Question

In a slider crank mechanism, the length of the crank and connecting rod are 100 mm 

and 400 mm respectively. The crank rotates uniformly at 600 r.p.m. clockwise. When the crank 

has turned through 45° from the inner dead centre, find, by analytical method : 1. Velocity and 

acceleration of the slider, 2. Angular velocity and angular acceleration of the connecting rod. 

Check your result by Klein’s construction. 

[Ans. 5.2 m/s; 279 m/s2; 11 rad/s; 698 rad/s2]

Answer 1

$CN = r sin \theta = l sin \phi \implies sin \phi = \frac{r}{l} sin \theta$

$sin \phi = \frac{sin \theta}{n} \implies n=\frac{l}{r}$

We know that $sin^2 \phi+cos^2 \phi=1 \implies \sqrt{1-\frac{sin^2 \phi}{n^2}}$

$X_p=r(1-cos \theta )+l(1-\sqrt{1-\frac{sin^2 \phi}{n^2}})$

$X_p=r(1-cos \theta )+r(n-\sqrt{n^2-sin^2 \theta})$

Velocity Since the velocity of the slider is rate of change of displacement with respect to time

$V_p= \frac{d(X_p)}{dt}= \frac{d}{d\theta} \frac{d \theta}{dt} (X_p)$

$V_p= \frac{d(X_p)}{dt}= \frac{d}{d\theta} \frac{d \theta}{dt} (r(1-cos \theta )+r(n-\sqrt{n^2-sin^2 \theta}))$

$V_p= \omega r \frac{d \theta}{dt} (r(1-cos \theta )+r(n-\sqrt{n^2-sin^2 \theta}))$

$V_p= \omega r [sin \theta+ \frac{sin \theta}{2* \sqrt{n^2-sin^2 \theta}}]$