T207: Engineering, Mechanics, Materials And Design Dissertation Hypothesis Examples

Type of paper: Dissertation Hypothesis

Topic: Collision, Conservation, Speed, Acceleration, Aircraft, Force, Earth, Gear

Pages: 5

Words: 1375

Published: 2020/09/09


1(a) The liner reciprocating motion of the piston is converted into rotary motion of the
crank shaft by a connecting rod, which connects the piston rod to the crank. As can be seen above, because of the nature of the linkage, at the start the crank and the connecting rod remain in a straight line. As the piston moves downwards during the power stroke, point b starts to move clockwise. This is obvious because the piston coming down means the connecting rod has to accommodate the travel of the piston. Finally, when the piston reaches the extreme end, the connecting rod has rotated 180o and will be in the position B as above.
This is how linear motion is converted into rotary motion in an I.C. Engine. In actual practice, there is not one but a number of cylinders which all operate at different timings so that there is continuous feed of power to the crankshaft which it requires to rotate at a particular speed. The special shape of the crankshaft is to facilitate the conversion of linear into rotary motion. For change in speed, the clutch is used to allow the gear on the crank to mesh with different gears having different numbers of teeth. This allows the gearbox shaft to rotate at the speed desired. To change the direction ( either ‘ clockwise to counter clockwise’ or vice versa), the gear meshing takes place through an idler gear in between, which reverses the direction of rotation.

Illustration from Block 3

Dynamics: Getting Moving
(E text-book)
1(b) Crankshaft speed at idle = 5000 RPM
Converting to radians, u = (5000*2π)/60 =52.33 rad/sec

Once shut off-the speed falls, acceleration becomes -30 rad/s-2

Therefore, speed after 4.0 s , or (v-52.33)/4= (-) 30 or, v= -67.67 rad/sec

As this does not make sense, the crankshaft comes to a halt before 4 seconds has passed.

Time required by crankshaft to rotate at zero speed =
Velocity diff/acceleration= 52.33/30=1.74seconds approximately
Therefore crankshaft will come to a halt at about 1.74 seconds from shut-off.

Angular velocity of crankshaft after 4 seconds will be zero.

1c. We assume that the acceleration of the airliner is constant and is a
Now if F=mxa, then m = 200 x 103 Kg
Now, 1 KN= 103N and 200 T= 200,000 Kg
(Force) Thrust= Mass x acceleration
[ 300 KN = 300 *103 N]
Therefore acceleration = Thrust/ Mass =1.5 m/s
We will use the formula v2= u2 +2aS, where S is the displacement
Now v=75 m/s,u=0, a=1.5 m/s-2
2*1.5*S= 752
S=1875 mts.
Therefore, the aircraft will need a runway length of 1.875 Kms.

In actual practice, things will not happen this way.

Acceleration will not be constant because of wind resistance against the aircraft flaps will slow down the acceleration.
The lifting up thrust of an airplane will depend upon air density. At places where altitude is high, air is less dense. Therefore the plane will need to taxi some more distance before it can take-off.

These are some differences between the assumed model and what will actually happen.

