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 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 2 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 3 |
The turning moment diagram of a flywheel fitted to a fictitious engine is shown in the
figure.
The mean turning moment is 2000 Nm. The average engine speed is 1000 rpm. For fluctuation in the speed to be within\pm 2 \% of the average speed, the mass moment of inertia of the flywheel is _________ kgm^2.
The mean turning moment is 2000 Nm. The average engine speed is 1000 rpm. For fluctuation in the speed to be within\pm 2 \% of the average speed, the mass moment of inertia of the flywheel is _________ kgm^2.
8.25 | |
5.56 | |
2.14 | |
3.58 |
Question 3 Explanation:
N = 1000 rpm
\begin{aligned} \omega &=\frac{2 \pi \times 1000}{60}=104.7195 \mathrm{rad} / \mathrm{s} \\ \Delta E_{\max } &=\left(\pi-\frac{\pi}{2}\right) \cdot(3000-2000)=\left(\frac{\pi}{2} \times 1000\right) \mathrm{J} \\ C_{s} &=\pm 2 \%=4 \%=0.04 \\ \Delta E_{\max } &=I \cdot \omega^{2} \cdot C_{s} \\ \frac{\pi}{2} \times 1000 &=I \times(104.7195)^{2} \times 0.04 \\ I &=\frac{1570.795}{(104.7195)^{2} \times 0.04}=3.58 \mathrm{kg}-\mathrm{m}^{2} \end{aligned}
\begin{aligned} \omega &=\frac{2 \pi \times 1000}{60}=104.7195 \mathrm{rad} / \mathrm{s} \\ \Delta E_{\max } &=\left(\pi-\frac{\pi}{2}\right) \cdot(3000-2000)=\left(\frac{\pi}{2} \times 1000\right) \mathrm{J} \\ C_{s} &=\pm 2 \%=4 \%=0.04 \\ \Delta E_{\max } &=I \cdot \omega^{2} \cdot C_{s} \\ \frac{\pi}{2} \times 1000 &=I \times(104.7195)^{2} \times 0.04 \\ I &=\frac{1570.795}{(104.7195)^{2} \times 0.04}=3.58 \mathrm{kg}-\mathrm{m}^{2} \end{aligned}
Question 4 |
A flywheel is attached to an engine to keep its rotational speed between 100 rad/s and
110 rad/s. If the energy fluctuation in the flywheel between these two speeds is 1.05 kJ
then the moment of inertia of the flywheel is _________________kg.m^2 (round off to 2
decimal places).
1 | |
2 | |
2.5 | |
4 |
Question 4 Explanation:
\begin{aligned} \Delta E_{\max } &=\frac{1}{2} I\left(\omega_{\max }^{2}-\omega_{\min }^{2}\right) \\ \therefore \qquad 1.05 \times 10^{3} \times 2 &=I\left(110^{2}-100^{2}\right) \\ I &=1 \mathrm{kg} . \mathrm{m}^{2} \end{aligned}
Question 5 |
A thin uniform rigid bar of length L and mass M is hinged at point O, located at a distance of \frac{L}{3} from one of its ends. The bar is further supported using springs, each of stiffness k, located at the two ends. A particle of mass m=\frac{M}{4} is fixed at one end of the bar, as shown in the figure. For small rotations of the bar about O, the natural frequency of the systems is
\sqrt{\frac{5k}{M}} | |
\sqrt{\frac{5k}{2M}} | |
\sqrt{\frac{3k}{2M}} | |
\sqrt{\frac{3k}{M}} |
Question 5 Explanation:
Taking Mass moment of inertia about point 0,
\begin{aligned} I &=\frac{M l^{2}}{12}+M\left(\frac{l}{2}-\frac{l}{3}\right)^{2}+m \times\left(\frac{2 l}{3}\right)^{2} \\ &=\frac{M l^{2}}{12}+\frac{M l^{2}}{36}+\frac{4 M l^{2}}{9} \\ &=\frac{M l^{2}}{9}+\frac{4 M l^{2}}{4 \times 9}=\frac{2 M l^{2}}{9} \end{aligned}
Now we will balance torque about 0
\begin{aligned} I \alpha &=k \times \frac{2 L}{3} \times\left(\frac{2 L}{3} \theta\right)+k \times \frac{L}{3} \times\left(\frac{L}{3} \theta\right) \\ \frac{2 M l^{2}}{9} \frac{d^{2} \theta}{d t^{2}} &=\frac{5 k}{2 M}=\omega_{n}^{2} \theta \\ \therefore \quad \omega_{n}&=\sqrt{\frac{5 k}{2 M}} \end{aligned}
\begin{aligned} I &=\frac{M l^{2}}{12}+M\left(\frac{l}{2}-\frac{l}{3}\right)^{2}+m \times\left(\frac{2 l}{3}\right)^{2} \\ &=\frac{M l^{2}}{12}+\frac{M l^{2}}{36}+\frac{4 M l^{2}}{9} \\ &=\frac{M l^{2}}{9}+\frac{4 M l^{2}}{4 \times 9}=\frac{2 M l^{2}}{9} \end{aligned}
Now we will balance torque about 0
\begin{aligned} I \alpha &=k \times \frac{2 L}{3} \times\left(\frac{2 L}{3} \theta\right)+k \times \frac{L}{3} \times\left(\frac{L}{3} \theta\right) \\ \frac{2 M l^{2}}{9} \frac{d^{2} \theta}{d t^{2}} &=\frac{5 k}{2 M}=\omega_{n}^{2} \theta \\ \therefore \quad \omega_{n}&=\sqrt{\frac{5 k}{2 M}} \end{aligned}
There are 5 questions to complete.
Sir,2nd and 5th questions not from the flywheel portion
Sir, question 8, either Cs in question is wrong or Cs in solution is wrong..plz correct it