Explanation Given information: The weight of the motor ( W ) is 400 lb . The weight of the rotor ( w ) is 1 oz . The spring constant ( k ) is 1200lb/in . . The distance between the weight of rotor to axis of rotation ( r ) is 8 in . or 0 .66667ft . The acceleration due to gravity ( g ) is 32.2 ft/s 2 . Calculation: Calculate the mass of the motor ( M ) using the formula: W = M g M = W g Substitute 400 lb for W and 32.2 ft/s 2 for g . M = 400 32.2 = 12.4224 lb ⋅ s 2 /ft Calculate the mass of the rotor ( m ) using the formula: w = m g m = w g Substitute 1 oz for w and 32.2 ft/s 2 for g . m = ( 1 oz 16 lb oz ) 32.2 = 0.001941 lb ⋅ s 2 /ft Calculate the natural frequency of the periodic force ( ω n ) using the formula: ω n = k M Substitute 1200lb/in . for k and 12.4224 lb ⋅ s 2 /ft for M . ω n = 1200 lb in . × 12 in . ft 12.4224 = 34.047 rad / s Calculate the amplitude of motion ( x m ) using the formula: x m = ( r m M ) ( ω f ω n ) 2 1 − ( ω f ω n ) 2 Substitute 0 .66667ft for r , 12.4224 lb ⋅ s 2 /ft for M , 0.001941 lb ⋅ s 2 /ft for m , and 34.047 rad / s for ω n . x m = ( 0.66667 × 0.001941 12.4224 ) ( ω f ω n ) 2 1 − ( ω f ω n ) 2 = ( 0.0001042 ft × 12 in ft ) ( ω f ω n ) 2 1 − ( ω f ω n ) 2 = 0.00125 in . ( ω f ω n ) 2 1 − ( ω f ω n ) 2 Consider in phase motion with the amplitude ( x m ) less than 0

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ISBN: 9781260663778

Author: BEER

Publisher: MCG

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Forced undamped vibrations

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Chapter 19.4, Problem 19.116P

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Find the range of allowable values of the motor speed (ωf) if the amplitude of the vibration is not to exceed 0.06in..

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Chapter 19.4, Problem 19.115P

Chapter 19.4, Problem 19.117P

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Find the range of allowable values of the motor speed ( ω f ) if the amplitude of the vibration is not to exceed 0.06 in . .

Solution Summary: The author calculates the range of allowable values of the motor speed using the formula: cW=Mg

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Find the range of allowable values of the motor speed (ωf) if the amplitude of the vibration is not to exceed 0.06in..

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Chapter 19 Solutions