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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.
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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.
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
x N Discuss the slipping Vs. Tipping If x > 1.5 ft tipping If x < 1.5 ft slipping
Slipping Vs. Freely rotating mg FA=msNA NA NB
“Wheely” Vs. Non-Wheely Wheely : lift the front wheel off the ground NB=0
Example 17-8 mBD=100kg mAB=mCD= Neglect q = 30o w = 6 rad/s TA= ? TB= ? aG=?
Problem 17-43 mBNB NA NB m= 80 kg mB=0.8 NA=? NB=? When rear wheel locks for break a =? Deceleration aG
Problem 17-43 a = 0 mBNB NA NB m= 80 kg mB=0.8 NA=? NB=? a =? When traveling at constant velocity and no break was applied
Problem 17-44 mkNA NA NB=0 m= 80 kg NA=? NB=? a =? mk=? minimum When rider applies the front break and back wheel start to lift off the ground
Ot Ox On Oy On Ot Pin Reaction Horizontal Reaction Vertical Reaction OR Normal Reaction Tangential Reaction
Example 17-9 Start from rest q=? Number w=20rad/s Pine reaction
Example 17-11 m=60 kg Radius of gyration kO=0.25 mb=20 kg a=? Drum a=?
Example 17-12 W=50 Ib kG=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 kG=0.3m. a=?.
Example 17-15 W= 50-lb radius of gyration kG=0.70ft. aG=?.
Example 17-16 m=100 kg IG=75 kg.m2 a=? The pole at rest