Determine the velocity and acceleration of the piston when the crank is @ $40°$ from $10\ celsius$. Also determine [1] the position of the crank for zero acceleration of piston [2] Angular velocity and angular acceleration of the connecting rod when crank is @ $40°$ from IDC.

$\rightarrow$ Given: $r = 0.3m = 300mm$

$L = 1.5m = 1500mm$ { $n = \frac{L}{r} = \frac{1500}{300} = 5$ }

$N = 180 rpm$

$\therefore$ $w = \frac{2 \pi (180)}{60} = 18.85$ rad/s

$\theta = 40°$ From IDC.

[1] $Vp = \omega .r. [sin \ \theta + \frac{sin 2 \ \theta}{2n}]$

$= 18.85 (0.3) [sin \ 40 + \frac{sin \ 2 \times 40}{2 \times 5}]$

$= 4.19 m/s$ - - - - 1st Ans.

[2] $a_p = \omega^2. r [ cos \ \theta + \frac{cos \ 2 \theta}{n}]$

$= (18.85)^2 (0.3) [cos \ 40 + \frac{cos \ (2 \times 40)}{5}]$

$= 85.35 \ m/s^2$ - - - - 2nd Ans.

[3] Position of the crank for zero acceleration of the piston.

$\therefore$ $a_p = \omega^2. r. [ cos \ \theta + \frac{cos \ 2 \theta}{n}]$

$\therefore$ $0 = [cos \ \theta + \frac{cos \ 2 \theta}{5}]$

$\therefore$ $5 \ cos \ \theta + cos \ 2 \ \theta = 0$

$5 \ cos \ \theta + 2 \ cos^2 \ \theta – 1 =0$

On solving,

$cos \ \theta = 0.186$ OR $cos \ \theta = -2.686$ x Invalid.

$\therefore$ $\theta \ = cos^{-1} \ (0.186)$

$\theta = 79.28°$

[4] Angular velocity of connecting rod.

$\omega_{pc} = \frac{w. \ cos \ \theta}{n} = \frac{18.85 \ cos \ 40}{5} = 2.88 rad/s$

[5] Angular acceleration of connecting rod.

$\alpha_{pc} = \frac{\omega^2. sin \ \theta}{n} = \frac{(18.85)^2 . sin \ 40}{5}$

$\alpha_{pc} = 45.68 \ rad/s^2$