Page 408 - Marks Calculation for Machine Design

P1: Naresh
January 4, 2005
Brown˙C09
Brown.cls
390
U.S. Customary 15:28 APPLICATION TO MACHINES SI/Metric
Step 2. Calculate the mass moment of inertia Step 2. Calculate the mass moment of inertia
(I shaft ) for the solid circular shaft using (I shaft ) for the solid circular shaft using
Eq. (9.53). Eq. (9.53).
1 4 1 4
I shaft = ρπ Lr i I shaft = ρπ Lr i
2 2
1 slug 1 kg
= 15.2 π(4ft) = 7,850 π(1.35 m)
2 ft 3 2 m 3
4
4
×[(0.125 ft) ] ×[(0.04 m) ]
slug 4 kg 4

= 95.5 [0.000244 ft ] = 16,650 [0.0000025 m ]
ft 2 m 2
= 0.02 slug · ft 2 = 0.04 kg · m 2
Step 3. Combine the mass moment of inertia Step 3. Combine the mass moment of inertia
oftheflywheel(I flywheel )foundinstep1withthe of the flywheel (I flywheel ) found in step 1 with
mass moment of inertia of the shaft (I flywheel ) the mass moment of inertia of the shaft (I shaft )
found in step 2 to give the total mass moment found in step 2 to give the total mass moment
of inertia (I total ) as of inertia (I total ) as
I total = I flywheel + I shaft I total = I flywheel + I shaft
2
2
= [(29.80) + (0.02) slug · ft ] = [(39.85) + (0.04) kg · m ]
= 29.82 slug · ft 2 = 39.89 kg · m 2
Notice that the contribution to the total mass Notice that the contribution to the total mass
moment of inertia from the shaft is almost neg- moment of inertia from the shaft is almost neg-
ligible. This is because mass farther away from ligible. This is because, mass farther away from
the axis counts more, in fact a function of the the axis counts more, in fact a function of the
distance squared. distance squared.
Step 4. Substitute the total mass moment of Step 4. Substitute the total mass moment of
inertia (I total ) found in step 3 and the given inertia (I total ) found in step 3, and the given
torque (T ) in Eq. (9.51). torque (T ), in Eq. (9.51).
T = I total α T = I total α
2
2
20 ft · lb = (29.82 slug · ft )α 30 N · m = (39.89 kg · m )α
Step 5. Solve for the angular acceleration (α) Step 5. Solve for the angular acceleration (α)
from step 4. from step 4.
20 ft · lb lb 30 N · m N
α = = 0.67 α = = 0.75
29.82 slug · ft 2 slug · ft 39.89 kg · m 2 kg · m
slug · ft /sec 2 rad kg · m/s 2 rad
= 0.67 = 0.67 = 0.75 = 0.75
slug · ft s 2 kg · m s 2

The inertial energy (E inertial ) of the flywheel and shaft assembly is given by the relation-
ship in Eq. (9.54) as
1 2
E inertial = I total ω (9.54)
2