Rigid Body Kinetics :: Force/Mass/Acc

Rigid Body Kinetics ForceMassAccGeneral Equations of Motion

G is the mass center of the body

ActionDynamic

Response

1ME101 - Division III Kaustubh Dasgupta

Rigid Body Kinetics ForceMassAccFixed Axis Rotationbull All points in body move in a circular path rotation axis

bull All lines of the body have the same ω and α

Accln comp of mass center an = r ω2 and at = r α

Two scalar comp of force eqns

ΣFn = m r ω2 and ΣFt = m r α

Moment of the resultants rotn axis O

Using parallel axis theorem


Rigid Body Kinetics ForceMassAccGeneral Plane Motion

bullCombines translation and rotation


Choice of moment eqn (a) mass centre G

(b) point P with known acc

Example (1) on general motionA metal hoop with radius r = 150 mm is released from rest on the 20o

incline Coefficients of static and kinetic friction are given Determine the

angular accln α of the hoop and the time t for the hoop to move a distance

of 3 m down the incline

Solution Draw the FBD and the KD


Example (1) on general motionIf the wheel slips when it rolls a ne rα

Check whether the hoop slips or not when it rolls

Assume that the hoop rolls without slipping

a = r α (Pure rolling)

Eliminating F and substituting a = rα a = 1678 ms2

a is independent of both m and r

Alternatively we may use

same relation

Check for the assumption of no slipping

Calculate F and N and compare F with its limiting value


Example (1) on general motionCheck for the assumption of no slipping

Maximum possible friction force is

Since the limiting value Fmax lt F

The assumption of pure rolling was incorrect The hoop slips as it rolls a ne rα

The friction force becomes the kinetic value

Repeat the calculations with this F

a = 225 ms2

α = 737 rads2 (α dependent on r but not on m)

at x = 3 m t = 1633 s


Example (2) on general motion



bull Solution FBD and KD

Horizontal component of acceleration of point D

Assuming rolling without slipping

Acceleration of mass centre G




Plane Kinetics of Rigid Bodies

Work and Energy

Advantages of Work Energy Method

bullThese principles are especially useful in describing motion resulting from

the cumulative effect of forces acting through distances

bullIf the forces are conservative velocity changes can be determined by

analyzing the energy conditions only at the beginning and at the end of

the motion interval

bullFor finite displacements no need to compute acceleration leads directly

to velocity changes as functions of forces which do work

bullInvolves only those forces which do work and thus produces change in

magnitudes of velocities

Simplifies calculations



Work and EnergyWork done by force F

or

ds is the magnitude of the vector displacement dr

Work done by couple MWork done by both forces (F) during translation

part of motion cancel each other (opposite dirn)

there4 Net work done will be the work done by the couple

(due to rotational part of motion)

During finite rotation work done



Work and EnergyKinetic Energy Three classes of rigid body plane motion

Translation

bull All particles will have same velocity

For entire body T = sumfrac12 miv2

(both rectilinear and curvilinear)

Fixed Axis Rotation

For a particle mi Ti = frac12 mi (riω)2

For entire body T = frac12 ω2 summi ri2 = frac12 ω2 Io

General Plane Motion

Third summation

since

T =

Also KE in terms of rotational vel

the instantaneous center C of zero vel



Work and EnergyPotential Energy and Work-Energy EquationWork-Energy relation for the motion of a general system of particles

(U1-2 is work done by all external forces)

(Ursquo1-2 is work done by all external forces other than

weight and spring forces which are taken care of

by means of potential energy rather than work)

bullIn case of interconnected system of rigid bodies the work-energy equations

include the effect of stored elastic energy in the connections

bullUrsquo1-2 includes the negative work of internal friction forces

bullWork-energy method is most useful for analyzing conservative systems of

interconnected bodies where energy loss due to -ve work of friction is negligible



Work and EnergyPowerbull Time rate at which work is performed

Work done at a given instant by a force F acting on a rigid body in plane motion

(v is the velocity at the point of application of the force)

Work done at a given instant by a couple M acting on a rigid body in plane motion

(ω is the angular velocity of the body)

bull M and ω have same senses +ve Power Energy is supplied to the body

bull M and ω have opposite senses -ve Power Energy is removed from the body

Total instantaneous power if both F and M acting simultaneously

P = Fv + M ω



Work and EnergyPowerPower can also be expressed as rate of change of total mechanical energy of the

system of rigid bodies

Work-energy relation for an infinitesimal displacement

dUrsquo = dT + dV

dUrsquo is the work done by the active forces and couples applied to the bodies

Dividing by dt

Total power developed by the active forces and couples equals the rate of

change of the total mechanical energy of the bodies or system of bodies

Since

R is resultant of all forces acting on body amp M is the resultant moment G of all forces

