Question 1 |
A cam with a translating flat-face follower is desired to have the follower motion
y(\theta ) = 4[2 \pi \theta - \theta ^2], 0 \leq \theta \leq 2 \pi
Contact stress considerations dictate that the radius of curvature of the cam profile should not be less than 40 mm anywhere. The minimum permissible base circle radius is _____ mm (round off to one decimal place)
y(\theta ) = 4[2 \pi \theta - \theta ^2], 0 \leq \theta \leq 2 \pi
Contact stress considerations dictate that the radius of curvature of the cam profile should not be less than 40 mm anywhere. The minimum permissible base circle radius is _____ mm (round off to one decimal place)
24 | |
12 | |
48 | |
32 |
Question 1 Explanation:
Flat face follower
Displacement equation:
\begin{aligned} y&=4\left(2 \pi \theta-\theta^{2}\right)\\ \frac{d y}{d \theta} &=V=4(2 \pi-2 \theta) \\ &=8(\pi-\theta) \\ \quad(\text { For } y \text { to be } \max \frac{d y}{d \theta}&=0 \Rightarrow \theta=\pi) \\ a &=\frac{d v}{d \theta}=-8 \\ \left(R_{\text {curvature }}\right)_{\operatorname{Min}} &=R_{\text {Base }}+(y+a)_{\min } \\ \quad(y_{min} & \text{ is 0 at 0},2\pi)\\ 40 &=R_{\text {Base }}+[0-8]_{\operatorname{Min}} \\ 40 &=R_{\text {Base }}+[-8] \\ R_{\text {Base }} &=40-(-8)=40+8=48 \mathrm{mm} \end{aligned}
Displacement equation:
\begin{aligned} y&=4\left(2 \pi \theta-\theta^{2}\right)\\ \frac{d y}{d \theta} &=V=4(2 \pi-2 \theta) \\ &=8(\pi-\theta) \\ \quad(\text { For } y \text { to be } \max \frac{d y}{d \theta}&=0 \Rightarrow \theta=\pi) \\ a &=\frac{d v}{d \theta}=-8 \\ \left(R_{\text {curvature }}\right)_{\operatorname{Min}} &=R_{\text {Base }}+(y+a)_{\min } \\ \quad(y_{min} & \text{ is 0 at 0},2\pi)\\ 40 &=R_{\text {Base }}+[0-8]_{\operatorname{Min}} \\ 40 &=R_{\text {Base }}+[-8] \\ R_{\text {Base }} &=40-(-8)=40+8=48 \mathrm{mm} \end{aligned}
Question 2 |
A flat-faced follower is driven using a circular eccentric cam rotating at a constant angular velocity \omega. At time t=0, the vertical position of the followeris y(0)=0, and the system is in the configuration shown below.
The vertical position of the follower face, y(t) is given by
The vertical position of the follower face, y(t) is given by
e \sin \omega t | |
e(1+ \cos 2\omega t) | |
e(1- \cos \omega t) | |
e \sin 2 \omega t |
Question 2 Explanation:
\begin{aligned} \mathrm{AA}_{1}=\mathrm{y} &=\mathrm{e}(1-\cos \theta) \\ &=\mathrm{e}(1-\cos \omega \mathrm{t}) \end{aligned}
Question 3 |
In a cam-follower,the follower rises by h as the cam rotates by \delta (radians) at constant angular velocity \omega ( radins/ s ). The follower is uniformly accelerating during the first half of the rise period and it is uniformly decelerating in the latter half of the rise period. Assuming that the magnitudes of the acceleration and deceleration are same, the maximum velocity of the follower is
\frac{4h\omega }{\delta } | |
h\omega | |
\frac{2h\omega }{\delta } | |
2h\omega |
Question 3 Explanation:
Here, outstroke angle
\begin{aligned} \theta_{0} &=\delta \\ \text { and stroke length } &=h \\ \text { angular velocity } &=\omega \end{aligned}
\begin{aligned} V &=u+a t \\ \left(V_{0}\right)_{\max } &=0+a \times \frac{t_{0}}{2}=a \times \frac{\delta}{2 \omega} \quad \ldots(i)\\ \frac{h}{2} &=0+\frac{1}{2} a\left(\frac{t_{0}}{2}\right)^{2} \\ h &=\frac{a t_{0}^{2}}{4}=\frac{a \delta^{2}}{4 \omega_{2}} \\ a &=\frac{4 \omega^{2} \cdot h}{\delta^{2}} \quad\ldots(ii)\\ &\text{ By(i) and (ii),} \\ \left(V_{0}\right)_{\max } &=\frac{a \delta}{2 \omega}=\frac{4 \omega^{2} \cdot h}{\delta^{2}} \times \frac{\delta}{2 \omega}=\frac{2 \omega \cdot h}{\delta} \end{aligned}
\begin{aligned} \theta_{0} &=\delta \\ \text { and stroke length } &=h \\ \text { angular velocity } &=\omega \end{aligned}
\begin{aligned} V &=u+a t \\ \left(V_{0}\right)_{\max } &=0+a \times \frac{t_{0}}{2}=a \times \frac{\delta}{2 \omega} \quad \ldots(i)\\ \frac{h}{2} &=0+\frac{1}{2} a\left(\frac{t_{0}}{2}\right)^{2} \\ h &=\frac{a t_{0}^{2}}{4}=\frac{a \delta^{2}}{4 \omega_{2}} \\ a &=\frac{4 \omega^{2} \cdot h}{\delta^{2}} \quad\ldots(ii)\\ &\text{ By(i) and (ii),} \\ \left(V_{0}\right)_{\max } &=\frac{a \delta}{2 \omega}=\frac{4 \omega^{2} \cdot h}{\delta^{2}} \times \frac{\delta}{2 \omega}=\frac{2 \omega \cdot h}{\delta} \end{aligned}
Question 4 |
Consider a rotating disk cam and a translating roller follower with zero offset. Which one of the following pitch curves, parameterized by t, lying in the interval 0 to 2\pi , is associated with the maximum translation of the follower during one full rotation of the cam rotating about the center at (x,y)=(0,0) ?
