Circles--Show that for all positions of the smaller circle, P remains on the x-axis.

[cccc]

Can anyone help me with this question?

My queries:

1. I don't really understand the question. 

2. So far, I've only got RL (radius of large circle)=2RS (radius of small circle)

[kw0573]

Question states (1) P is fixed on small circle (2) small circle is free to move around O in the big circle (3) Prove that P does not leave the line through O and where P is right now when the small circle travels in rolling without slipping motion. This is a close enough video clip, assume P is one of the eight white dots. If I were to solve the problem I need to make use of vector calculus so it's very very unlikely it's going to appear on the IB exam.

1) Rolling without slipping: tangential velocity has the same magnitude as velocity of the center of the small circle, A. Not quite the definition but this is the useful fact we need. Let's call the constant tangential speed = v.The velocity is constant in magnitude but variant in direction. If we assume the small circle to rotate clockwise around the big circle (the small circle *falls down* in a sense), then P actually rotates counter-clockwise around A. This is in reverse direction as the above clip.

2) Note the vector relationship OP = OA + AP. Assume small circle radius 1 big radius 2. We are going to use position vector AP = AP(t) (or AP as a vector function of t) = (cos(vt), sin(vt)). |AP| = 1 = radius of small circle. note that cos(vt) and sin(vt) are analogous from unit circle going counterclockwise, except vt is used as the angle so when we take the derivative the speed is v. The derivative of AP is AP'(t) = (-vsin(vt), vcos(vt)), with magnitude v. This is from vector calculus; it's basically each component taking derivative with respect to t. Note that from 1) |OA'| = |AP'|, speed of centre A is tangential speed of P.

3) OA'(t) = (-vsin(vt), -vcos(vt)), with magnitude v (tangential speed = centre speed) This is mostly seeing x component of OA' velocity from 0 (all -y direction initial OA' velocity) goes to -v (when small circle is at the bottom, centre is going all left), to 0 (at half way, going straight up) to v (at 3/4 of way, going to the right) and back to 0 when complete. This is modeled by -vsin(vt) function. Similar argument goes for y. I state without proof that trig functions perfectly describe the component behaviour of uniform circular motion. 

4) OA (by taking integral dt) = (cos(vt), -sin(vt)). All +C constants are incidentally, all 0. Same thing, each component is taken the integral in vector calculus.
Finally OP = OA + AP = (cos(vt), -sin(vt)) + (cos(vt), sin(vt)) = (2 cos(vt), 0). The y-component is 0, hence the particle P always stay at x-axis.

As a final note, it just so happens that OA has the same magnitude as AP by details in the questions. If say it was 5:1 instead of 2:1 radius ratio then OA'(t) would instead be (-vsin(v/4 t), -vcos(v/4 t)), just so when we take the antiderivative the magnitude is 4, and not 1. 

Edited April 4, 2016 by kw0573

[ultimateone]

Sorry for the stupid question but would this be applicable to SL?

[kw0573]

2 hours ago, ultimateone said:

Sorry for the stupid question but would this be applicable to SL?

Well I don't know for sure if I solved it the most efficient way but I made use of vector calculus which is completely outside of the HL content so I wouldn't worry if I were you.

[cccc]

Thank you! @kw0573Was getting worried!