Dot product accounts for the case of curvilinear motion of the mass center where v and a

are not in the same direction



Work and EnergyExample

Solution

bull Wheel has a general plane motion

bull Draw the FBD of the wheel



Work and EnergyExampleSolution


bull Only 100 N and 40x981 = 392 N forces do work

C is the instantaneous center of zero velocity

Vel of point A vA = [(200+100)100]v = 3v

Point A on the cord moves a dist of

(200+100)100 = 3 times that of O = 3x3 = 9m

Including the effect of weight in U term

The work done by the wheel

U1-2 = 100(9) ndash 392sin15(3) = 595 J

For general plane motion KE

T1 = 0 T2 = frac12 (40)(01ω)2 + frac12 (40)(015)2ω2 = 065 ω2

since vel of the center of the wheel v = r ω = 01 ω




Work-energy eqn

0 + 595 = 065 ω2

ω = 303 rads

We may also calculate the KE of the wheel using

KE in terms of rotational vel the instantaneous

center C of zero vel

Same relation

Power input from 100 N force when ω = 303 rads

P = Fv

P = 100(03)(303) = 908 W

Ic = I + m|OC|2 and I = Io = mko2




Solution

bull Plane Motion

bull Conservative since friction forces

can be neglected

bull Choosing the datum for zero

gravitational potential energy vg

through O




Let us define three states

at θ = 45o θ = 0o amp max spring deflection

At θ = 45o the wheel starts from rest

At θ = 0o the wheel momentarily comes to rest

For the interval from θ = 45o to θ = 0o the initial and final KE of the wheels = 0

At State 2 Each link is rotating O Total KE

During this interval the collar B drops a distance 0375radic2 = 0265 m

PE

V2 is zero since the collar and the links reaches the datum

There are no active forces that are doing work (other than the weights which are

included in PE) Ursquo1-2 = 0 Work-Energy equation




At the third state (max deflection of spring)

all parts of the system are momentarily at rest

KE T3 = 0

Using the Work-Energy equation between states 1 and 3

Maximum deformation of the spring

x = 601 mm (positive root)

Try solving this problem without using the work-energy equation


Rigid Body Kinetics ForceMassAccFixed Axis Rotationbull All points in body move in a circular path rotation axis

