(\omega_2 t + \phi \right)[/math][math]\dot{y} = \omega_1 R \cos \left(\omega_1 t \right) + \omega_2 r \cos \left(\omega_2 t + \phi \right)[/math]Use [math]\dot{x} = 0[/math] and [math]\dot{y} = 0[/math] to solve for [math]\phi[/math] and [math]r[/math][math]\omega_2 r \sin \left(\omega_2 t + \phi \right) = -\omega_1 R \sin \left(\omega_1 t \right)[/math][math]\omega_2 r \cos \left(\omega_2 t + \phi \right) = -\omega_1 R \cos \left(\omega_1 t \right)[/math][math]\tan \left(\omega_2 t + \phi \right) = \tan \left(\omega_1 t\right)[/math][math]\phi = \left(\omega_1 - \omega_2 \right) t[/math]Substituting into the first equation for [math]\phi[/math][math]\omega_2 r \sin \left(\omega_2 t + \omega_1 t - \omega_2 t \right) = -\omega_1 R \sin \left(\omega_1 t \right)[/math][math]\omega_2 r \sin \left(\omega_1 t \right) = -\omega_1 R \sin \left(\omega_1 t \right)[/math][math]\left|r\right| = \frac{\omega_1}{\omega_2} R[/math]So there are some practical limits for the position where the velocity of a point on the small circle can even be zero.In the limit where [math]\omega_1 = \omega_2[/math] for example, it's only possible if the radius of the small circle is equal to its distance from the center of the big circle.
>>16451744>|r|=ω1/ω2*RWell that much I obtained much easier but what would be the angular velocity around IAR positioned in that point?
>>16452095>I already got that muchWell then why did you ask for help? You’ve already got r and the phase, and you’ve got the Cartesian velocities now, just convert it to an angular velocity for a fixed r.
>>16452101But would it be around that point that's r away from the center of the small disc?
>>16451744Hm interesting
>>16451744I wonder if there's a way to do this specifically through axes and not kinematics like this
>>16455975Not sure
