Question 1 |

The torque provided by an engine is given by T(\theta ) = 12000 + 2500 sin(2\theta ) N.m, where \theta is the angle turned by the crank from inner dead center. The mean speed of the engine is 200 rpm and it drives a machine that provides a constant resisting torque. If variation of the speed from the mean speed is not to exceed \pm 0.5%, the minimum mass moment of inertia of the flywheel should be _______ kg.m^2 (round off to the nearest integer).

245 | |

360 | |

570 | |

640 |

Question 1 Explanation:

\begin{aligned} \omega &=\frac{\pi \times 200}{30}=20.9439 \mathrm{rad} / \mathrm{s} \\ \Delta E &=\int_{0}^{\frac{\pi}{2}}\left(T-T_{\text {mean }}\right) d \theta=2500 \int_{0}^{\frac{\pi}{2}} \sin 2 \theta d \theta \\ &=2500 \times 1=2500 \mathrm{~J} \\ \Delta E &=I \omega^{2} C_{s} \\ 2500 &=I \times 20.9439^{2} \times 0.01 \\ I &=569.934 \mathrm{kgm}^{2} \simeq 570 \mathrm{~kg} . \mathrm{m}^{2} \end{aligned}

Question 2 |

The wheels and axle system lying on a rough surface is shown in the figure.

Each wheel has diameter 0.8 m and mass 1 kg. Assume that the mass of the wheel is concentrated at rim and neglect the mass of the spokes. The diameter of axle is 0.2 m and its mass is 1.5 kg. Neglect the moment of inertia of the axle and assume g=9.8 m/s^2. An effort of 10 N is applied on the axle in the horizontal direction shown at mid span of the axle. Assume that the wheels move on a horizontal surface without slip. The acceleration of the wheel axle system in horizontal direction is ________m/s^2(round off to one decimal place).

Each wheel has diameter 0.8 m and mass 1 kg. Assume that the mass of the wheel is concentrated at rim and neglect the mass of the spokes. The diameter of axle is 0.2 m and its mass is 1.5 kg. Neglect the moment of inertia of the axle and assume g=9.8 m/s^2. An effort of 10 N is applied on the axle in the horizontal direction shown at mid span of the axle. Assume that the wheels move on a horizontal surface without slip. The acceleration of the wheel axle system in horizontal direction is ________m/s^2(round off to one decimal place).

2.8 | |

3.6 | |

5 | |

6.4 |

Question 2 Explanation:

\begin{aligned} I_G&=2 \times (M \times r^2) \\&=2 \times 1 \times 0.4^2=0.32\; kg/m^2 \\ \Sigma F&=m\cdot a\\ &\Rightarrow 10-f=3.5 \times a\;...(i)\\ \Sigma T&=I\cdot \alpha \\ &\Rightarrow 10 \times 0.1 -f\times 0.4=0.32 \times \frac{a}{0.4} \;...(ii)\\ \end{aligned}

[As there is no slip \therefore \;\;a=r\alpha ]

Solving (i) and (ii),

\therefore \;\;a=5\; m/s^2

Question 3 |

Consider the system shown in the figure. A rope goes over a pulley. A mass, m, is hanging from the rope. A spring of stiffness, k, is attached at one end of the rope. Assume rope is inextensible, massless and there is no slip between pulley and rope.

The pulley radius is r and its mass moment of inertia is J. Assume that the mass is vibrating harmonically about its static equilibrium position. The natural frequency of the system is

The pulley radius is r and its mass moment of inertia is J. Assume that the mass is vibrating harmonically about its static equilibrium position. The natural frequency of the system is

