Page 407 - Marks Calculation for Machine Design

P1: Naresh
15:28
January 4, 2005
Brown.cls
Brown˙C09
MACHINE ENERGY
The torque (T ) can vary over time; therefore, the angular acceleration (α) and angular
velocity (ω) must also vary over time. 389
The relationship between the torque (T ) and the angular acceleration (α) for a flywheel
and shaft assembly rotating about a fixed axis is given by Eq. (9.51) as
T = I total α = (I flywheel + I shaft )α (9.51)
where (I total ) is the total mass moment of inertia, which is the sum of the mass moment of
inertia of the flywheel (I flywheel ) and the mass moment of inertia of the shaft (I shaft ), both
calculated about the axis of rotation.
For a solid disk flywheel with an outside radius (r o ) and inside radius (r i ) mounted on a
shaft with an outside radius equal to the inside radius of the flywheel, the mass moments
of inertia (I flywheel ) and (I shaft ) are given by the following two formulas as
1 2 2 2
I flywheel = ρπt r − r i (9.52)
o
2
1 4
I shaft = ρπ Lr i (9.53)
2
where (t) is the thickness of the flywheel and (L) is the length of the shaft, and where the
density (ρ) of the flywheel and shaft are assumed to be the same.

U.S. Customary SI/Metric
Example 1. Calculate the angular accelera- Example 1. Calculate the angular accelera-
tion (α) produced by a torque (T ) on a steel tion (α) produced by a torque (T ) on a steel
solid disk flywheel and shaft assembly, where solid disk flywheel and shaft assembly, where
T = 20 ft · lb T = 30 N · m
3
3
ρ = 15.2 slug/ft (steel) ρ = 7,850 kg/m (steel)
r o = 18 in = 1.5 ft r o = 45 cm = 0.45 m
r i = 1.5 in = 0.125 ft r i = 4cm = 0.04 m
t = 3in = 0.25 ft t = 8cm = 0.08 m
L = 4ft L = 1.35 m
solution solution
Step 1. Calculate the mass moment of iner- Step 1. Calculate the mass moment of iner-
tia (I flywheel ) for a solid disk flywheel using tia (I flywheel ) for a solid disk flywheel using
Eq. (9.52). Eq. (9.52).
1 2 2 2 1 2 2 2
I flywheel = ρπt r − r i I flywheel = ρπt r − r i
o
o
2 2
1 slug 1 kg
= 15.2 π(0.25 ft) = 7,850 π(0.08 m)
2 ft 3 2 m 3
2 2
2 2
2
2
×[(1.5ft) − (0.125 ft) ] ×[(0.45 m) − (0.04 m) ]
slug kg

= 5.97 = 986.5
ft 2 m 2
2 2
2 2
×[(2.25 − 0.0156) ft ] ×[(0.2025 − 0.0016) m ]

slug 4 kg 4
= 5.97 [4.99 ft ] = 986.5 [0.0404 m ]
ft 2 m 2
= 29.80 slug · ft 2 = 39.85 kg · m 2