Energy Conservation (Ch 18)
All types of energy are measured
in Joules
Potential Energy
PE = mgh
m = mass (kg)
g = gravity accel m/s2
h = height (m)
Kinetic Energy (Linear)
KE = 0.5mv2
m = mass (kg)
v = velocity (m/s)
Work (Linear)
W=FS
F = Force (N)
S = displacement (m)
Kinetic Energy (Rotational)
KE = 0.5I2
I = Mass moment of Inertia (kgm2)
 = velocity (rad/s)
Spring Energy
SE = 0.5kx2
k = Spring constant (N/m)
x = displacement or deformation (m)
Work (Rotation)
W = T
T = Torque (Nm)
 = Angular displacement (rad)
Conservation of Energy.
(Energy at point 1) = (Energy at point 2)
PE1 + KE1 + SE1 +/- W = PE2 + KE2 + SE2
+W = Positive work = Force moved the object (motor)
-W = Negative work = Brakes (the motor is working as a generator).
Friction is always negative.
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Rollercoaster
Tuesday, 17 May 2011
5:33 PM
Example 18.2: Roller Coaster.
Mass = 500kg, starting height A=18m, lowest point B=0m, up to C=13m.
PEA + KEA = PEC + KEC
Straight from A to C
PEA + KEA = PEC + KEC
PEA = PEC + KEC
mgha = mghc + 0.5*m*vc2
500*9.81*18 = 500*9.81*13 + 0.5*500*v2
88290 = 63765 + 250*v2
250*v2 = 88290 - 63765 = 24525 J
v2 = 24525/250 = 98.1
v = 98.1 ^ 0.5 = 9.9045 m/s (9.9045*3.6 = 35.6562 km/h)
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Rail-Car Buffer
Tuesday, 8 May 2012
8:16 PM
PE1 + KE1 + SE1 +/- W = PE2 + KE2 + SE2
PE1= SE2
mgh = 0.5*k*x^2
x = (2mgh/k)^0.5
= (2*9400*9.81*2.6/640000)^0.5 = 0.8656 m
Note: We could not do this using linear motion & F=ma
because the spring does not give a constant force (therefore not
constant accel)
e.g. velocity before impact..(2*9.81*2.6)^0.5 = 7.1423 m/s
Assume this stops in 0.8656m, by linear motion...
v2-vo2 = 2as, so a = (0-7.1423^2)/(2*0.8656) = -29.4665 m/s2
From F=ma, F = 9400*29.4665 = 276.98 kN from spring
How many mm? F=kx x=F/k = 276980/640 = 432.7813 mm
This shows the average is half of the maximum.
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Rolling Inertia
Tuesday, 17 May 2011
6:02 PM
Example 18.3. Cylinder 100mm,
Mass = 30kg, height difference 1.2m
Find linear/rotary speed at bottom.
PE1 + KE1 + SE1 +/- W = PE2 + KE2 + SE2
PE1 = KE2
mgh = 0.5mv2 + 0.5I2
Calculate mass moment of inertia I = 0.5mr2
= 0.5*30*0.05^2
= 0.0375 kgm2
30*9.81*1.2 = 0.5*30*v2 + 0.5*0.0375*2
Oh no! We have 2 variables in 1 equation!!!!
No probs. Use v=r
30*9.81*1.2 = 0.5*30*v2 + 0.5*0.0375* (v/r)2
353.16 = 15v2 + 0.5*0.0375* (v/0.05)2
353.16 = 15v2 + 0.5*0.0375* (v2/0.052)
353.16 = v2(15 + 0.5*0.0375/0.052)
v2 = 353.16/ 22.5 = 15.696
v= 15.696 ^0.5 = 3.9618 m/s
=v/r 3.9618/0.05 = 79.236 rad/s
Note: Compare to not spinning. v=(2gh)^0.5
= (2*9.81*1.2)^0.5 = 4.8522 m/s
(or with a zero mass moment of inertia I)
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Weight falls on a spring.
Tuesday, 17 May 2011
6:26 PM
Example 18.5. Mass 15kg, falls 350mm to spring.
Modulus 10N/mm. Find max compression.
PE1 + KE1 + SE1 +/- W = PE2 + KE2 + SE2
PE1 = PE2 + KE2
PE1 - PE2 = KE2
mg(h2-h1) = 0.5mv2
15*9.81*0.35 = 0.5 *15*v2
v2 = 51.5025 /(0.5*15) = 6.867
v = 6.867 ^0.5 = 2.6205 m/s (Check (2*9.81*0.35)^0.5 = 2.6205m/s)
PE2 + KE2 + SE2 +/- W = PE3 + KE3 + SE3
PE2 + KE2 = SE3
mgx + 0.5mv2 = 0.5kx2
Quadratic equation...
0.5kx2 - mgx - 0.5mv2 = 0
0.5*10*1000x2 - 15*9.81x - 0.5*15*2.62052 = 0
5000 x2 - 147.15 x - 51.5027 = 0
a = 5000, b = -147.15, c = - 51.5027
Quadratic equation solution…
Read the book. Page 258
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So answers are x = 117mm or x = -87mm (????)
This quadratic formula assumes the mass and spring are stuck
together. Imagine dropping the mass, then it latches onto the spring
for good.
-87mm is how high the stuck mass will pull the spring ABOVE the
initial start position of the spring.
If it doesn't latch, the mass will bounce back to the start (350mm).
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Multiple Bodies
Tuesday, 17 May 2011
7:03 PM
PE1 + KE1 + SE1 +/- W = PE2 + KE2 + SE2
PE1 - W = PE2 + KE2
PE1 = (1.3*9.81*1.6+3.3*9.81*1.6) = 72.2016 J
PE2 = (1.3*9.81*3.2+3.3*9.81*0) = 40.8096 J
W = T
Find angle: s=r  so = s/r = 1.6/0.04 = 40.0 rad
W = 0.340*40 = 13.6 J
PE1 - W - PE2 = KE2
72.2016 - 13.6 - 40.8096 = 17.792
KE = 0.5*mv^2
v = (2KE/m)^0.5 = (2*17.792/(1.3+3.3) )^0.5 = 2.7813 m/s
Check: Freefall drop 1.6m v = (2*9.81*1.6)^0.5 = 5.6029
Our answer is slower than freefall, so looks good.
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Charpy Energy
Tuesday, 8 May 2012
6:44 PM
Job for CAD...
1341.4078 mm
PE = mgh
= 33*9.81*1.3414
= 434.2514 J
Velocity of hammer before impact?
v = (2gh)^0.5 = (2*9.81*1.3414)^0.5
= 5.1301 m/s
Velocity of hammer after impact?
PE after: PE = mgh = 33*9.81*
0.32253 = 104.41264 J
KE after = 104.41264
v = (2KE/m)^0.5 = (2*
104.41264/33)^0.5 = 2.51556 m/s
Check: v = (2gh)^0.5 = (2*9.81*0.32253)^0.5 =
2.51556 m/s
Finding height by
CAD.
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