Question

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 $\alpha$ (with initial angular velocity zero).

Answer 1

Here,

Motion of solid can be imagined to be in pure rotation about a point (say I) at a certain Instant. The instantaneous axis whose positive sense is directer ding $w$ of the solid and which passes through the point $I$ is Instantaneous axis of Rotation.

velocity vector of an point (P) of the solid

is $\overrightarrow{V_p}=\overrightarrow{W}\times \overrightarrow{r_{PL}}$...........(1)

on the basis of eq (1) for the (M.Cc)

of the disc:-

$\overrightarrow{V_c}=\overrightarrow{W}\times \overrightarrow{r_d}$............(2)

According to the given problem : $\overrightarrow{V_c}\uparrow i$

and $\overrightarrow{w}\uparrow \uparrow \hat{x}$ i,e $\overrightarrow{w}\perp x-y$ plane, so to satisfy the Eqn (2), $\overrightarrow{r_C}$ is directed along $\left(\overrightarrow{-j}\right)$. Hence point I is at distance $\overrightarrow{r_{cy}}=y$

Using above : equation (2) becomes:-

$V_c=wy$ or $y=\frac{v_C}{w}$..........(3)

(b) From the angular kinematics Equation:-

$W_2=W_{O_2}+{\beta }_{2}t({\beta }_2$ is acc along axis)

$w=|3tbx|$

on other hande, $x=vt$ (where 'x' is the ordinate of the O.M)

$\therefore t=\frac{x}{v}$...........(5)

From (4) and (5) $cv=\frac{\beta x}{v}$

From (3)

and $y=\frac{v_c}{w}=\frac{v}{\beta x/v}=\frac{v^2}{\beta x}$ (Hyper bola)

Given

$\beta =\alpha$

$\therefore y=\frac{v^2}{\alpha x}$

(a) and As centre ${}^{\prime }{C}^{\prime }$ moves with constant acceleration (a), with

zero initial velocity:-

$x=\frac{}{}a{t}^2$ and $v_c=at$

$\therefore V_c=a\sqrt{\frac{2x}{a}}=\sqrt{2xa}$

Hence $y=\frac{v_c}{w}=\frac{2x}{w}$