Machinery's Handbook, 31st Edition
188 Energy Solution: From page 253 the moment of inertia of a solid cylinder with respect to a gravity axis at right angles to the circular cross section is given as 1 ∕ 2 Mr 2 . From page 184 , 100 rpm = 10.47 rad/sec, hence an acceleration of 100 rpm per second = 10.47 rad/sec 2 . Therefore, using the first of the preceding formulas, . . T J 2 1 3216 1000 2 3 10 47 366 ft-lbs 2 # = M = α = a a k
o k Using metric SI units, the formulas are: T o = J M α = Mk o newton-meters; J M = the moment of inertia in kg·m
2 α , where T
o = torque in
2 , and α = the angular accelera
tion in rad/s 2 . Example: A flywheel has a diameter of 1.5 m and a mass of 800 kg. What torque is needed to produce an angular acceleration of 100 rpm/s? As in the preceding ex ample, α = 10.47 rad/s 2 . Thus: . J Mr kg 2 1 800 075 225 m M 2 2 2 # # $ = = = Therefore: T o = J M α = 225 3 10.47 = 2356 N·m. Energy.— A body is said to possess energy when it is capable of doing work or overcoming resistance. The energy may be either mechanical or non-mechanical, the latter including chemical, electrical, thermal, and atomic energy. Mechanical energy includes kinetic energy (energy possessed by a body because of its motion) and potential energy (energy possessed by a body because of its position in a field of force and/or its elastic deformation). 2 1 Kinetic Energy: The motion of a body may be one of pure translation, pure rotation, or a combination of rotation and translation. By translation is meant motion in which every line in the body remains parallel to its original position throughout the motion; that is, no rotation is associated with the motion of the body. The kinetic energy of a translating body is given by the formula (3a) where M = mass of body ( M = W / g ); V = velocity of the center of gravity of the body in ft/sec; W = weight of body in lbs; and g = acceleration due to gravity = 32.16 ft/sec 2 . The kinetic energy of a body rotating about a fixed axis O is expressed by the formula: (3b) where J MO is the moment of inertia of the body about the fixed axis O in lbs-ft-sec 2 , and ω = angular velocity in rad/sec. For a body that is moving with both translation and rotation, the total kinetic energy is given by the following formula as the sum of the kinetic energy due to translation of the center of gravity and the kinetic energy due to rotation about the center of gravity: (3c) E MV J Total Kinetic Energy in ft-lbs MG 2 2 ~ = = + T E MV g WV 2 1 2 2 = = Kinetic Energy in ft-lbs due to translation = KT 2 E J 2 1 KR = Kinetic Energy in ft-lbs due to rotation = MO 2 ~
2 1
2 1
g WV J 2 2 1 2
2 2 1 2
g WV
Wk
g W V k 2 + ^
2 2 ~
2
2 2 2 ~
h
= +
= +
=
~
g
MG
where J MG is the moment of inertia of the body about its gravity axis in lbs-ft-sec 2 , k is the radius of gyration in feet with respect to an axis through the center of gravity, and the other quantities are as previously defined. In the metric SI system, energy is expressed as the joule (J). One joule = 1 newton- meter. The kinetic energy of a translating body is given by the formula E KT = 1 ∕ 2 MV 2 ,
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