{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–5. Sign up to view the full content.

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 ~ θ , yields: θ m t

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
!! ! ( 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 1 " # 2 = 2 1 " 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:

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 !
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern