Question

A rotating disc (figure) moves in the positive direction of the $x$ -axis. Find the equation $y (x)$ describing the position of the instantaneous axis of rotation, if at the initial moment the axis $C$ of the disc was located at the point $O$ after which it moved.

(a) With a constant velocity $v$ , while the disc started rotating counter clockwise with a constant angular acceleration $β$ (the initial angular velocity is equal to zero);

(b) With a constant acceleration a (and the zero initial velocity), while the disc rotates counterclockwise with a constant angular velocity $ω$ .

Accepted Answer

Rotating disc moves along the $x$ -axis in plane motion in $x - y$ plane. A motion of a solid can be imagined to be in pure rotation about a point (say $C$ ) at a certain instant known as an instantaneous centre of rotation. The instantaneous axis whose positive sense is directed along vector $O$ of the solid and which passes through the point $I$ is known as the instantaneous axis of rotation.

Therefore, the velocity vector of an arbitrary point ( $P$ ) of the solid can be represented as

$\vec{V_{P}} = \vec{ω} \times \vec{r_{PI}} \dots (1)$

On the basis of equation $(1)$ for centre of mass of the disc,

$\vec{V_{C}} = \vec{ω} \times \vec{r_{d}} \dots (2)$

It is given that $\vec{V_{c}} ∥ \hat{i} and \vec{ω} ∥ \hat{k}$

So to satisfy the equation $(2)$ ,

$\vec{r_{cl}}$ is directed along $(- \vec{j})$ . Hence, point $I$ is at a distance $r_{cl} = y,$ above the centre of the disc along $y$ -axis. Using all values in equation $(2),$

$v_{c} = ω y or y = \frac{v_{c}}{ω} \dots (3)$

$(a)$ From the angular kinematic equation

$ω_{z} = ω_{0 z} + β_{z} t ω = β t \dots (4)$

We know $x = v t$ ,

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

From equation $(4) and (5)$ ,

$ω = \frac{β x}{v}$

Using the value of $ω$ in equation $(3)$ ,

$y = \frac{v_{c}}{ω} = \frac{v}{\frac{β x}{v}} = \frac{v^{2}}{β x}$

The trajectory is hyperbola.

$(b)$ As centre $C$ moves with constant acceleration $a$ with zero initial velocity

$\Rightarrow$ $x = \frac{1}{2} ω t^{2} and v_{c} = ω t$

$v_{c} = ω \sqrt{\frac{2 x}{ω}} = \sqrt{2 x ω}$

Hence, $y = \frac{v_{c}}{ω} = \frac{\sqrt{2 x ω}}{ω}$

Trajectory is parabola.

Important Questions on Rotational Mechanics

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