bull All lines of the body have the same ω and α

Accln comp of mass center an = r ω2 and at = r α

Two scalar comp of force eqns

ΣFn = m r ω2 and ΣFt = m r α

Moment of the resultants rotn axis O

Using parallel axis theorem




















same relation










a = 225 ms2


at x = 3 m t = 1633 s













Work and Energy






the motion interval









or










Translation




Fixed Axis Rotation




Third summation

since

T =

























P = Fv + M ω






dUrsquo = dT + dV


Dividing by dt



Since







Solution











(200+100)100 = 3 times that of O = 3x3 = 9m



U1-2 = 100(9) ndash 392sin15(3) = 595 J


T1 = 0 T2 = frac12 (40)(01ω)2 + frac12 (40)(015)2ω2 = 065 ω2





Work-energy eqn

0 + 595 = 065 ω2

ω = 303 rads




Same relation


P = Fv

P = 100(03)(303) = 908 W





Solution

bull Plane Motion


can be neglected



through O











PE









KE T3 = 0
























same relation










a = 225 ms2


at x = 3 m t = 1633 s













Work and Energy






the motion interval









or










Translation




Fixed Axis Rotation




Third summation

since

T =

























P = Fv + M ω






dUrsquo = dT + dV


Dividing by dt



Since







Solution











(200+100)100 = 3 times that of O = 3x3 = 9m



U1-2 = 100(9) ndash 392sin15(3) = 595 J


T1 = 0 T2 = frac12 (40)(01ω)2 + frac12 (40)(015)2ω2 = 065 ω2





Work-energy eqn

0 + 595 = 065 ω2

ω = 303 rads




Same relation


P = Fv

P = 100(03)(303) = 908 W





Solution

bull Plane Motion


can be neglected



through O











PE









KE T3 = 0



















same relation










a = 225 ms2


at x = 3 m t = 1633 s













Work and Energy






the motion interval









or










Translation




Fixed Axis Rotation




Third summation

since

T =

























P = Fv + M ω






dUrsquo = dT + dV


Dividing by dt



Since







Solution











(200+100)100 = 3 times that of O = 3x3 = 9m



U1-2 = 100(9) ndash 392sin15(3) = 595 J


T1 = 0 T2 = frac12 (40)(01ω)2 + frac12 (40)(015)2ω2 = 065 ω2





Work-energy eqn

0 + 595 = 065 ω2

ω = 303 rads




Same relation


P = Fv

P = 100(03)(303) = 908 W





Solution

bull Plane Motion


can be neglected



through O











PE









KE T3 = 0













same relation










a = 225 ms2


at x = 3 m t = 1633 s













Work and Energy






the motion interval









or










Translation




Fixed Axis Rotation




Third summation

since

T =

























P = Fv + M ω






dUrsquo = dT + dV


Dividing by dt



Since







Solution











(200+100)100 = 3 times that of O = 3x3 = 9m



U1-2 = 100(9) ndash 392sin15(3) = 595 J


T1 = 0 T2 = frac12 (40)(01ω)2 + frac12 (40)(015)2ω2 = 065 ω2





Work-energy eqn

0 + 595 = 065 ω2

ω = 303 rads




Same relation


P = Fv

P = 100(03)(303) = 908 W





Solution

bull Plane Motion


can be neglected



through O











PE









KE T3 = 0












a = 225 ms2


at x = 3 m t = 1633 s













Work and Energy






the motion interval









or










Translation




Fixed Axis Rotation




Third summation

since

T =

























P = Fv + M ω






dUrsquo = dT + dV


Dividing by dt



Since







Solution











(200+100)100 = 3 times that of O = 3x3 = 9m



U1-2 = 100(9) ndash 392sin15(3) = 595 J


T1 = 0 T2 = frac12 (40)(01ω)2 + frac12 (40)(015)2ω2 = 065 ω2





Work-energy eqn

0 + 595 = 065 ω2

ω = 303 rads




Same relation


P = Fv

P = 100(03)(303) = 908 W





Solution

bull Plane Motion


can be neglected



through O











PE









KE T3 = 0

















Work and Energy






the motion interval









or










Translation




Fixed Axis Rotation




Third summation

since

T =

























P = Fv + M ω






dUrsquo = dT + dV


Dividing by dt



Since







Solution











(200+100)100 = 3 times that of O = 3x3 = 9m



U1-2 = 100(9) ndash 392sin15(3) = 595 J


T1 = 0 T2 = frac12 (40)(01ω)2 + frac12 (40)(015)2ω2 = 065 ω2





Work-energy eqn

0 + 595 = 065 ω2

ω = 303 rads




Same relation


P = Fv

P = 100(03)(303) = 908 W





Solution

bull Plane Motion


can be neglected



through O











PE









KE T3 = 0















Work and Energy






the motion interval









or










Translation




Fixed Axis Rotation




Third summation

since

T =

























P = Fv + M ω






dUrsquo = dT + dV


Dividing by dt



Since







Solution











(200+100)100 = 3 times that of O = 3x3 = 9m



U1-2 = 100(9) ndash 392sin15(3) = 595 J


T1 = 0 T2 = frac12 (40)(01ω)2 + frac12 (40)(015)2ω2 = 065 ω2





Work-energy eqn

0 + 595 = 065 ω2

ω = 303 rads




Same relation


P = Fv

P = 100(03)(303) = 908 W





Solution

bull Plane Motion


can be neglected



through O











PE









KE T3 = 0









Work and Energy






