King Fahd University of Petroleum & Minerals Mechanical

King Fahd University of Petroleum & Minerals Mechanical

King Fahd University of Petroleum & Minerals Mechanical Engineering Dynamics ME 201 BY Dr. Meyassar N. Al-Haddad Lecture # 30 CHAPTER 17 :Planar Kinetics of a Rigid Body Force and Acceleration Objective Moment of Inertia of a body Parallel Axis Theorem Radius of Gyration Moment of Inertia of Composite Bodies Moment and Angular Acceleration

When 0, rigid body experiences angular acceleration Relation between M and is analogous to relation between F and a F ma, M I Mass = Resistance Moment of Inertia Moment of Inertia This mass analog is called the moment of inertia, I, of the object I r 2 dm m

r = moment arm SI units are kg m2 Using dm dV , where is the volume density : I r 2 dV I r 2 dx dy dz Shell Element dV (2 y ) z dy Disk Element 2 dV ( y ) dz Example 17-1

dm dV (2 r dr h) R 4 1 2 I r dm 2 h r dr R h R ( R 2 h) 2 2 m 0 2 3 m R 2 h 1 I z m R2 2

Moments of inertia for some common geometric solids L R2 L R2 R 1 I ML2 3 Thin Rod (axis at end) 1 I ML2 12

Thin Rod a 1 2 2 I M ( R1 R2 ) 2 Hollow Cylinder 1 I MR 2 2 Solid Disk a R b

b 1 I M (a 2 b 2 ) 12 Rectangular Plate (through center) 1 I Ma 2 3 Thin Rectangular Plate (about edge) R 2 I MR 2 5 Solid Sphere R

2 I MR 2 3 Thin Walle d Hollow Sphere I MR 2 Thin Walle d Hollow Cylinder 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: I O I G md 2 Where

IG= moment of inertia for mass centre G m = mass of the body d = perpendicular distance between the parallel axes. Radius of Gyration Frequently tabulated data related to moments of inertia will be presented in terms of radius of gyration. I mk 2 or I k m Mass Center

y ~ ym m Example y ~ ym 1(10 / 32.2) 2(10 / 32.2) 1.5 ft m (10 / 32.2) (10 / 32.2)

Ib 10 Moment of Inertia of Composite bodies 1. Divide the composite area into simple body. 2. Compute the moment of inertia of each simple body about its centroidal axis from table. 3. Transfer each centroidal moment of inertia to a parallel reference axis 4. The sum of the moments of inertia for each simple body about the parallel reference axis is the moment of inertia of the composite body. 5. Any cutout area has must be assigned a negative moment; all others are considered positive. Moment of inertia of a hollow cylinder Moment of Inertia of a solid cylinder A hollow cylinder M

= I = 1/2 mR2 m1 R1 - m2 R2 I = 1/2 m1R12 - 1/2 m2R22 = 1/2 M (R12 - R22 ) Example 17-3 md dVd 8000

kg 2 [ ( 0 . 25 m ) (0.01 m)] 15.71 kg 3 m 1 2 ( I d ) O md rd md d 2 2 1

(15.71kg )(0.25m) 2 (15.71kg )(0.25m) 2 2 1.473 kg.m 2 kg mh hVh 8000 3 [ (0.125 m) 2 (0.01 m)] 3.93 kg m 1 2 ( I h ) O mh rh mh d 2 2 1 (3.93kg )(0.125m) 2 (3.93kg )(0.25m) 2 2 0.276 kg.m 2 1 IG m r 2 2

I O ( I d ) O ( I h )O 1.473 0.276 1.2 kg.m 2 1 3 2 2 2 I zz md rd ( mh rh mh d ) 2 2 kg 2 [ ( 0 . 25

m ) (0.01 m)] 15.71 kg 3 m kg mh hVh 8000 3 [ (0.125 m) 2 (0.01 m)] 3.93 kg m md dVd 8000 3 1 I zz (15.71)(0.25) 2 ( (3.93kg )(0.125m) 2 (3.93kg )(0.25m) 2 ) 2 2 Example 17-4 1 2 1 10Ib ( I OA ) O ml (

)(2ft ) 2 0.414 slug.ft 2 3 3 32.2ft/s 1 1 10 10 2 2 2 ( I BC ) O ml md ( )(2) ( )(2) 2 12 12 32.2 32.2 1.346 slug.ft 2 I O 0.414 1.346 1.76 slug.ft 2

y y ~ ym ~ ym m 1(10 / 32.2) 2(10 / 32.2) 1.5 ft m (10 / 32.2) (10 / 32.2) I O I G md 2 20 1.76 I G (

)(1.5) 2 32.2 I G 0.362 slug.ft 2

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