Question 1 |
A rigid body in the X-Y plane consists of two point
masses (1 kg each) attached to the ends of two
massless rods, each of 1 cm length, as shown in the
figure. It rotates at 30 RPM counter-clockwise about
the Z-axis passing through point O. A point mass of
\sqrt{2} kg, attached to one end of a third massless rod,
is used for balancing the body by attaching the free
end of the rod to point O. The length of the third rod
is ________ cm.
1 | |
\sqrt{2} | |
\frac{1}{\sqrt{2}} | |
\frac{1}{2\sqrt{2}} |
Question 1 Explanation:
m_1=1kg, m_2=1kg, r_1=1cm, r_2=1cm
Balancing mass, m_b=\sqrt{2}kg
Making force polygon for complete balance
From the right angle triangle,
\begin{aligned} (m_br_b)^2&=(m_1r_1)^2+(m_2r_2)^2 \\ (\sqrt{2}r_b)^2&=(1\times 1)^2+(1\times 1)^2 \\ 2r_b^2&=2 \\ r_b&=1 \end{aligned}
Question 2 |
Consider a reciprocating engine with crank radius R and connecting rod of length L. The secondary unbalance force for this case is equivalent to primary unbalance force due to a virtual crank of _____
radius \frac{L^2}{4R} rotating at half the engine speed | |
radius \frac{R}{4} rotating at half the engine speed | |
radius \frac{R^2}{4L} rotating at twice the engine speed | |
radius \frac{L}{2} rotating at twice the engine speed |
Question 2 Explanation:
Unbalanced secondary force,
F_s=mr\omega \frac{\cos 2\theta }{n}=m \times \left ( \frac{R}{4n} \right )\times (2\omega )^2 \times \cos 2\theta
Balancing radius =\frac{R}{4n}=\frac{R}{4 \times \frac{L }{R}}=\frac{R^2}{4L }
Balancing crank speed =2\omega
F_s=mr\omega \frac{\cos 2\theta }{n}=m \times \left ( \frac{R}{4n} \right )\times (2\omega )^2 \times \cos 2\theta
Balancing radius =\frac{R}{4n}=\frac{R}{4 \times \frac{L }{R}}=\frac{R^2}{4L }
Balancing crank speed =2\omega
Question 3 |
Two masses A and B having mass m_a \; and \; m_b, respectively, lying in the plane of the figure shown, are rigidly attached to a shaft which revolves about an axis through O perpendicular to the plane of the figure. The radii of rotation of the masses m_a \; and \; m_b are r_a \; and \; r_b, respectively. The angle between lines OA and OB is 90^{\circ}. If m_a =10 kg, m_b=20kg, r_a=200mm, r_b=400mm, then the balance mass to be placed at a radius of 200 mm is ______ kg (round off to two decimal places).
21.25 | |
28.56 | |
35.55 | |
41.23 |
Question 3 Explanation:
\begin{array}{l} \mathrm{m}_{\mathrm{A}}=10 \mathrm{kg} \\ \mathrm{m}_{\mathrm{B}}=20 \mathrm{kg} \\ \mathrm{r}_{\mathrm{A}}=200 \mathrm{mm} \\ \mathrm{r}_{\mathrm{B}}=400 \mathrm{mm} \\ \mathrm{r}=200 \mathrm{mm} \\ \mathrm{mr} \omega^{2}=\sqrt{\left(\mathrm{m}_{\mathrm{A}} \mathrm{r}_{\mathrm{A}} \omega^{2}\right)^{2}+\left(\mathrm{m}_{\mathrm{B}} \mathrm{r}_{\mathrm{B}} \omega^{2}\right)^{2}} \\ \Rightarrow \mathrm{m} \times 200=\sqrt{(10 \times 200)^{2}+(20 \times 400)^{2}} \\ \Rightarrow \mathrm{m}=41.23 \mathrm{kg} \end{array}
Question 4 |
Three masses are connected to a rotating shaft supported on bearings A and B as shown in the figure. The system is in a space where the gravitational effect is absent. Neglect the mass of shaft and rods connecting the masses.for m_{1}=10 kg,m_{2}=5 kg and m_{3}=2.5 kg and for a shaft angular speed of 1000 radian/s, the magnitude of the bearing reaction (in N) at location B is_____
1 | |
2 | |
3 | |
0 |
Question 4 Explanation:
It means all three masses are in same plane
let us calculates the net force
\begin{array}{c} \Sigma F_{x}= \Sigma \mathrm{m} \mathrm{r} \omega^{2} \cos \theta \\ =\left(10 \times 0.1 \times \omega^{2}\right)-\left(5 \times 0.2 \times 6^{2} \cos 60^{\circ}\right) \\ -\left(2.5 \times 0.4 \times \omega^{2} \cos 60^{\circ}\right) \\ =\left[1-\left(5 \times 0.2 \times \frac{1}{2}\right)-\left(2.5 \times 0.4 \times \frac{1}{2}\right)\right] \omega^{2}=0 \\ \Sigma F_{y}= \Sigma \mathrm{mr} \omega^{2} \sin \theta \\ =\left(5 \times 0.2 \times \omega^{2} \sin 60^{\circ}\right) \\ -\left(2.5 \times 0.4 \times \omega^{2} \sin 60^{\circ}\right) \\ =\left[\left(1 \times \frac{\sqrt{3}}{2}\right)-\left(1 \times \frac{\sqrt{3}}{2}\right)\right] \omega^{2} \\ =\left[\left(\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\right)\right] \omega^{2}=0 \end{array}
Net force,
=\sqrt{\Sigma F_{x}^{2}+\Sigma F_{y}^{2}}=0
Therefore reaction at B is zero.
R_{B}=0
Question 5 |
Two masses m are attached to opposite sides of a rigid rotating shaft in the vertical plane. Another pair of equal masses m1 is attached to the opposite sides of the shaft in the vertical plane as shown in figure. Consider m = 1 kg, e = 50 mm, e1 = 20 mm, b = 0.3 m, a = 2 m and a1 = 2.5 m. For the system to be dynamically balanced, m1 should be ________ kg.
1kg | |
2kg | |
3kg | |
4kg |
Question 5 Explanation:
Balance moment of all forces
\begin{array}{c} \omega^{2}\left(m_{1} \times 0.02 \times 2.5\right)+(1 \times 0.05 \times 0.3) \omega^{2}=\left(m_{1} \times\right. \\ 0.02 \times 0) \omega^{2}+(1 \times 0.05 \times 2.3) \omega^{2} \\ m_{1}=2 \mathrm{kg} \end{array}
There are 5 questions to complete.