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}
Question 6 |
In a cam design, the rise motion is given by a simple harmonic motion (SHM) s=\frac{h}{2}(1-cos\frac{\pi \theta }{\beta }) where h is total rise,
\theta is camshaft angle, \beta is the total angle of the rise interval. The jerk is given by
\frac{h}{2}(1-cos\frac{\pi \theta }{\beta }) | |
\frac{\pi h}{\beta 2}sin(\frac{\pi \theta }{\beta }) | |
\frac{\pi^2 h}{\beta^2 2}cos(\frac{\pi \theta }{\beta }) | |
\frac{-\pi^3 h}{\beta^3 2}sin(\frac{\pi \theta }{\beta }) |
Question 6 Explanation:
The jerk is given by \frac{d^{3} s}{d t^{3}}
\begin{array}{l} \text { if } S=\frac{h}{2}\left(1-\cos \frac{\pi \theta}{\beta}\right) \\ \therefore \quad \frac{d^{3} s}{d t^{3}}=-\frac{\pi^{3}}{\beta^{3}} \frac{h}{2} \sin \left(\frac{\pi \theta}{\beta}\right)\\ \text{where } \theta=\omega t \end{array}
\begin{array}{l} \text { if } S=\frac{h}{2}\left(1-\cos \frac{\pi \theta}{\beta}\right) \\ \therefore \quad \frac{d^{3} s}{d t^{3}}=-\frac{\pi^{3}}{\beta^{3}} \frac{h}{2} \sin \left(\frac{\pi \theta}{\beta}\right)\\ \text{where } \theta=\omega t \end{array}
Question 7 |
In a cam-follower mechanism, the follower needs to rise through 20 mm during 60^{\circ} of cam rotation, the first 30^{\circ} with a constant acceleration and then with a deceleration of the same magnitude. The initial and final speeds of the follower are zero. The cam rotates at a uniform speed of 300 rpm. The maximum speed of the follower is
0.60 m/s | |
1.20 m/s | |
1.68 m/s | |
2.40 m/s |
Question 7 Explanation:
Angular velocity,
\omega=\frac{2 \pi N}{60}=\frac{2 \pi 300}{60}=10 \pi
Time taken to move 30^{\circ}
=\frac{\frac{\pi}{180} \times 30}{\omega}=\frac{\frac{\pi}{6}}{10 \pi}=\frac{1}{60} \mathrm{s}
During this time, follower moves by distance of 10mm with initial velocity
\begin{aligned} u&=0\\ \text{Now;} \quad S &=u t+\frac{1}{2} a t^{2} \\ 0.01 &=0+\frac{1}{2} \times a \times\left(\frac{1}{60}\right)^{2} \\ \therefore \quad a &=0.01 \times 2 \times(60)^{2}\\ &=72 \mathrm{m} / \mathrm{s}^{2} \\ V_{\max } &=U+a t\\ &=0+72 \times \frac{1}{60}=1.20 \mathrm{m} / \mathrm{s} \end{aligned}
\omega=\frac{2 \pi N}{60}=\frac{2 \pi 300}{60}=10 \pi
Time taken to move 30^{\circ}
=\frac{\frac{\pi}{180} \times 30}{\omega}=\frac{\frac{\pi}{6}}{10 \pi}=\frac{1}{60} \mathrm{s}
During this time, follower moves by distance of 10mm with initial velocity
\begin{aligned} u&=0\\ \text{Now;} \quad S &=u t+\frac{1}{2} a t^{2} \\ 0.01 &=0+\frac{1}{2} \times a \times\left(\frac{1}{60}\right)^{2} \\ \therefore \quad a &=0.01 \times 2 \times(60)^{2}\\ &=72 \mathrm{m} / \mathrm{s}^{2} \\ V_{\max } &=U+a t\\ &=0+72 \times \frac{1}{60}=1.20 \mathrm{m} / \mathrm{s} \end{aligned}
There are 7 questions to complete.
In question number 7 has some correction in question follower moves by distance of 20mm and solution has 10mm.