Subject: Kinematics of Machinery

Topic: Velocity Analysis of Mechanism (Mechanism up-to 6 links)

Difficulty: High

It is seen that the acceleration of a moving point relative to the fixed body (fixed coordinate system) may have two components of acceleration; the centripetal and the tangential. However, in some cases. The point may have its motion relative to a moving body(moving coordinate) system. For example, the motion of a slider on a rotating link. The following analysis is made to investigate the acceleration at that point.

Let a link AR rotate about a fixed point A on it (fig.). P is a point on a slider on the link.

At given instant,

$\omega$ = Angular velocity of the link

$\alpha$ = Angular acceleration of the link

$v$ = Linear velocity of the slider on the link $f$ = Linear acceleration of the slider on the link $r$ = Radial distance of point P on the slider. In short interval of time $\delta t$, let $\delta\theta$ be the angular displacement of the link and $\delta r$, the radial displacement of the slider in the outward direction. ![enter image description here][1] After the short interval of time $\delta t$, let $\omega'$ = $\omega$ + $\alpha\delta t$ = Angular velocity of the link $v'$ = $v$ + $f\delta t$ = Linear velocity ot the slider on the link $r'$ = $r$ + $\delta r$ = Radial distance of the slider **Acceleration of P parallel to AR** Initial velocity of P along AR = $v=v_{pq}$ Final velocity of P along AR = $ν’\cos\delta\theta - \omega’r’\sin\delta\theta$ Change of velocity along AR = $(ν’\cos\delta\theta- \omega’r’\sin\delta\theta) – ν$ Acceleration of P along AR = $\frac{(ν + f \delta t)\cos\delta\theta-(\omega + \alpha\delta t)(r + \delta r)\sin \delta\theta}{\delta t}$ In the limit, as P along AR = $f – \omega r \frac{\text{d}\theta}{\text{dt}}$ = $f – \omega rw$ = $f – \omega^{2} r$ = Acc. Of slider – cent. Acc. This is the acceleration of P along AR in the radially outward direction. $f$ will be negative if the slider has deceleration while moving in the outward direction or has acceleration while moving in the inward direction. **Acceleration of P Perpendicular to AR:** Initial velocity of $P\bot\space to\space AR=\omega r$ Final velocity of $P\bot\space to\space AR=ν’ \sin\delta\theta+ \omega’ r’\cos\delta\theta$ Change of velocity $\bot\space to\space AR= (ν’ \sin\delta\theta+ \omega’r’\cos\delta\theta) – \omega r$ Acceleration of $P\bot\space to\space AR=\frac{(ν + f \delta t)\cos\delta\theta-(\omega + \alpha\delta t)(r + \delta r)\cos\delta\theta–\omega r}{\delta t}$ In the limit as $\delta t\rightarrow\delta\theta$ $\cos\delta\theta\rightarrow1$ and $\sin\delta\theta\rightarrow\delta\theta$ Acceleration of $P\bot\space to \space AR=v\frac{\partial \theta}{dt}+\frac{\partial r}{dt}+r\alpha$ = $ν\omega+ \omega ν+r\alpha$ = $2 \omega ν+ r\alpha$ = $2 \omega ν$ + Tangential acc. This is the acceleration of P perpendicular to AR. The component $2\omega v$ is known as the Coriolis acceleration component. ![enter image description here][2] It may be noted that the direction of coriolis component of acceleration changes sign, if either $\omega$ or $v$ is reversed in direction. But the direction of coriolis component of acceleration will not be changed in sign if both $\omega$ and $v$ are reversed in direction. It is concluded that the direction of coriolis component of acceleration is obtained by rotating $v$ ,at $90^{\circ}$, about its origin in the same direction as that of $\omega$. The direction of coriolis component of acceleration ($2\omega v$)for all four possible cases, is shown in fig.11. The directions of $\omega$ and $v$ are given.