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Angular velocity and acceleration with applications to mechanics

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Angular Velocity and Acceleration with Application to Mechanics

The diagram shows a point P moving in a plane XOY. The length OP = r and the $\angle{POX}\;=\;\theta$

Then

$x\;=\;r\:cos\,\theta$

(1)

and

$y\;=\;r\:sin\,\theta$

(2)

Differentiating with respect to time

$\dot{x}\;=\;\frac{dx}{dt}\;=\:\dot{r}\:cos\:\theta\:-\:r\:sin\,\theta\;\dot{\theta}$

(3)

$\dot{y}\;=\;\frac{dy}{dt}\;=\;\dot{r}\:sin\,\theta\:+\:r\:cos\:\theta\;\dot{\theta}$

(4)

The radial component of velocity ( i.e. perpendicularr to OP in the direction of $\theta$ increasing

$=\;\dot{y}\:cos\,\theta\:-\:\dot{x}\:sin\:\theta$

(5)

$=\;r\:\dot{\theta}\;=\;r\:\omega$

(6)

$(\omega\;=\;\dot{\theta}\;=\:the\:angular\:velocity\:of\:OP)$

(7)

Differentiating equations 4 and 5

$\ddot{x}\:=\:\ddot{r}\:cos\,\theta\:-\:\dot{r}\:sin\:\theta\:-\:r\:cos\,\theta\;\ddot{\theta}^2\:-\:r\:sin\,\theta\;\ddot{\theta}$

(8)

$\ddot{y}\:=\:\ddot{r}\:sin\:\theta\:+\:\dot{r}\:cos\theta\;\dot{\theta}\:+\:r\:sin\:\theta\;\dot{\theta}^2\:+\:r\:cos\:\theta\;\ddot{\theta}$

(9)

Radial components of acceleration

$=\;\ddot{x}\:cos\,\theta\:+\:\ddot{y}\:sin\:\theta$

(10)

$=\;\ddot{r}\;-\;r\:\dot{\theta}^2\;\:\:\;from\:above$

(11)

$=\;\dot{v}\;-\;r\,\omega^2$

(12)

Tangential components of acceleration

$=\;\ddot{y}\:cos\,\theta\;-\;\ddot{x}\:sin\,\theta$

(13)

$=\;r\:\ddot{\theta}\;+\;2\:\dot{r}\:\dot{\theta}\;\;\;\;by\;substitution$

(14)

$=\;r\:\alpha\;=\;2\:v\:\omega\;\;\;(\alpha\:=\:angular\:acceleration\:of\:OP\:=\,\ddot{\theta})$

(15)

Of these four terms in equations ( 14 ) and ( 17 )

$\dot{v}$ = the rate of change of radial velocity.
$r\:\omega^2$ = the centripetal acceleration due to the rotation of OP.
$r\:\alpha$ = due to the change in angular velocity.
$2\:v\:\omega$ is called the compound supplementary acceleration or "Coriolis component. Notice that the direction of the latter is the same as $r\:\omega$ when v is radially outwards.

Velocity and Acceleration of a Piston

From the diagram it can be seen that

$x\;=\;r\:cos\,\theta\:+\:l\:cos\,\phi$

(16)

but

$r\:sin\,\theta\;=\;l\:sin\,\phi$

(17)

$\therefore\:\;\;cos\,\phi\;=\;\sqrt[]{(1\:-\:sin^2\:\frac{\theta}{n^2})}\;\;\;where \;n\:=\:\frac{l}{r}$

(18)

and

$x\;=\;r\;[cos\,\theta\:+\:\sqrt[]{(n^2\:-\:sin^2\,\theta)}}]$

(19)

Piston velocity

$=\;\dot{x}\;=\,-\:r\:\omega\:\left( sin\,\theta\:+\:\frac{sin\,2\theta}{2\:\sqrt[]{(n^2\:-\:sin\,^2\,\theta}} \right)$

(20)

Piston Acceleration

$=\:\ddot{x}\:=\:-\:r\,\left( cos\,\theta\:+\:\frac{n^2\,cos\,2\theta\:+\:sin^4\,\theta}{(n^2\:-\:sin^2\,\theta)^\frac{3}{2}} \right)$

(21)

where $\omega$ is the uniform angular velocity of the crank and the positive direction of velocity and acceleration is away from the crankshaft.

Normally $sin^2\:\theta$ can be neglected in comparison to $n^2$ and the above equations can be reduced to

$\dot{x}\;=\;-\:r\,\omega\left( sin\,\theta\:+\:\frac{sin\,2\theta}{2\,n} \right)$

(22)

$\ddot{x}\;=\;-\:r\,\omega^2\left(cos\,\theta\:+\:\frac{cos\:2\theta}{n} \right)$

(23)

Example 1

In the crank and slotted -lever mechanism shown, the crank OP is driven at a uniform speed of $\omega\:radians\:per\:second$ . If OL is the perpendicular from O on WQ the centre-line of the slotted lever, prove that the angular acceleration of the slotted lever is given by

$\phi\;=\;\omega^2\:.\frac{OL}{XP}\:\left(\frac{2\,PL}{XP}\:-\:1 \right)$

(24)

Hence or otherwise find the acceleration of the point Q in magnitude and direction when the crank angle

$\theta\:=\:60^0,\:given\:that\:\omega\:=\:10\,rad./sec.$

(25)

OP = 3in. and XQ = 18in.

$OX\:sin\,\phi\:=\:OL\:=\,OP\:sin\,(\theta\,-\:\phi)$

(26)

Differentiate with respect to t

$OX\:cos\,\phi\:.\:\dot{\phi}\:=\:OP\:cos(\theta\:-\:\phi)\:.\:(\omega\:-\:\phi)$

(27)

$\left[OX\:cos\,\phi\:+\:OP\:cos\:(\theta\:-\:\phi) \right]\phi\:=\:OP\:cos(\theta\:-\:\phi)\:.\,\omega$

(28)

i.e. $XP\:.\:\phi\:=\:PL\:.\:\omega$ or $\phi\;=\:\frac{PL}{XP}\:\omega$

Differentiating ( ),

$[OX\:cos\,\phi\:+\:OP\:cos(\theta\:-\:\phi)]\ddot{\phi}\:-\:OX\:sin\,\phi\:.\:\dot{\phi}^2\:-\:OP\:sin\,(\theta\:-\:\phi)\:.\:\dot{\phi}(\omega\:-\:\phi)\:=\:-\:OP\:sin(\theta\:-\:\phi)\:.\:\omega(\omega\:-\:\phi)$

(29)

Substituting for $\phi$ from ( ) and making use of the geometry of the mechanism, this reduces to:-

$XP\:.\:\ddot{\phi}\:-\:OL\left(\frac{PL^2}{XP^2} \right)\omega^2\,-\:OL\left( \frac{PL}{XP} \right)\left( 1\:-\:\frac{PL}{XP \right)}\:\omega^2\:=\:-\:OL\left(1\:-\:\frac{PL}{XP} \right)\:\omega^2$

(30)