question 1 1.Figure Q1 shows a wheel of 300 mm radius rotating about the fixed pin O. Point A on the...

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question 1 1.Figure Q1 shows a wheel of 300 mm radius rotating about the fixed pin O. Point A on the rim of the wheel is connected to the fixed pin C via rods AB and BC. Find: a) the angular velocities of both rods;(14 marks)b) the angular acceleration of both rods.(6 marks)Figure Q1question 2 2.Figure Q2 shows point A on the centre of the wheel, when point A is moving to the left at 3 m/s and decelerating at 6 m/s2. The mass of the wheel is 10 kg and its radius of gyration about point A is 80 mm. The mass of the piston C and the mass of rod BC are negligible. Point B can be considerd to be on the circumference of the wheel. The wheel does not slip. For the position shown: 	Draw the free-body diagram of the system.   (10 marks) 	Find the value of the force F acting on the piston C.   (10 marks)Figure Q2question 3  	A crank AB is driven by a constant moment of MAB = 3 N.m, Figure Q3. The mass per unit length of each of the bar is 20 kg/m. The system starts from rest at the position shown in the figure. Using the work and energy method, determine:     	the angular velocity of member CD after the crank AB has revolved through an angle ?;  (14 marks)b) the angular velocity of member CD after the crank has completed one revolution.  (6 marks)figure q3question 4 The damped spring-mass oscillator in Figure Q4 has a mas m = 2 kg, spring constant k = 8 N/m, and damping constant c = 1 N.s/m. At time t = 0, the mass is released from rest in the position x = 0.1 m.  	 Determine the position x as a function of time t.  (14 marks) 	Plot a graph of x = x(t) the first ten seconds of motion.   (6 marks)Figure Q4

Robert · Accepted Answer

Answer
1. Angle of Climb is defined as, 
sinγ = (Ta-Tr)/W 
Where Ta is available thrust at 100% RPM and Tr is minimum thrust required and W is the 
weight of the aircraft 
Climb of aircraft is visible when (Ta-Tr) is maximum, so from the graph given above we have, 
Ta (max) = 4200lb and 
Tr (min) = 830lb 
W = 12,000lb so, Angle of climb will be, 
sinγ = (Ta-Tr)/W 
= (4200-830)/12000 =0.2808 
Angle of climb is 
sinγ = (Ta-Tr)/W = (4200-830)/12000 =0.2808 
γ =16.31
0
 (Answer)
2. Rate of climb is defined as , 
Rate of Climb = 101.3*Vk* Sinγ 
Given that, 
At 300 KTAS from the graph we can determine its Ta and Tr 
 Thrust Available is Ta =4200lb while thrust required is Tr =1000lb 
So, Rate of climb is, 
Rate of Climb = 101.3*Vk* Sinγ = 101.3*300*(4200-1000)/12000  
Rate of Climb =8104 ft/min (Answer)
3. Now the question asks about 95% RPM from the graph it is clearly visible that the 
maximum level air speed is , 
So, 
Vmax = 500 KTAS (Answer)
4. Now we need to calculate two different speeds maximum endurance and maximum 
Range speed and this can be calculated from the graph with the use of their definitions, 
 Max endurance airspeed V (BE) occurs when minimum thrust is required so at 
Tr is minimum the speed is  
V (BE) = 240 KTAS (Answer) 
While for the Maximum Range speed we have, 
Maximum range speed is equal to the value where the straight line drawn from the origin 
intersects the graph. (Look for the graph) 
So, 
V (BR) will be 300 KTAS (Answer)
Answer
5.   As done previously Max range airspeed occurs V(BR) occurs when the straight line 
drawn from the origin intersects the graph given so if we give a look to the graph we 
have,

question 1 1. Figure Q1 shows a wheel of 300 mm radius rotating about the fixed pin O. Point A on the rim of the wheel is connected to the fixed pin C via rods AB and BC. Find: a) the angular...

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