Modern Engineering Thermodynamics

1.11 Potential and Kinetic Energies 21
Table 1.7 continued
x
x
Solid circular cylinder
I x = mR 2 /2
I y = I z = m(3R 2 + L 2 )/12
L
y
R
z
Thin circular cylinder
I x =mR 2
R
x
x
R 1
Hollow circular cylinder
L
I x = mR 2 1 + R R2 2 /2 2
Solid circular cone
I x =3mR 2 /10
I y = I z =3m(R 2 /4 + L 2 )/5
R
L
y
z
x
x
R
R
Solid sphere
I x = I y = I z =2mR 2 /5
z
Hollow sphere
I x = I y = I z =2mR 2 /3
z
y
y
In the equations that follow in this chapter and throughout the text, the phrase kinetic energy can mean either
translational or rotational. You must be alert to the conditions present in the problem to choose the correct
form. Later in the text, general equations for the first law of thermodynamics are developed that include kinetic
energy terms. Usually only the translational kinetic energy expression is written out in these equations, but be
alert to the fact that this could change to rotational kinetic energy or a combination of both translational and
rotational kinetic energy in any problem.
The following examples illustrate the calculation of kinetic and potential energies in both SI and Engineering
English units systems.
EXAMPLE 1.7
Determine the potential energy of an automobile weighing 2000. lbf when it is 8.00 ft off the floor on a hoist in a repair
shop (Figure 1.14). Express the result in both SI and Engineering English units.
Solution
The formula for the potential energy (PE) of an object of mass m at a distance Z above the reference height is given by Eq. (1.10) as
PE = k 1 mgZ = mgZ
g c
We first calculate the automobile’s mass from its weight using Newton’s second law, m = Fg c /g = Weight g c /g. In the SI units
system (see Table 1.2), we find that g c = 1 (dimensionless), so
m = Fg c
g
ð2000: lbfÞ
1N
ð1Þ
=
0:2248 lbf
9:81 m=s 2 = 907 N .s 2
m = 906:9kg (Continued )