x(t)=cost,y(t)=sint | |
x(t)=cost,y(t)=2sint | |
x(t)=1/2+cost,y(t)=2sint | |
x(t)=1/2+cost,y(t)=sint |
Question 4 Explanation:
Maximum translation of the follower during one full rotation about center will be
d=\sqrt{x^2+y^2}\text{ where }d: \text{translation of follower}
and d_{max} \text{ at } \frac{d\;d(t)}{dt}=0
d_{max}: Maximum translation of follower
\begin{aligned} x(t)&=\frac{1}{2} + \cos t\\ y(t)&=2 \sin t \\ d &=\sqrt{\left (\frac{1}{2} + \cos t \right )^2 +(2 \sin t)^2} \\ &= \sqrt{\frac{1}{4}+ \cos t+ \cos ^2 t +4 \sin ^2 t}\\ &= \sqrt{\frac{5}{4}+\cos t +3 \sin ^2 t} \end{aligned}
\begin{aligned} \text{For }d_{max}:\\ \frac{d(\cos t + 3 \sin ^2 t)}{dt}&=0\\ - \sin t +6 \sin t \cos t&=0\\ \cos t&=0\\ t=\frac{\pi}{2},\frac{3 \pi}{2}&\\ d_{max}=\sqrt{\frac{5}{4}+0+3}\\ =\sqrt{\frac{5+12}{4}}\\ =\sqrt{\frac{17}{4}}=2.06\\ d_{max}=2.06 \end{aligned}
d=\sqrt{x^2+y^2}\text{ where }d: \text{translation of follower}
and d_{max} \text{ at } \frac{d\;d(t)}{dt}=0
d_{max}: Maximum translation of follower
\begin{aligned} x(t)&=\frac{1}{2} + \cos t\\ y(t)&=2 \sin t \\ d &=\sqrt{\left (\frac{1}{2} + \cos t \right )^2 +(2 \sin t)^2} \\ &= \sqrt{\frac{1}{4}+ \cos t+ \cos ^2 t +4 \sin ^2 t}\\ &= \sqrt{\frac{5}{4}+\cos t +3 \sin ^2 t} \end{aligned}
\begin{aligned} \text{For }d_{max}:\\ \frac{d(\cos t + 3 \sin ^2 t)}{dt}&=0\\ - \sin t +6 \sin t \cos t&=0\\ \cos t&=0\\ t=\frac{\pi}{2},\frac{3 \pi}{2}&\\ d_{max}=\sqrt{\frac{5}{4}+0+3}\\ =\sqrt{\frac{5+12}{4}}\\ =\sqrt{\frac{17}{4}}=2.06\\ d_{max}=2.06 \end{aligned}
Question 5 |
In the mechanism given below, if the angular velocity of the eccentric circular disc is 1 rad/s, the angular velocity (rad/s) of the follower link for the instant shown in the figure is
Note: All dimensions are in mm
Note: All dimensions are in mm
0.05 | |
0.1 | |
5 | |
10 |
Question 5 Explanation:
\begin{aligned} \Delta P Q O &\sim \Delta S R O \\ \therefore \quad \frac{P Q}{S R}&=\frac{P O}{S O} \qquad\cdots(1) \end{aligned}
Applying Pythagoras theorem in \Delta PQO
\begin{aligned} (P O)^{2}&=(P Q)^{2}+(O Q)^{2}\\ \therefore \quad (P Q)&=\sqrt{(P O)^{2}-(O Q)^{2}} \\ &=\sqrt{(50)^{2}-(25)^{2}}=43.3 \mathrm{mm} \end{aligned}
Putting value of P Q in Eq. (1), we get
\begin{aligned} \frac{43.3}{S R} &=\frac{45+5}{5} \\ S R &=\frac{43.3 \times 5}{50}=4.33 \mathrm{mm} \end{aligned}
Velocity of Q is equal to velocity of R
\begin{aligned} V_{Q} &=V_{R} \\ &=SR \times \omega \\ &=4.33 \times 1=4.33 \mathrm{mm} / \mathrm{s} \\ \omega_{P Q} &=\frac{V_{P Q}}{P Q}=\frac{4.33}{43.3}=0.1 \mathrm{rad} / \mathrm{s} \end{aligned}
There are 5 questions to complete.
In question number 7 has some correction in question follower moves by distance of 20mm and solution has 10mm.