Unit Conversion Examples
3.1
Unit Conversion Examples
1. An M=4.09 kg is attached to a spring with spring constant, K. The natural frequency
K
ωn =
of the system is 2.38 Hz (or 2.38 cycles/sec). What is the spring constant, K?
M
ω n = 2.38
ωn =
cycles ⎛ 2π rad ⎞
rad
⎜⎜
⎟⎟ = 14.96
sec ⎝ cycle ⎠
sec
K
rad ⎞
⎛
→ K = Mω n2 = (4.09kg )⎜14.96
⎟
M
sec ⎠
⎝
2
⎛ 1 N − sec 2 ⎞
N ⋅ rad 2
N
⎜⎜
⎟⎟ → K = 915
= 915
m
m
⎝ 1 kg ⋅ m ⎠
2. A mass that weighs 4.09 lb is attached to a spring with spring constant, K. The natural
K
frequency ω n =
of the system is 2.38 Hz (or 2.38 cycles/sec). What is the spring
M
constant, K?
W = Mg → M =
ωn =
W
4.09lb
,
=
ft
g
32.2
sec 2
⎛
⎜ 4.09lb
K
2
→ K = Mω n = ⎜
M
⎜ 32.2 ft
⎜
sec 2
⎝
ω n = 2.38
cycles ⎛ 2π rad ⎞
rad
⎟⎟ = 14.96
⎜⎜
sec ⎝ cycle ⎠
sec
⎞
2
⎟
2
⎞
⎛
⎟⎛⎜14.96 rad ⎞⎟ ⎜ 1 ft ⎟ → K = 2.37 lb ⋅ rad = 2.37 lb
sec ⎠ ⎜⎝ 12 in ⎟⎠
in
in
⎟⎝
⎟
⎠
3. A 60.0 lbm mass experiences a 9300 N force. Find the acceleration, a (f = ma) in units of
ft/sec2 and m/sec2
a=
f
9300 N ⎛ 32.2 ft ⋅ lbm ⎞⎛ 1lbf ⎞
ft ⎛ 1m ⎞
m
⎟⎟ = 342
⎟
⎜⎜
=
⎟ = 1120 2 ⎜⎜
2 ⎟⎜
m 60.0 lbm ⎝ lbf ⋅ sec ⎠⎝ 4.448 N ⎠
sec ⎝ 3.2808 ft ⎠
sec 2
4. The kinetic energy of a point mass is given by KE = ½ mV2, where m = mass, and
V=velocity. Kinetic energy is typically expressed in US Engineering units of ft-lbf, or the
SI units of joules (J). With a 56 lbm mass and a velocity of 37 ft/sec,
⎛1⎞
⎛ ft ⎞
KE = ⎜ ⎟(56 lbm)⎜ 37 ⎟
⎝2⎠
⎝ s⎠
2
⎛
⎞
lbf − s 2
⎜⎜
⎟⎟ = 1190 ft − lbf
32.178ft
lbm
−
⎝
⎠
⎛ 1 m ⎞⎛ 4.448 N ⎞
⎛ 1J ⎞
KE = 1190 ft − lbf ⎜
⎟⎜
⎟ = 1610N − m ⎜
⎟ = 1610J
⎝ 3.2808 ft ⎠⎝ 1 lbf ⎠
⎝ 1N - m ⎠
Unit Conversion Examples
3.2
5. The acceleration of gravity is 39 ft/s2 on Planet Zordar. Find the force of attraction (what
we call weight, w = mg) of the gravitational field on a 25 lbm object in units of:
⎞
lbf − s 2
⎛ ft ⎞⎛
⎟⎟
w = mg = (25lbm )⎜ 39 2 ⎟⎜⎜
⎝ s ⎠⎝ 32.178ft − lbm ⎠
⎞
lbf − s 2
⎛ 4.448N ⎞
⎛ ft ⎞⎛
⎟⎟ = 30.3lbf ⎜
w = mg = (25lbm )⎜ 39 2 ⎟⎜⎜
⎟ = 135N
⎝ lbf ⎠
⎝ s ⎠⎝ 32.178ft − lbm ⎠
6. The kinetic energy of a rotating mass is given by KE = ½ Iω2, where I = mass moment of
inertia, and w=angular velocity. With a mass moment of inertia of I=6.50 lbm-ft2 and an
angular velocity of 1260 RPM,
rev ⎞⎛ 2π rad ⎞⎛ 1 min ⎞
rad
⎛
⎟⎟⎜⎜
⎟⎟ = 132
ω = ⎜1260
⎟⎜⎜
min ⎠⎝ 1 rev ⎠⎝ 60 sec ⎠
sec
⎝
rad ⎞
⎛1⎞
⎛
KE = ⎜ ⎟(6.50 lbm - ft 2 )⎜132
⎟
s ⎠
⎝2⎠
⎝
2
⎛
⎞
lbf − s 2
⎜⎜
⎟⎟ = 1760 ft − lbf
⎝ 32.178ft − lbm ⎠
7. The speed of sound in an ideal gas at temperature T is given by: c = (kRT)½
For air, k = 1.40 (dimensionless) and R = 53.3 ft-lbf/lbm-°R. Find the speed of sound at
T = 70°F = 530°R in units of ft/sec and mi/hr
c = (kRT )
0.5
c = 1130
ft − lbf ⎞
⎛
⎛
⎛ 32.178ft − lbm ⎞ ⎞
o
= ⎜ (1.4 )⎜ 53.3
⎟(530 R )⎜
⎟⎟
o
lbm − R ⎠
lbf − s 2
⎝
⎝
⎠⎠
⎝
0.5
= 1130
ft
s
ft ⎛ 1mi ⎞⎛ 60s ⎞⎛ 60min ⎞
⎟⎜
⎟
⎜
⎟⎜
s ⎝ 5280ft ⎠⎜⎝ 1min ⎟⎠⎜⎝ 1hr ⎟⎠
c = 1130
mi
ft ⎛ 1mi ⎞⎛ 60s ⎞⎛ 60min ⎞
⎟⎟⎜⎜
⎟⎟ = 769
⎜
⎟⎜⎜
s ⎝ 5280ft ⎠⎝ 1min ⎠⎝ 1hr ⎠
hr
8. The units for exhaust emissions from internal combustion engines are frequently given in
g/(hp-hr). Convert the following to the equivalent in the SI system, kg/(kW-day)
0.0376
0.0376
g
g ⎛ 1kg ⎞⎛ 1hp ⎞⎛ 24hr ⎞
= 0.0376
⎜
⎟⎜
⎟
⎟⎜
hp − hr
hp − hr ⎜⎝ 1000g ⎟⎠⎝ .746kW ⎠⎜⎝ 1day ⎟⎠
g
g ⎛ 1kg ⎞⎛ 1hp ⎞⎛ 24hr ⎞
kg
⎜⎜
⎟⎟⎜
⎟⎟ = 1.21x10 −3
= 0.0376
⎟⎜⎜
hp − hr
hp − hr ⎝ 1000g ⎠⎝ .746kW ⎠⎝ 1day ⎠
kW − day