4DA00: Dynamics
S6/6 WORK-ENERGY RELATIONS
Erik Steur (e.steur@tue.nl) and Rob Fey (r.h.b.fey@tue.nl)
Department of Mechanical Engineering, Dynamics and Control Group
Work-Energy equation
“integration of Newton’s second law over positions”
Same concept as in S3/6+S3/7 (particles) and S4/3 (systems of particles):
• Work done by forces and couples
• Translational and rotational kinetic energy
• Gravitational and elastic potential energy
• Power and efficiency
Advantages:
• avoids calculation of accelerations
• Direct calculation of changes in velocity as function of forces which do
work
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Work done by forces and couples
• Work done by a force (see S3/6): dU= F ⋅ dr
• Work done by a force over a finite displacement:
U= ∫ ⋅ dr =
F ∫ ( F cos α )ds = ∫ F ds
t
• Work done by a couple M = Fb :
=dU F= (bdθ ) M dθ
due to rotation about A '
• Work done by a couple over a finite rotation:
θ2
U = ∫ M dθ
θ1
is positive if M is in the direction of rotation, negative otherwise
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Kinetic energy: translation Kinetic energy: rotation
T = 1/ 2∑ mi vi2 = 1/ 2∑ mi vG2 T = 1/ 2∑ mi vi2 = 1/ 2∑ mi (riω ) 2
= 1 / 2vG2 ∑ mi 2ω 2 ∑ mi ri 2 1/ 2ω 2 ∫ r 2 dm
= 1/=
1/ 2mvG2
⇒ T= 1/ 2 I Oω 2
⇒ T=
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Kinetic energy: general plane motion
Sum the kinetic energy over all mass elements: i
i
T 1/ 2∑ m=
v 1/ 2∑ mi (v + ( ρiω ) + 2vG ρiω cos θi )
G
2 2 2
= i i G
G
(use the law of cosines, p. 305)
• First term: 1/ 2∑ mi vG2 = 1/ 2mvG2
• Second term: 1/ 2∑ mi ( ρiω ) 2 = 1/ 2 I Gω 2 Kinetic energy expressed in
• Third term: ∑ mi vG ρiω cos θi = vgω ∑ mi yi = 0 relation to instantaneous
center of zero velocity C :
1/ 2mvG2 + 1/ 2 I Gω 2
⇒ T= T = 1/ 2 I Cω 2
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Gravitational potential energy Elastic potential energy
Vg = 0 Translation Rotation
Force/Moment F = kx [ N ] M = kT θ [ Nm]
• Particle: Vg = mgh Stiffness k [ N / m] kT [ Nm / rad ]
• Rigid Body: Vg = mghG Energy (in [ J ] ) Ve = 1 / 2k x 2 Ve = 1 / 2kT θ 2
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Work-Energy relation for (coupled) rigid bodies
T1 + U1− 2 =
T2
or
T1 + (Vg )1 + (Ve )1 + U1′− 2 =T2 + (Vg ) 2 + (Ve ) 2 or U1′− 2 = ∆T + ∆Vg + ∆Ve
where
• Ti is the kinetic energy in position i
• U1− 2 is the work done by all external forces/moments acting on the rigid body
• (Vg )i and (Ve )i are the gravitational and elastic potential energy in
position i , respectively
• U1′− 2 is the work done by all external forces/moments acting on the rigid body
except for gravitational and elastic forces/moments
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Problem 6/141 (7th edition)
g = 9.81[m / s 2 ]
Determine the angular velocity θ [rad / s ] at the
moment the lid closes ( θ = π / 2 [rad ] )
Given:
• Identical torsion springs in hinges A and B:
• torsional stiffness: K [ Nm / rad ]
• springs unstretched at θ = 0 [rad ]
• Lid has mass m [kg ] and length l [m]
• Lid is released from rest at θ = 0 [rad ]
• No friction
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Problem 6/141 (7th edition)
Determine the angular velocity θ [rad / s ] at the
moment the lid closes ( θ = π / 2 [rad ] )
Solution
• Kinetic energy:
∆T = T2 − T1 = 1/ 2 I Oθ 2 − 0 = 1/ 6 ml 2θ 2
• Potential energy
∆Vg= (Vg ) 2 − (Vg )1 = 0 − mg (l / 2) = −1/ 2mgl
∆V=e (Ve ) 2 − (Ve )=
1 1/ 2(2 K )(π / 2) 2 − 0 = 1/ 4 K π 2
• Work-Energy equation
2 2
U1′− 2 =∆T + ∆Vg + ∆Ve ml=
= θ / 6 − mgl / 2 + K π 2 / 4 0
6 mgl − 3 K π 2
⇒ θ =
2ml 2
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Questions/comments: e.steur@tue.nl
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