Dynamic Analysis of Slider-crank

As Disc (Link 2) roll without slipping on Link 1 (surface) so their I-centre of rotation lies on point of contact to avoid relative motion in between them. So \mathrm{I}_{12} lies on point of contact.
Link 2 and 3 have instantaneous center of rotation at \mathrm{I}_{13} due to fixed surface of Link 1.

As per point Plies on link 2 and 3 both so, point P has same velocity for Link 2 and 3 both.
As per Kennedy's theorem, if three plane bodies have relative motion among themselves, their I-centre must lie on a straight line. So, \mathrm{I}_{23} must be on line joining \mathrm{I}_{13} and \mathrm{I}_{12} at point P.
By similar \triangle \mathrm{PI}_{12} \mathrm{~B} and \Delta \mathrm{Pl}_{13} \mathrm{~A}, we have
\frac{\mathrm{l}_{12} \mathrm{P}}{\mathrm{l}_{12} \mathrm{~B}}=\frac{\mathrm{Pl_{13 }}}{\mathrm{PA}}
or, \quad \frac{\mathrm{I}_{12} \mathrm{I}_{23}}{2}=\frac{\mathrm{I}_{23} \mathrm{l}_{13}}{8}
or, \mathrm{I}_{12} \mathrm{l}_{13}=0.25 \mathrm{I}_{23} \mathrm{l}_{13} \;\;\;...(i)
Velocity of point P is same w.r.t. all instantaneous rotation so,
\omega_{1} \times I_{12} I_{23}=\omega_{2} \times I_{23} l_{13}
By equation (i), we get
\omega_{2}=0.25 \omega_{1}

\begin{aligned} \cos \frac{\alpha}{2} &=\frac{150}{300}=\frac{1}{2} \\ \Rightarrow \qquad \frac{\alpha}{2} &=60^{\circ} \\ \alpha &=120^{\circ} \\ \beta &=\left(360^{\circ}-120^{\circ}\right)=240^{\circ} \\ \mathrm{QRR} &=\left(\frac{\beta}{\alpha}\right)=\frac{240}{120}=2 \end{aligned}

since the velocity of the point A and B are parallel \omega_{A B}=0
\begin{array}{l} \overrightarrow{\mathrm{a}}_{\mathrm{B}}=\overrightarrow{\mathrm{a}}_{\mathrm{A}}+\overrightarrow{\mathrm{a}}_{\mathrm{AB}} \\ \overrightarrow{\mathrm{a}}_{\mathrm{B}}=\mathrm{a}_{\mathrm{B}} \hat{\mathrm{j}} \\ \overrightarrow{\mathrm{a}}_{\mathrm{A}}=-\omega^{2} \mathrm{r} \hat{\mathrm{i}} \\ \overrightarrow{\mathrm{a}}_{\mathrm{AB}}=-\alpha \mathrm{r}_{\mathrm{AB}} \sin 45 \hat{\mathrm{i}}-\alpha \mathrm{r}_{\mathrm{AB}} \cos 45 \hat{\mathrm{j}} \\ \quad\left(\because \omega^{2} \mathrm{r}_{\mathrm{AB}} \text { along link } \mathrm{AB}=0\right) \\ \mathrm{a}_{\mathrm{B}} \hat{\mathrm{j}}=-\omega^{2} \mathrm{r} \hat{\mathrm{r}}-(\alpha \hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}) \quad=-\left(\omega^{2} \mathrm{r}+\alpha\right) \hat{\mathrm{i}}-\alpha \hat{\mathrm{j}} \\ \omega^{2} \mathrm{r}+\alpha=0 \\ \alpha=-\omega^{2} \mathrm{r} \\ \mathrm{a}_{\mathrm{B}}=-\alpha=-\left(-\omega^{2} \mathrm{r}\right)=\omega^{2} \mathrm{r} \end{array}

\begin{aligned} 30 \sin 45^{\circ} &=70 \sin \beta \\ \sin \beta &=\frac{30 \sin 45}{70}=0.303045 \\ \beta &=17.6406^{\circ} \\ F_{C} \cos \beta &=40 \\ F_{C} &=\frac{40}{\cos \beta}=41.9737 \mathrm{N} \\ F_{t} &=F_{C} \sin (\theta+\beta) \\ &=41.9737 \times \sin \left(17.6406^{\circ}+45^{\circ}\right) \\ F_{t} &=37.2785 \mathrm{N} \\ T &=F_{t} \times r=47.2785 \times 0.030 \\ &=1.1183 \mathrm{N}-\mathrm{m}=1.118 \mathrm{N}-\mathrm{m} \\ &=1.12 \mathrm{N}-\mathrm{m} \end{aligned}

a=r \omega^{2}\left[\cos \theta+\frac{\cos 2 \theta}{n}\right]
\text{at} \quad \theta=0
\begin{aligned} a &\Rightarrow a_{\max } \\ a_{\max } &=r \omega^{2}\left[1+\frac{1}{n}\right] \\ n &=\frac{l}{r}=1 \\ a_{\max } &=2 r \omega^{2}=2 \times 0.3 \times 14^{2} \\ &=117.6 \mathrm{m} / \mathrm{s}^{2} \end{aligned}