1d) 2501000
The data given to us includes OD and ID of disc as shown in
the adjacent figure
The MI of the disc I= ½ *M*(0.5-0.125)2
I= 0.5*150*0.3752 = 10.547 Kg-m2 -------------------------(1)
Speed of disc =10,000 RPM
Converting into radians , ω = [10000/60]*2π = 333.33π rad/s ------------ (2).From derivations (1) and (2) and KE formula
Therefore, Kinetic energy (angular)= ½ Iω2 =0.5*10.547*333.33*(22/7) =5524.56 J
Or, KE of the disc = 5.524KJ.
Answer 1e) Collisions can be of two types-Inelastic Collision and Elastic Collision. If the sum of the Kinetic Energy of the objects before collision remains the same post-collision too, then this type of collision is called elastic collision. However, if the Kinetic energy sum of the objects differ pre and post collision, then the collision is an inelastic collision. In real life, elastic collisions too suffer small energy loss due to conversion to Sound and Heat energy. An important point is that irrespective of collision type, momentum is conserved in all collisions though Kinetic energy may or may not be.
(i)Kinetic Energy is conserved in case of two pool balls striking each other and then bouncing off, taking new positions on the pool board.
(ii) A collision between a very heavy truck and a car both from opposite directions. Due to the impact, the car gets stuck on the front of the truck and this combined system starts moving in the original direction of the truck.
(iii) The short kinetic energy is converted into sound energy and heat energy due to the collision. (The kinetic energy of the truck with car attached to its front will be less than the sum of the individual kinetic energies of the car and the truck).
2(a) Sphere Volume can be calculated by ; Vol(V) = (4/3)*π * (D/2)3
Given D = 1000 mt
Hence V=(4/3)*(22/7)* (103/2)3 =( 1.33*3.142*109)/8
Or, V= 0.5223 x 109 Mts/sec
Asteroid density = 2500 Kg/m3
Therefore, Mass = Vol x density
Or, M = 1305.89 x 109 Kgs.

Hence, Kinetic energy of the asteroid

=1/2* 1305.89 x 109 x (2 x 104)2
=2611.78 J x 1017
Or, Asteroid KE = 2.61 x 1020

2(b) In the case of Bruce Willis,

Earth’s relative velocity is the same i.e. 20000 m/s
Earth Mass = 6 x 1024 kg
Therefore, Earth’s KE = (1/2) * 6 x 1024 * (2 x 104)2
= 3 x 1024 x4 x 108=12 x 1032
Or, KE earth = 1.2 x 1033 J
3c. We will use the conservation of momentum here.

Let us assume that the suffix 0 denotes pre-collision state and 1 denotes the post-collision state,

And m denote the asteroid Mass, M denote earth mass and Mx denote the combined mass post collision, then the conservation of momentum says that pre and post collision momentums will be equal.

Now since the collision is inelastic, post collision both bodies will move as one body.

Pre-collision state:
Momentum of the asteroid= movo
Or, m = (1305.89 x 109) x (2 x 104) Kg-m (values taken from 2(a) above)
Momentum of Asteroid = 2611.78 x 1013 =2.61x 1016 Kg-m
Momentum of Earth =Movo
= 6 x 1024 x 2 x 104= 12 x 1028 Kg-m

Post Collision Mass

Mx =[ (6 x 10 ^24) + ( 1305.89 x109)] kg-m

Equating the pre and post collision momentum

Vx = [2.61x 1016 + 12 x 1028] / [ (6 x 10 ^24) + ( 1.3 x1012)]= approx. 20,000 mts/sec
The Kinetic Energy post collision can also be found out which will be lower than the total Kinetic energies as found out in 2(a) and 2(b).

The difference will be the energy spent during the collision.

(the best way to find this out will be to ignore the relatively smaller values. For example 10^12 is by itself a big number but if one has to add 10^12 and 10^28, the former can be safely ignored)
= 6 x 1024 x 2 x 104
Momentum is conserved in all collisions. However, we can examine collisions under two titles if we consider conservation of energy. For example, if the objects collide and momentum and kinetic energy of the objects are conserved than we call this collision “elastic collision”. On the other hand if the momentum of the object is conserved but kinetic energy is not conserved than we call this type of collision “inelastic collision”. In an in-elastic collisions some of the kinetic energy is converted to heat and sound.
2 d. We have to consider the entire Kinetic Energy of the asteroid as the explosive energy as it is now in the pre-collision state. In reality, the impact will not be elastic and cannot be termed as inelastic either. To simplify matters, we will assume that it is an inelastic collision.
Now, 1.0 Kt = 4.2 x 1012 J
Hence, 1 Mt = 4.2 x 109 J
Therefore, asteroid explosive Energy = (2.61 x 1020 ) / (4.2 x 109)
=0.621 x 1011
Or, Explosive energy in Mt = 6.21 x 1010 J.
2e.> Now, F=Mass x acceleration
Or. 103 = 1305.89 x 109 x a = 7.67 x 105 x a
Or, a(acceleration) = 7.67 x 10-10 mts
Now, S=u t + ½ a t2
Since initial lateral velocity is zero or U=0, we have
S = 127 x 106 = 0.5 * 7.67 x 10-10. * t2
Or, t^2 = 33.11 x 10 16
Or, t = 5.75 x 108 seconds
(somewhere in between 19 and 20 years)

