King Fahd University of Petroleum Minerals Mechanical Engineering

King Fahd University of Petroleum & Minerals Mechanical Engineering Dynamics ME 201 BY Dr. Meyassar N. Al-Haddad Lecture # 35

Objective • • Moment of Inertia of a body Parallel Axis Theorem Radius of Gyration Moment of Inertia of Composite Bodies

Parallel Axis Theorem • The moment of inertia about any axis parallel to and at distance d away from the axis that passes through the centre of mass is: • Where – IG= moment of inertia for mass centre G – m = mass of the body – d = perpendicular distance between the parallel axes.

Radius of Gyration

Mass Center Example

Objectives • Apply the Equation of Translation motion – Rectilinear Translation – Curvilinear Translation • Apply the equation of Rotation motion • • The roll of the center of mass G Discuss the slipping Vs. Tipping Discuss the slipping Vs. freely rotating Discuss the “wheely” Vs. Non-wheely

General Application of the Equations of Motion Summation of moment in FBD = summation of the kinetic moment in K. D

Rectilinear Translation

Curvilinear Translation

Discuss the slipping Vs. Tipping If x > 1. 5 ft tipping If x < 1. 5 ft slipping x N

Slipping Vs. Freely rotating mg FA=ms. NA NA NB

“Wheely” Vs. Non-Wheely : lift the front wheel off the ground NB=0

Example 17 -5

Example 17 -8 m. BD=100 kg m. AB=m. CD= Neglect q = 30 o w = 6 rad/s TA= ? TB= ? a. G=?

Problem 17 -43 m= 80 kg m. B=0. 8 NA=? NB=? When rear wheel locks for break a =? Deceleration a. G m. BNB NA NB

Problem 17 -43 m= 80 kg m. B=0. 8 NA=? NB=? a =? When traveling at constant velocity and no break was applied a=0 m. BNB NA NB

Problem 17 -44 m= 80 kg NA=? NB=? a =? mk=? minimum When rider applies the front break and back wheel start to lift off the ground mk NA NA NB=0

Rotation About a Fixed Axis

Rotation About a Fixed Axis

Horizontal Reaction Vertical Reaction OR Normal Reaction Tangential Reaction Pin Reaction Ox On Oy On Ot Ot

Example 17 -9 Start from rest q=? Number w=20 rad/s Pine reaction

Example 17 -10

Example 17 -11 m=60 kg Radius of gyration k. O=0. 25 mb=20 kg a=? Drum a=?

Example 17 -12 W=50 Ib k. G=0. 6 ft w= 8 rad/s Pin reaction =?

Frictional Rolling Problems - rolls without slipping - slides as it rolls

Example 17 -14 m = 8 Kg radius of gyration k. G=0. 3 m. a=? .

Example 17 -15 W= 50 -lb radius of gyration k. G=0. 70 ft. a. G=? .

Example 17 -16 m=100 kg IG=75 kg. m 2 a=? The pole at rest

Example 17 -17