Rigid Body :: Virtual Work- relation for an infinitesimal : dU’ = dT + dV (dU’ :: total work done by all active ) For interconnected systems, differential change in KE for the entire system:

For each body:

dsi̅ :: infinitesimal linear disp of the center of dθi :: infinitesimal angular disp of the body in the plane of

ME101 - Division III Kaustubh Dasgupta 1 Kinetics :: Virtual Work

Now, a̅ i ·ds̅ i is identical to (a̅ i )t ds̅ i

αi :: angular accln θ̈i of the body

(a̅ i )t :: component of a̅ i along tangent to the curve described by mass center of the body 

 th Ri :: resultant and acting on i body th MGi :: resultant couple acting on i body

dθi :: dθi k  Differential change in = Differential work done by the resultant forces and couples

ME101 - Division III Kaustubh Dasgupta 2 Rigid Body Kinetics :: Virtual Work dU’ = dT + dV (dU’ :: total work done by all active forces) dV :: differential change in total Vg and Ve

hi :: vertical distance of the center of mass mi above a convenient datum plane xi :: deformation of elastic member (spring of stiffness kj) of system

(+ve for same dirn. of accn and disp)

 Direct relation between the and the active forces

Virtual Work

ME101 - Division III Kaustubh Dasgupta 3 Rigid Body Kinetics :: Virtual Work • – Virtual work eqn

• Kinetics

• If a rigid body is in equilibrium • total virtual work of external forces acting on the body is zero for any virtual displacement of the body

ME101 - Division III Kaustubh Dasgupta 4 Example (1) on virtual work

Solution: : Work done by the weight and spring will be accounted for in the PE terms : No other external force except 80 N do any work For an infinitesimal upward disp dx of rack A, work done on the system: dU’ = 80 dx dU’ = dT + dV

ME101 - Division III Kaustubh Dasgupta 5 Example (1) on virtual work

Since disp and accln of mass center of gear will be half that of rack A

Substituting in the work-energy equation:

 a = 4.43 m/s2

ME101 - Division III Kaustubh Dasgupta 6 Example (2) on virtual work

Solution:

Consider virtual displacements s1 and s2 respectively for mass centres of the bars from the assumed natural position during the accln  δU’ = 0

ME101 - Division III Kaustubh Dasgupta 7 Example (2) on virtual work

For virtual displacements in steady state, α = 0 for both bars

Choosing horz line through A as the datum for zero :  PE of the links:

 Virtual change in PE:

Substituting in the work-energy eqn for virtual changes:

ME101 - Division III Kaustubh Dasgupta 8 Rigid Body Kinetics :: / Impulse-Momentum equations are useful . when the applied forces are expressed as functions of . when interaction between particles or rigid bodies occur during short periods of time, such as, impact Linear Momentum Linear momentum of any mass system, rigid or non-rigid: v̅ is the if mass center

F dt = Linear Impulse

ME101 - Division III Kaustubh Dasgupta 9 Rigid Body Kinetics :: Impulse/Momentum

ρi̇ = of mi wrt G and its magnitude = ρi ω 2 Magnitude of ρi x mi ρi̇ = ρi ωmi With ω = ωk

Moment-Angular Momentum Relation:

and

M dt = Angular Impulse

ME101 - Division III Kaustubh Dasgupta 10 Rigid Body Kinetics :: Impulse/Momentum Angular Momentum

•G and HG have vector properties analogous to those of the resultant force and couple  Angular momentum @ any point O:

 Applicable at any particular instant of time @ O, which may be a fixed or moving point on or off the body When the body rotates @ a fixed point O on the body or body extended, v = r̅ ω and d = r̅ 

Similarly,

and

ME101 - Division III Kaustubh Dasgupta 11 Rigid Body Kinetics :: Impulse/Momentum Interconnected Rigid Bodies

For a single rigid body,

O :: fixed reference point for the entire system For the system,

Over a finite time interval,

ME101 - Division III Kaustubh Dasgupta 12 Rigid Body Kinetics :: Impulse/Momentum Conservation of Momentum For a single rigid body or a system of interconnected rigid bodies:

if ∑F = 0 for a given interval of time  linear-momentum vector undergoes no change in the absence of a resultant linear impulse. For interconnected rigid bodies, linear momentum may change in individual parts during the interval, but there will be no resultant momentum change for the whole system if there is no resultant linear impulse

Similarly, if ∑MO = 0 or ∑MG = 0 for a given interval of time

angular-momentum either @ a fixed point or @ mass center undergoes no change in the absence of a corresponding resultant angular impulse For interconnected rigid bodies, ang momentum may change in individual parts, but no change for the whole system if there is no resultant angular impulse

ME101 - Division III Kaustubh Dasgupta 13 Rigid Body Kinetics :: Impulse/Momentum Impact of Rigid Bodies

•Complex phenomenon if central impact is not considered.

•Will not be discussed in ME101

•Principles of conservation of Linear and Angular Momentum can be used when they are applicable when discussing impact and other interactions of rigid bodies.

ME101 - Division III Kaustubh Dasgupta 14 Example (1) on impulse/momentum A force P, applied to the cable wrapped around the wheel, is increased slowly according to P = 6.5t, where P is in N and t is time in sec after P is first applied. At time t = 0, the wheel is rolling (without slipping) to the left with a velocity of its center of 0.9 m/s. Wheel has a mass of 60 kg and radius of gyration @ its center is 250 mm. Determine the of the wheel 10 seconds after P is applied.

Solution Draw the impulse-momentum diagrams of the wheel at initial and final stages

ME101 - Division III Kaustubh Dasgupta 15 Example (1) on impulse/momentum Solution

Apply linear and angular impulse-momentum eqns over the entire interval

10 60(0.9)  (6.5t  F)dt  60(0.45 ) 0 2

2  0.9  10 2 60(0.25)     [0.45F  0.225(6.5t)]dt  60(0.25) 2  0.45  0 Eliminating F by multiplying 2nd eqn by 1/0.45 and adding the two eqns.

Integrating and solving  ω2 = 2.6 rad/s (clockwise)

ME101 - Division III Kaustubh Dasgupta 16