\sqrt{\frac{kr^2}{J-mr^2}} | |

\sqrt{\frac{kr^2}{J+mr^2}} | |

\sqrt{\frac{k}{m}} | |

\sqrt{\frac{kr^2}{J}} |

Question 3 Explanation:

\begin{aligned} E &=\frac{1}{2} m \dot{x}^{2}+\frac{1}{2} k x^{2}+\frac{1}{2} I \omega^{2} \\ &=\frac{1}{2} m r^{2} \dot{\theta}^{2}+\frac{1}{2} k r^{2} \theta^{2}+\frac{1}{2} J \dot{\theta}^{2} \\ &=\frac{1}{2}\left[\left(J+m r^{2}\right) \dot{\theta}^{2}+k r^{2} \theta^{2}\right]=0 \\ \frac{d E}{d t} &=0 \\ \left(J+m r^{2}\right) \times 2 \dot{\theta} \ddot{\theta}+k r^{2} 2 \theta \dot{\theta} &=0 \\ \left(\ddot{\theta}+\frac{k r^{2}}{J+m r^{2}}\right) \theta &=0 \\ \omega_{n} &=\sqrt{\frac{k r^{2}}{J+m r^{2}}} \end{aligned}

Question 4 |

A machine of mass 100 kg is subjected to an external harmonic force with a frequency of 40 rad/s. The designer decides to mount the machine on an isolator to reduce the force transmitted to the foundation. The isolator can be considered as a combination of stiffness (K) and damper (damping factor, \xi) in parallel. The designer has the following four isolators:

1) K = 640 kN/m, \xi = 0.70

2) K = 640 kN/m, \xi = 0.07

3) K = 22.5 kN/m, \xi = 0.70

4) K = 22.5 kN/m, \xi = 0.07

Arrange the isolators in the ascending order of the force transmitted to the foundation.

1) K = 640 kN/m, \xi = 0.70

2) K = 640 kN/m, \xi = 0.07

3) K = 22.5 kN/m, \xi = 0.70

4) K = 22.5 kN/m, \xi = 0.07

Arrange the isolators in the ascending order of the force transmitted to the foundation.

1-3-4-2 | |

1-3-2-4 | |

4-3-1-2 | |

3-1-2-4 |

Question 4 Explanation:

Transmitted force problem:

\begin{aligned} \omega&=40 \mathrm{rad} / \mathrm{s} \text { (force frequency) }\\ m&=100 \mathrm{~kg} \\ \epsilon&=\frac{\sqrt{1+\left\{\frac{2 \xi \omega}{\omega_{n}}\right\}^{2}}}{\sqrt{\left\{1-\left(\frac{\omega}{\omega_{n}}\right)^{2}+\left\{\frac{2 \xi \omega}{\omega_{n}}\right\}^{2}\right\}}} \\ \omega_{n}&=\sqrt{\frac{640 \times 10^{3}}{100}}=80 \Rightarrow \frac{\omega}{\omega_{n}}=0.5\\ \omega_{n}&=\sqrt{\frac{22.5 \times 10^{3}}{100}}=47.434164 \Rightarrow \frac{\omega}{\omega_{n}}=\frac{40}{15}=2.666 \end{aligned}

\begin{aligned} \omega&=40 \mathrm{rad} / \mathrm{s} \text { (force frequency) }\\ m&=100 \mathrm{~kg} \\ \epsilon&=\frac{\sqrt{1+\left\{\frac{2 \xi \omega}{\omega_{n}}\right\}^{2}}}{\sqrt{\left\{1-\left(\frac{\omega}{\omega_{n}}\right)^{2}+\left\{\frac{2 \xi \omega}{\omega_{n}}\right\}^{2}\right\}}} \\ \omega_{n}&=\sqrt{\frac{640 \times 10^{3}}{100}}=80 \Rightarrow \frac{\omega}{\omega_{n}}=0.5\\ \omega_{n}&=\sqrt{\frac{22.5 \times 10^{3}}{100}}=47.434164 \Rightarrow \frac{\omega}{\omega_{n}}=\frac{40}{15}=2.666 \end{aligned}

Question 5 |

A power transmission mechanism consists of a belt drive and a gear train as shown in the figure.

