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Lecture 3:

example: rolling no slip • rotational equation of of • solving problems • dynamics example: pulley with mass

Thursday, April 11, 13 Rigid Body Kinematics v = v +(! r ) A B ⇥ A/B a = a +(↵ r )+(! (! r )) A B ⇥ A/B ⇥ ⇥ A/B Useful Shortcuts for 2D planar motion ↵ = ↵k • ! = !k r = rxi + ryj

! r = r !i + r !j ⇥ y x ↵ r = r ↵i + r ↵j ⇥ y x ! (! r)= r !2i r !2j ⇥ ⇥ x y

Thursday, April 11, 13 Rigid Body Dynamics

Linear Motion: sum of the is the d(mv) F = ma = rate of change of dt linear

Works for particles - and also works for rigid bodies if the is at the !

F = maG

Thursday, April 11, 13 Rigid Body Dynamics

Rotational Motion: d(H ) sum of the moments is M = G = H˙ G dt G the time rate of change of about the center of mass

Thursday, April 11, 13 Rigid Body Dynamics

Rotational Motion: sum of the moments is d(HG) MG = = IG↵ the time rate of change dt of angular momentum Mass :

2 2 IG = (miri )= (⇢r )dV i V X Z “Rotational Inertia” Resistance to

Thursday, April 11, 13 Mass Moment of Inertia L m m Two spherical connected by a massless bar

thin plate, height b, solid cylinder z length c, rotating length L radius R, around y-axis rotating around z- y axis

Thursday, April 11, 13 Thursday, April 11, 13 Thursday, April 11, 13 Parallel Axis Theorem

G O object rotating around axis d “O” NOT through the CG

Find parallel axis through CG

2 IO = IG + md

IG can read from tables

Thursday, April 11, 13 Composite Bodies O

d1

1 2 3 d2

d2

2 2 2 IO = IG1 + m1d1 + IG2 + m2d2 + IG3 + m3d3

IG can read from tables

Thursday, April 11, 13 rigid body dynamics problems: 2D planar motion • ! • 3 : Fx = max

Fy = may

• problem constraints Mz = I↵ • mass moment of inertia calculation • can we solve? if not, need more eqns: • kinematics equations: connection between ↵, ! AND v, a

Thursday, April 11, 13 refer to 2012 HW#7 for more details and solution

Thursday, April 11, 13