the motion interval









or










Translation




Fixed Axis Rotation




Third summation

since

T =

























P = Fv + M ω






dUrsquo = dT + dV


Dividing by dt



Since







Solution











(200+100)100 = 3 times that of O = 3x3 = 9m



U1-2 = 100(9) ndash 392sin15(3) = 595 J


T1 = 0 T2 = frac12 (40)(01ω)2 + frac12 (40)(015)2ω2 = 065 ω2





Work-energy eqn

0 + 595 = 065 ω2

ω = 303 rads




Same relation


P = Fv

P = 100(03)(303) = 908 W





Solution

bull Plane Motion


can be neglected



through O











PE









KE T3 = 0







Work and Energy






the motion interval









or










Translation




Fixed Axis Rotation




Third summation

since

T =

























P = Fv + M ω






dUrsquo = dT + dV


Dividing by dt



Since







Solution











(200+100)100 = 3 times that of O = 3x3 = 9m



U1-2 = 100(9) ndash 392sin15(3) = 595 J


T1 = 0 T2 = frac12 (40)(01ω)2 + frac12 (40)(015)2ω2 = 065 ω2





Work-energy eqn

0 + 595 = 065 ω2

ω = 303 rads




Same relation


P = Fv

P = 100(03)(303) = 908 W





Solution

bull Plane Motion


can be neglected



through O











PE









KE T3 = 0








or










Translation




Fixed Axis Rotation




Third summation

since

T =

























P = Fv + M ω






dUrsquo = dT + dV


Dividing by dt



Since







Solution











(200+100)100 = 3 times that of O = 3x3 = 9m



U1-2 = 100(9) ndash 392sin15(3) = 595 J


T1 = 0 T2 = frac12 (40)(01ω)2 + frac12 (40)(015)2ω2 = 065 ω2





Work-energy eqn

0 + 595 = 065 ω2

ω = 303 rads




Same relation


P = Fv

P = 100(03)(303) = 908 W





Solution

bull Plane Motion


can be neglected



through O











PE









KE T3 = 0








Translation




Fixed Axis Rotation




Third summation

since

T =

























P = Fv + M ω






dUrsquo = dT + dV


Dividing by dt



Since







Solution











(200+100)100 = 3 times that of O = 3x3 = 9m



U1-2 = 100(9) ndash 392sin15(3) = 595 J


T1 = 0 T2 = frac12 (40)(01ω)2 + frac12 (40)(015)2ω2 = 065 ω2





Work-energy eqn

0 + 595 = 065 ω2

ω = 303 rads




Same relation


P = Fv

P = 100(03)(303) = 908 W





Solution

bull Plane Motion


can be neglected



through O











PE









KE T3 = 0



























P = Fv + M ω






dUrsquo = dT + dV


Dividing by dt



Since







Solution











(200+100)100 = 3 times that of O = 3x3 = 9m



U1-2 = 100(9) ndash 392sin15(3) = 595 J


T1 = 0 T2 = frac12 (40)(01ω)2 + frac12 (40)(015)2ω2 = 065 ω2





Work-energy eqn

0 + 595 = 065 ω2

ω = 303 rads




Same relation


P = Fv

P = 100(03)(303) = 908 W





Solution

bull Plane Motion


can be neglected



through O











PE









KE T3 = 0















P = Fv + M ω






dUrsquo = dT + dV


Dividing by dt



Since







Solution











(200+100)100 = 3 times that of O = 3x3 = 9m



U1-2 = 100(9) ndash 392sin15(3) = 595 J


T1 = 0 T2 = frac12 (40)(01ω)2 + frac12 (40)(015)2ω2 = 065 ω2





Work-energy eqn

0 + 595 = 065 ω2

ω = 303 rads




Same relation


P = Fv

P = 100(03)(303) = 908 W





Solution

bull Plane Motion


can be neglected



through O











PE









KE T3 = 0










dUrsquo = dT + dV


Dividing by dt



Since







Solution











(200+100)100 = 3 times that of O = 3x3 = 9m



U1-2 = 100(9) ndash 392sin15(3) = 595 J


T1 = 0 T2 = frac12 (40)(01ω)2 + frac12 (40)(015)2ω2 = 065 ω2





Work-energy eqn

0 + 595 = 065 ω2

ω = 303 rads




Same relation


P = Fv

P = 100(03)(303) = 908 W





Solution

bull Plane Motion


can be neglected



through O











PE









KE T3 = 0








Solution











(200+100)100 = 3 times that of O = 3x3 = 9m



U1-2 = 100(9) ndash 392sin15(3) = 595 J


T1 = 0 T2 = frac12 (40)(01ω)2 + frac12 (40)(015)2ω2 = 065 ω2





Work-energy eqn

0 + 595 = 065 ω2

ω = 303 rads




Same relation


P = Fv

P = 100(03)(303) = 908 W





Solution

bull Plane Motion


can be neglected



through O











PE









KE T3 = 0













(200+100)100 = 3 times that of O = 3x3 = 9m



U1-2 = 100(9) ndash 392sin15(3) = 595 J


T1 = 0 T2 = frac12 (40)(01ω)2 + frac12 (40)(015)2ω2 = 065 ω2





Work-energy eqn

0 + 595 = 065 ω2

ω = 303 rads




Same relation


P = Fv

P = 100(03)(303) = 908 W





Solution

bull Plane Motion


can be neglected



through O











PE









KE T3 = 0








Work-energy eqn

0 + 595 = 065 ω2

ω = 303 rads




Same relation


P = Fv

P = 100(03)(303) = 908 W





Solution

bull Plane Motion


can be neglected



through O











PE









KE T3 = 0








Solution

bull Plane Motion


can be neglected



through O











PE









KE T3 = 0















PE









KE T3 = 0










KE T3 = 0






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Rigid Body Kinetics :: Force/Mass/Acc

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