Diameters of pulleys of belt drive and number of teeth (T) on the gears 2 to 7 are indicated in the figure. The speed and direction of rotation of gear 7, respectively, are

Diameters of pulleys of belt drive and number of teeth (T) on the gears 2 to 7 are indicated in the figure. The speed and direction of rotation of gear 7, respectively, are

255.68 rpm; clockwise | |

255.68 rpm; anticlockwise | |

575.28 rpm; clockwise | |

575.28 rpm; anticlockwise |

Question 5 Explanation:

\begin{aligned} T_{3} &=44 \\ T_{6} &=36 \\ T_{2} &=18 \\ T_{4} &=15 \\ \frac{N_{1}}{N_{0}} &=\frac{d_{0}}{d_{1}} \\ \Rightarrow\qquad N_{1} &=\frac{150}{250} \times 2500\\ N_{1}&=15,00=N_{2} \\ N_{3}&=\frac{N_{2} \times T_{2}}{T_{3}}=\frac{1500 \times 18}{44}=613.63636=N_{4} \\ N_{6}&=\frac{N_{4} \times T_{4}}{T_{6}}=\frac{613.63636 \times 15}{36} \\ &=255.6818=N_{7}(\text { Clockwise }) \end{aligned}

Gear 5 is idler.

Gear 5 is idler.

Question 6 |

The controlling force curves P, Q and R for a spring controlled governor are shown in the figure, where r_1 and r_2 are any two radii of rotation.

The characteristics shown by the curves are

The characteristics shown by the curves are

P - Unstable; Q - Stable; R - Isochronous | |

P - Unstable; Q - Isochronous; R - Stable | |

P- Stable; Q - Isochronous; R - Unstable | |

P - Stable; Q - Unstable; R - Isochronous |

Question 6 Explanation:

F(r)|_P=ar+b \quad \rightarrow \text{Unstable}

F(r)|_Q=ar+b \quad \rightarrow \text{Isochronous}

F(r)|_R=ar-b \quad \rightarrow \text{Stable}

F(r)|_Q=ar+b \quad \rightarrow \text{Isochronous}

F(r)|_R=ar-b \quad \rightarrow \text{Stable}

Question 7 |

Consider the mechanism shown in the figure. There is rolling contact without slip between the disc and ground.

Select the correct statement about instantaneous centers in the mechanism.

Select the correct statement about instantaneous centers in the mechanism.

Only points P, Q, and S are instantaneous centers of mechanism | |

Only points P, Q, S and T are instantaneous centers of mechanism | |

Only points P, Q, R, S, and U are instantaneous centers of mechanism | |

All points P, Q, R, S, T and U are instantaneous centers of mechanism |

Question 7 Explanation:

Points P, Q, R, S, T and U are instantaneous centers of mechanism.

Question 8 |

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).

The quick return ratio for the mechanism is ________(round off to one decimal place).

1 | |

2 | |

2.5 | |

3.5 |

Question 8 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 9 |

A tappet valve mechanism in an IC engine comprises a rocker arm ABC that is hinged at B as shown in the figure. The rocker is assumed rigid and it oscillates about the hinge B. The mass moment of inertia of the rocker about B is 10^{-4}\;kg.m^2. The rocker arm dimensions are a = 3.5 cm and b = 2.5 cm. A pushrod pushes the rocker at location A, when moved vertically by a cam that rotates at N rpm. The pushrod is assumed massless and has a stiffness of 15 N/mm. At the other end C, the rocker pushes a valve against a spring of stiffness 10 N/mm. The valve is assumed massless and rigid.

Resonance in the rocker system occurs when the cam shaft runs at a speed of ______rpm (round off to the nearest integer).

Resonance in the rocker system occurs when the cam shaft runs at a speed of ______rpm (round off to the nearest integer).

496 | |

4739 | |

790 | |

2369 |

Question 9 Explanation:

I=10^{-4} \mathrm{~kg}-\mathrm{m}^{2}

By D'Alembert Principle

I \ddot{\theta}+\left[10000 \times(0.025)^{2}+15000 \times(0.035)^{2}\right] \theta=0

\begin{aligned} \left(10^{-4}\right) \ddot{\theta}+(24.625) \theta&=0 \\ \ddot{\theta}+\left(\frac{24.625}{10^{-4}}\right) \theta&=0\\ \Rightarrow \qquad\qquad \omega_{n}^{2}&=(246250) \\ \omega_{n}&= 496.2358\; \text{rad/s} \\ \Rightarrow \qquad\qquad N_C&=\frac{496.2358 \times 60}{2 \pi} \\ &=4738.70 \text{ rpm}\end{aligned}

Question 10 |

Consider a two degree of freedom system as shown in the figure, where PQ
is a rigid uniform rod of length, b and mass, m.

