Work-Energy (WE) for Rigid Bodies

From last class: The WE equation for a system of particles also applies TU11 + -2 = T2 to a system of rigid bodies.

Work terms (ΣU1-2): The same ones for particles (force, weight, spring) also apply to rigid bodies. But there is one new term, the work of a couple. (Rotation is not defined for particles.)

Work of a force: Work of weight:

UForce = (P cos )(d) UWeight = (mg)( h )

Work of friction: Work of a spring: 1 UFriction = -F (d) U ks2 s2 Spring = 2 2 1

Work of a couple: M = couple or torque, lb-ft or N-m = angular displacement, + UM = M in radians M Work-Energy (WE) for Rigid Bodies More on the work of a couple: If a couple, M, is a function of θ, like the torsional spring on a mouse or rat trap, the energy stored in the spring is the area under the M vs. θ curve.

Work of a couple: M = couple or torque, lb-ft or N-m = angular displacement, + UM = M in radians M

If the couple, M, varies with like for a torsional spring:

Area under M( ) curve is energy: UM = M d Examples: M Mouse trap Some hinges

1 2 Work-Energy (WE) for Rigid Bodies

Rigid bodies in general plane motion store kinetic energy in both translation AND rotation, so they have two KE terms.

1 1 mv2 + I 2 T = 2 G 2 G G vG KE stored in KE stored in translation rotation of mass center

Examples: Translation, FA Rotation Fixed Axis Rotation

1 I 2 T = 2 G vG O r G vG = 0 at O.

Translation T = 1 mv2 Examples: Bench grinder 2 G Hand-held grinder Spin awhile after turned off (neglecting tire rotation) due to stored KE. Examples of general plane motion

Bar sliding down wall Rolling Wheel (Slip or no slip) A

G vG G vG

1 1 B mv2 + I 2 T = 2 G 2 G 1 1 mv2 + I 2 T = 2 G 2 G

Kinetic energy is stored in both the translation of the mass center and the rotation of the body. Kinematics can be a

challenge because you need to relate vG’s and ω’s. Example of fixed axis rotation where CG is not at pin:

Fixed Axis Rotation Fixed Axis Rotation c not at the pin. G cG not at the pin. A r

G G r v A G vG

1 1 1 2 1 mv2 + I 2 T = mv + I 2 T = 2 G 2 G 2 G 2 G

Kinematics: v = r Kinematics: vG = r G 1 1 I 2 T = I 2 T = 2 Pin 2 Pin

These show that a body in fixed axis rotation whose cG is not at the pin will have both vG and ω terms. However, there is a simpler 2 equation, T = ½IPinω , which can be used for this case. From where do we get the 1 I 2 T = 2 Pin equation? A r

Consider the slender bar: G Its stored KE is given by: vG 1 mv2 + 1 I 2 = r T = 2 G 2 G

Sub in v = r kinematics: G 1 I + 2 2 T = 2 G mr Factor out 1 and 2 2 = Parallel IPin ! Axis Thm Result: 1 I 2 T = 2 Pin

Only use this equation for fixed axis rotation where vG = rω applies.