Question 3 :

A permanent magnet DC motor is required to emit a torque of 5 N-m at 35 RPM
Hence , Power = 5*35*2π/60 = 18.3 W.
Power Supply = 12V
Stall Current = 15 Amps
Therefore, Power at stall = 12*15 =180 Watts
Rotor friction losses =V/I= 12/15= 0.8 Ohms
Back Emf at no load = 12 V- (voltage drop across rotor)
Now, Voltage drop across rotor =0.8*0.5=0.4V
Hence back emf at no load = 12-0.4 =11.6 V.
Hence, we see that a power output of 180 watts is available and 18.3 Watts is all that is required.
We know that generally for this type of DC Motors, the center is the most efficient point i.e. the

Center between stalling speed and say zero RPM.

Let’s consider 3000 RPM
We will follow this thumb rule and find the gear ratio for 3000 RPM
Gear Train ratio = 3000/35=85.71 0r, 1: 86. This will probably not be feasible in a single stage.

A two stage gear system is recommended therefore.

Question 4 Solution)
As the aircraft is flying, there will be downward force acting on it, which is gravity.
If FG is the gravity force then FG is the Weight of the aircraft acting downwards.
The engines of the aircraft convert chemical energy of the fuel to mechanical energy which propels the aircraft forwards. This is called Thrust.
Again as per the third law, a force acts on the aircraft ( air friction etc, acting opposite to the propulsion force. This is called the Drag force.

These are the four forces which affect the stability and smooth flying of an aircraft in flight.

4b In actual practice , we are only assuming that the horizontal forces act along a single point. In actual practice, this can never be so. What happens that instead of a force, an unbalanced moment acts on the aircraft at a point. This is opposed by a balancing moment which is aided by the tail plane in creating a opposite yawing moment. In case of lift and weight, the weight which always acts at right angles to the earth’s surface creates a compensating moment.

Illustration from Block 3; Parts 3 and 4

Balanced Moment Equation
Fw * L + Fl * H =0 (for the weight vs lift moments).
In the Concorde, approximately 95 tons of fuel is kept in fuel tanks at the front and rear and across the fuselage. Therefore, when the center of mass shifts, the fuel fills up a different part of the tank and nullifies the moment to be zero.


Anglia Ruskin University, 2013, “Guide to the Harvard Style of Referencing (Fifth Edition)”. Accessed online at http://libweb.anglia.ac.uk/referencing/harvard.htm
Block 3, the Open University, 2009. Dynamics Getting Moving- Parts 3 and 4, 3rd(Ed), Chennai: Alden Preset Services
Block 3, The Open University, 2009. Dynamics Getting Moving-Parts 1 and 2, 3rd(Ed), Chennai : Alden Preset Services
Karad, A.A. et. al. 2009, Basic Mechanical Engineering, Pune: Technical Publications.

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WePapers. (2020, September, 09) T207: Engineering, Mechanics, Materials And Design Dissertation Hypothesis Examples. Retrieved November 29, 2022, from https://www.wepapers.com/samples/t207-engineering-mechanics-materials-and-design-dissertation-hypothesis-examples/
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"T207: Engineering, Mechanics, Materials And Design Dissertation Hypothesis Examples." WePapers, Sep 09, 2020. Accessed November 29, 2022. https://www.wepapers.com/samples/t207-engineering-mechanics-materials-and-design-dissertation-hypothesis-examples/
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T207: Engineering, Mechanics, Materials And Design Dissertation Hypothesis Examples. Free Essay Examples - WePapers.com. https://www.wepapers.com/samples/t207-engineering-mechanics-materials-and-design-dissertation-hypothesis-examples/. Published Sep 09, 2020. Accessed November 29, 2022.

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