A rotating disc moves in the positive direction of `x`-axis as shown. Find the equation `y(x)` describing the position of the instantaneous axis of rotation if at the initial moment the centre `C` of the disc was located at origin after which (a) it moved with constant acceleration a (initial velocity zero) while the disc rotating anticlockwise with constant angular velocity `omega`. (b) it moved with constant velocity `v` while the disc started rotating anticlockwise with a constant angular acceleration a (with initial angular velocity zero).

Explanation

Rotating disc moves along the x-axis, in plane motion in `x-y` plane. Plane motion of a solid can be imagined to be in pure rotation about a point (say I) at a certain instant known as instantaneous centre of rotation. The instantaneous axis whose positive sense is directed along `vecomega` of the solid and which passes through the point I, is known as instantaneous axis of rotation.
Therefore the velocity vector of an arbitrary point `(P)` of the solid can be represented as:
`vecv_P=vecomegaxxvecr_(PI)` (1)
On the basis of Eq. (1) for the C.M. (C) of the disc
`vecv_c=vecomegaxxvecr_d` (2)
According to the problem `vecv_cuarruarrveci` and `vecomegauarruarrveck` i.e., `vecomega_|_x-y` plane, so to satisfy the Eqn. (2) `vecr_(CI)` is directed along `(-vecj)`. Hence point I is at a distance `r_(CI)=y`, above the centre of the disc along y-axis. Using all these facts in Eq. (2), we get
`v_C=omegay` or `y=(v_c)/(omega)` (3)
(a) From the angular kinematical equation
`omega_z=omega_(0z)+beta_(z)t` (4)
`omega=betat`.
On the other hand `x=vt`, (where x is the x coordinate of the C.M.)
or, `t=x/v` (5)
From Eqs. (4) and (5), `omega=(betax)/(v)`
Using this value of `omega` in Eq. (3) we get `y=(v_c)/(omega)=(v)/(betax//v)=(v^2)/(betax)` (hyperbola)
(b) As centre C moves with constant acceleration `w`, with zero initial velocity
So, `x=1/2wt^2` and `v_c=wt`
Therefore, `v_c=wsqrt((2x)/(w))=sqrt(2xw)`
Hence `y=(v_c)/(omega)=(sqrt(2wx))/(omega)` (parabola)