SolSec1_4 - Problems and Solutions Section 1.4 (problems...

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Problems and Solutions Section 1.4 (problems 1.52 through 1.65) 1.52 Calculate the frequency of the compound pendulum of Figure 1.20(b) if a mass m T is added to the tip, by using the energy method. Solution Using the notation and coordinates of Figure 1.20 and adding a tip mass the diagram becomes: If the mass of the pendulum bar is m , and it is lumped at the center of mass the energies become: Potential Energy: U = 1 2 ( ! ! ! cos " ) mg + ( ! ! ! cos ) m t g = ! 2 (1 ! cos )( mg + 2 m t g ) Kinetic Energy: T = 1 2 J ˙ ! 2 + 1 2 J t ˙ 2 = 1 2 m ! 2 3 ˙ 2 + 1 2 m t ! 2 ˙ 2 = ( 1 6 m + 1 2 m t ) ! 2 ˙ 2 Conservation of energy (Equation 1.52) requires T + U = constant: ! 2 (1 ! cos )( mg + 2 m t g ) + ( 1 6 m + 1 2 m t ) ! 2 ˙ 2 = C Differentiating with respect to time yields: ! 2 (sin )( mg + 2 m t g ) " + ( 1 3 m + m t ) ! 2 " "" = 0 " ( 1 3 m + m t ) ! + 1 2 ( mg + 2 m t g )sin = 0 Rearranging and approximating using the small angle formula sin θ ~ , yields: θ m t
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!! ! ( t ) + m 2 + m t 1 3 m + m t g " " # $ $ $ % & ( t ) = 0 ( ) n = 3 m + 6 m t 2 m + 6 m t g " rad/s Note that this solution makes sense because if m t = 0 it reduces to the frequency of the pendulum equation for a bar, and if m = 0 it reduces to the frequency of a massless pendulum with only a tip mass. 1.53 Calculate the total energy in a damped system with frequency 2 rad/s and damping ratio ζ = 0.01 with mass 10 kg for the case x 0 = 0.1 and v 0 = 0. Plot the total energy versus time. Solution: Given: ω n = 2 rad/s, ζ = 0.01, m = 10 kg, x 0 = 0.1 mm, v 0 = 0. Calculate the stiffness and damped natural frequency: k = m n 2 = 10(2) 2 = 40 N/m d = n " # 2 = " 0.01 2 = 2 rad/s The total energy of the damped system is E ( t ) = 1 2 m ˙ x 2 ( t ) + 1 2 kx ( t ) where x ( t ) = Ae ! 0.02 t sin(2 t + " ) ˙ x ( t ) = ! 0.02 Ae ! 0.02 t sin(2 t + ) + 2 Ae ! 0.02 t cos(2 t + ) Applying the initial conditions to evaluate the constants of integration yields: x (0) = 0.1 = A sin ˙ x (0) = 0 = " 0.02 A sin + 2 A cos # = 1.56 rad/s, A = 0.1 m Substitution of these values into E ( t ) yields:
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1.54 Use the energy method to calculate the equation of motion and natural frequency of an airplane's steering mechanism for the nose wheel of its landing gear. The mechanism is modeled as the single-degree-of-freedom system illustrated in Figure P1.54. The steering wheel and tire assembly are modeled as being fixed at ground for this calculation. The steering rod gear system is modeled as a linear spring and mass system ( m , k 2 ) oscillating in the x direction. The shaft-gear mechanism is modeled as the disk of inertia J and torsional stiffness k 2 . The gear J turns through the angle θ such that the disk does not slip on the mass. Obtain an equation in the linear motion x . Solution: From kinematics: x = r ! , " ˙ x = r ˙ Kinetic energy: 2 2 2 1 2 1 x m J T ! ! + = Potential energy: 2 1 2 2 2 1 2 1 k x k U + = Substitute r x = : 2 2 1 2 2 2 2 2 2 1 2 1 2 1 2 1 x r k x k x m x r J U T + + + = + ! ! Derivative: ( ) 0 = + dt U T d J r 2 ˙ ˙ x ˙ x + m ˙ ˙ x ˙ x + k 2 x ˙ x + k 1 r 2 x ˙ x = 0 J r 2 + m !
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SolSec1_4 - Problems and Solutions Section 1.4 (problems...

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