Planar Mechanisms

\begin{array}{l} \Rightarrow \mathrm{V}_{\mathrm{R}}=(\mathrm{IR}) \omega \\ \Rightarrow \omega=\frac{\mathrm{V}_{\mathrm{R}}}{(\mathrm{IR})} \\ \Rightarrow \omega \times \frac{\mathrm{V}_{\mathrm{R}}}{\frac{1}{2}} \\ \Rightarrow \omega=2 \mathrm{V}_{\mathrm{R}} \end{array}

\begin{array}{l} \mathrm{AB}=5 \mathrm{cm} \\ \mathrm{AD}=4 \mathrm{cm} \\ \mathrm{DC}=2 \mathrm{cm} \\ \omega_{\mathrm{AB}}=10 \mathrm{rad} / \mathrm{s} \\ \because A \mathrm{B} \| \mathrm{DC} \\ \therefore A B \cdot \omega_{\mathrm{AB}}=\mathrm{DC} \cdot \omega_{\mathrm{DC}} \\ 5 \times 10=2 \times \omega_{\mathrm{DC}} \\ \omega_{\mathrm{DC}}=25 \mathrm{rad} / \mathrm{s} \end{array}

After plotting I-centres

Here, I_{23} and I_{24} will come at same point
Applying angular Velocity Theorem at I_{24}
\begin{aligned} \therefore \quad \omega_{2}\left(I_{24} I_{12}\right) &=V_{4}=V_{B} \\ V_{B} &=\omega_{2}\left(I_{24} I_{12}\right)=10 \times 0.1 \\ \left(I_{24} I_{12}=100 \mathrm{mm}\right.&=0.1 \mathrm{m}=A B) \\ V_{B} &=1 \mathrm{m} / \mathrm{s} \end{aligned}

\begin{aligned} \text { Here } & V_{B}=1 \mathrm{m} / \mathrm{s} \\ & \omega_{\text {crank }}=? \end{aligned}
Let us draw the I-centres


Applying angular velocity theorem at I_{24}
\begin{array}{l} \therefore \omega_{2}\left(I_{24} I_{12}\right)=V_{4}=V_{B}=1 \\ \omega_{2}\left(I_{24} I_{12}\right)=1 \end{array}


\begin{aligned} \theta &=\tan ^{-1}(4 / 3)=53.13010 \\ \alpha &=90^{\circ}-\theta=36.869 \\ \cos \alpha &=\frac{3}{I_{12} I_{24}} \\ I_{12} I_{24} &=\frac{3}{\cos 36.869}=3.7499 \mathrm{meter} \\ \text{Now}, & \omega_{2}(3.7499)=1 \\ \omega_{2} &=\frac{1}{3.7499}=0.266 \mathrm{rad} / \mathrm{s} \end{aligned}

\begin{aligned} F &=3(6-1)-2 \times 7-0 \\ &=15-14=1 \end{aligned}

\begin{aligned} \tau(\text { torque }) &=I \ddot{\theta} \\ \tau &=\text { constant } \\ I &=\text { constant } \\ \tau &=I \frac{d^{2} \theta}{d t^{2}} \\ \frac{d^{2} \theta}{d t^{2}} &=\frac{\tau}{I} \\ \frac{d \theta}{d t} &=\frac{\tau}{I} t+C_{1} \\ \theta &=\frac{\tau}{I} \frac{t^{2}}{2}+C_{1} t+C_{2}\\ \theta=f(t)&= \text{ parabolic function}\\ \end{aligned}
But practically graph will be fluctuating about ideal parabolic curve.

A simple truss is formed by enlarging the basic truss element which contains three members and three joints, by adding two additional members for each additional joint, so the total number of members m in a simple truss is given by
m=3+2(j-3)=2 j-3
where m= number of members
j= total number of joints (including those attached to the supports)

S+L \lt P+Q
In this case, shortest link is fixed and hence, the
resulting mechanism is double crank mechanism.

\begin{aligned} \therefore \; \omega _{PQ}&=\frac{V_P}{PI_{PQ}}=\frac{V_Q}{QI_{PQ}} \\ PI_{PQ}&=\text{Distance between P and }I_{PQ} \\ &= 2 \cos 70^{\circ}+ 2 \sin 70^{\circ} \times \tan 45^{\circ}\\ QI_{PQ} &= \text{Distance between Q and }I_{PQ} \\ &=\sqrt{2}\times 2 \times \sin 70^{\circ} \\ &= 2.66m\\ \therefore \;\; \frac{V_P}{2.56} &=\frac{1}{2.66}\\ V_P&=\frac{2.56}{2.66}=0.96\; m/s \end{aligned}

\begin{array}{l} P Q=2.0 \mathrm{m}, Q R=3.0 \mathrm{m}, R S=2.5 \mathrm{m} \\ S P=2.7 \mathrm{m} \end{array}
(i) If shortest line PQ=2.0m is fixed, then we will give double crank.
(ii) If link adjacent to shortest is fixed, then we will get crank-rocker mechanism.
(iii) If we fix coupler 'RS', we will get double rocker mechanism.