Determine:

A] The crank angle @ which the maximum velocity occurs.

B] Maximum velocity of the piston.

Solution:

Given: $r = 300 mm$

$L = 1m = 1000 mm$

$N = 200 rpm$

$\therefore$ $\omega = \frac{2\pi(200)}{60} = 20.944 \ rad/s$

To find : $\theta_{max} \ and \ V_{max}$

$\rightarrow$ $n = \frac{L}{r} = \frac{1000}{300} = 3.33$

$\rightarrow$ To get maximum $\theta$,

$\frac{dVp}{d \theta} = 0$

$\therefore$ $\frac{d}{d \theta} [ \omega r. (sin \ \theta + \frac{sin \ 2 \theta}{2n})] = 0$

$\therefore$ $cos \ \theta \ + \frac{cos \ 2 \theta \ \times (2)}{2n} = 0$

$\therefore$ $cos \ \theta + \frac{cos \ 2 \ \theta}{n} = 0$

[ Hint : $1 + cos \ 2 \ \theta \ = 2 \ cos^2 \ \theta]$

[ $\therefore$ $cos \ 2 \ \theta = 2 \ cos^2 \ \theta \ - 1$]

$\eta \ cos \ \theta + 2 \ cos^2 \ \theta \ – 1 = 0$

$3.33 \ cos \ \theta \ + 2 cos^2 \ \theta \ – 1 = 0$

On solving,

$cos \ \theta = 0.26$ OR $cos \ \theta = -1.92$ X Invalid.

$\therefore$ $\theta \ = cos^{-1} \ (0.26)$

$\theta_{max} = 74.93°$ - - - -1st Ans.

To get maximum velocity put $\theta$ max in Vp.

$\therefore$ $(vp)_{max} = \omega r. [sin \ \theta + \frac{sin \ 2 \theta}{2n}]$

$= (20.944) (0.3) [sin \ (74.93) + \frac{sin \ (2 \times 74.93)}{2 \times 3.33}]$

$= 6.54 m/s$ - - - 2nd Ans.