# Dynamic Analysis of Slider-crank

 Question 1
The figure shows a wheel rolling without slipping on a horizontal plane with angular velocity $\omega _1$. A rigid bar PQ is pinned to the wheel at P while the end Q slides on the floor.
What is the angular velocity $\omega _1$ of the bar PQ? A $\omega _2=2\omega _1$ B $\omega _2=\omega _1$ C $\omega _2=0.5\omega _1$ D $\omega _2=0.25\omega _1$
GATE ME 2023   Theory of Machine
Question 1 Explanation: 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
$\omega_{2}=0.25 \omega_{1}$
 Question 2
The Whitworth quick return mechanism is shown in the figure with link lengths as follows: OP = 300 mm, OA = 150 mm, AR = 160 mm, RS = 450 mm. The quick return ratio for the mechanism is ________(round off to one decimal place).
 A 1 B 2 C 2.5 D 3.5
GATE ME 2021 SET-1   Theory of Machine
Question 2 Explanation:  \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}

 Question 3
The crank of a slider-crank mechanism rotates counter-clockwise (CCW) with a constant angular velocity $\omega$, as shown. Assume the length of the crank to be r. Using exact analysis, the acceleration of the slider in the y direction, at the instant shown, where the crank is parallel to x-axis, is given by
 A $-\omega^2 r$ B $2 \omega^2 r$ C $\omega^2 r$ D $-2 \omega^2 r$
GATE ME 2019 SET-2   Theory of Machine
Question 3 Explanation: since the velocity of the point A and B are parallel $\omega_{A B}=0$
$\begin{array}{l} \overrightarrow{\mathrm{a}}_{\mathrm{B}}=\overrightarrow{\mathrm{a}}_{\mathrm{A}}+\overrightarrow{\mathrm{a}}_{\mathrm{AB}} \\ \overrightarrow{\mathrm{a}}_{\mathrm{B}}=\mathrm{a}_{\mathrm{B}} \hat{\mathrm{j}} \\ \overrightarrow{\mathrm{a}}_{\mathrm{A}}=-\omega^{2} \mathrm{r} \hat{\mathrm{i}} \\ \overrightarrow{\mathrm{a}}_{\mathrm{AB}}=-\alpha \mathrm{r}_{\mathrm{AB}} \sin 45 \hat{\mathrm{i}}-\alpha \mathrm{r}_{\mathrm{AB}} \cos 45 \hat{\mathrm{j}} \\ \quad\left(\because \omega^{2} \mathrm{r}_{\mathrm{AB}} \text { along link } \mathrm{AB}=0\right) \\ \mathrm{a}_{\mathrm{B}} \hat{\mathrm{j}}=-\omega^{2} \mathrm{r} \hat{\mathrm{r}}-(\alpha \hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}) \quad=-\left(\omega^{2} \mathrm{r}+\alpha\right) \hat{\mathrm{i}}-\alpha \hat{\mathrm{j}} \\ \omega^{2} \mathrm{r}+\alpha=0 \\ \alpha=-\omega^{2} \mathrm{r} \\ \mathrm{a}_{\mathrm{B}}=-\alpha=-\left(-\omega^{2} \mathrm{r}\right)=\omega^{2} \mathrm{r} \end{array}$
 Question 4
A slider crank mechanism is shown in the figure. At some instant, the crank angle is $45^{\circ}$ and a force of 40 N is acting towards the left on the slider. The length of the crank is 30 mm and the connecting rod is 70 mm. Ignoring the effect of gravity, friction and inertial forces, the magnitude of the crankshaft torque (in Nm) needed to keep the mechanism in equilibrium is _________ (correct to two decimal places). A 0.12 B 0.56 C 1.12 D 2.52
GATE ME 2018 SET-1   Theory of Machine
Question 4 Explanation: \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}
 Question 5
In a certain slider-crank mechanism, lengths of crank and connecting rod are equal. If the crank rotates with a uniform angular speed of 14 rad/s and the crank length is 300 mm, the maximum acceleration of the slider (in m/$s^{2}$) is ___________
 A 117.6m/s B 215.4m/s C 125.2m/s D 458.6m/s
GATE ME 2015 SET-2   Theory of Machine
Question 5 Explanation:
$a=r \omega^{2}\left[\cos \theta+\frac{\cos 2 \theta}{n}\right]$
$\text{at} \quad \theta=0$
\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}

There are 5 questions to complete.