Assume that the spring deflects only horizontally and force F is applied horizontally at Q. For this system, the Lagrangian, L is

Assume that the spring deflects only horizontally and force F is applied horizontally at Q. For this system, the Lagrangian, L is

\frac{1}{2}\left ( M+m \right )\dot x^{2}+\frac{1}{6}mb^{2}\dot\theta ^{2}-\frac{1}{2}kx^{2}+mg\frac{b}{2}\cos\theta | |

\frac{1}{2}\left ( M+m \right )\dot x^{2}+\frac{1}{2}mb\dot{\theta }\dot{x}\cos\theta +\frac{1}{6}mb^{2}\dot\theta ^{2}-\frac{1}{2}kx^{2}+mg\frac{b}{2}\cos\theta | |

\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}mb\dot{\theta }\dot{x}\cos\theta +\frac{1}{6}mb^{2}\dot \theta ^{2}-\frac{1}{2}kx^{2} | |

\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}mb\dot{\theta }\dot{x}\cos\theta +\frac{1}{6}mb^{2}\dot \theta ^{2}-\frac{1}{2}kx^{2}+mg\frac{b}{2}\cos\theta +Fb\sin\theta |

Question 10 Explanation:

\begin{aligned} \text { For mass, } T&=\frac{1}{2} M \dot{x}^{2}\\ V&=\frac{1}{2} K x^{2} \end{aligned}

For mass, m

\begin{aligned} d m &=\frac{m}{b} d y \\ \text { Displacement } &=x+y \sin \theta \\ \text { Velocity } &=\dot{x}+y \cos \theta \dot{\theta} \\ d T & =\frac{1}{2} d M V e l^{2}=\frac{1}{2} d m\left[\dot{x}^{2}+y^{2} \dot{\theta}^{2} \cos ^{2} \theta+2 \dot{x} \dot{\theta} y \cos \theta\right] \\ T &=d T=\frac{1}{2} \frac{m}{b} \int\left(\dot{x}^{2}+y^{2} \dot{\theta}^{2} \times 1+2 \dot{x} \dot{\theta} y \cos \theta\right) d y \\ T &=\frac{1}{2} \frac{m}{b}\left(\dot{x}^{2} b+\frac{b^{3}}{3} \dot{\theta}^{2}+2 \dot{x} \dot{\theta} \cos \theta \frac{b^{2}}{2}\right) \\ T &=\frac{1}{2} \frac{m}{b}\left(\dot{x}^{2} b+\frac{b^{3}}{3} \dot{\theta}^{2}+\dot{x} \dot{\theta} b^{2} \cos \theta\right) \\ T &=\frac{1}{2} m \dot{x}^{2}+\frac{m b^{2}}{6} \dot{\theta}^{2}+\frac{m}{2} \dot{x} \dot{\theta} b \cos \theta \\ V &=-m g \frac{b}{2} \cos \theta \end{aligned} For both masses, M and m

\begin{aligned} T &=\frac{1}{2} M \dot{x}^{2}+\frac{1}{2} m \dot{x}^{2}+\frac{m b^{2}}{6} \dot{\theta}^{2}+\frac{m}{2} b \dot{\theta} \dot{x}^{2} \cos \theta \\ V &=\frac{1}{2} k x^{2}-m g \frac{b}{2} \cos \theta \\ L &=T-V \\ &=\frac{1}{2}(M+m) \dot{x}^{2}+\frac{1}{2} m b \dot{x} \dot{c} \cos \theta+\frac{m b^{2} \dot{\theta}^{2}}{6}-\frac{1}{2} k x^{2}+m g \frac{b}{2} \cos \theta \end{aligned}

There are 10 questions to complete.

figures of questions are missing,and in some question statement is given wrong..

Thank you prakash for your suggestions.

Can you please share the questions numbers for the above suggestions.

question 8

We have review the question 8 with original published paper by IIT. There is only one figure which is there in our question -8. It might possible due to slow internet image takes time to load.

Thank you for your support.

Please check figures of Q28 &29. They are wrong.

what is wrong with q28 and q29?can you say that?

Dear Franklin,

q28 and q29 are correct.

Answer to Q 184 is wrong, it should be option B- 3.

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