Question 3–12

1 105 Question 3 12A particle of massmis attached to a linear spring with spring constantKandunstretched lengthr0as shown in Fig. P3-12. The spring is attached at its otherend at pointPto the free end of a rigid massless arm of lengthl. The armis hinged at its other end and rotates in a circular path at a constant angularrate . Knowing that the angle is measured from the downward directionand assuming no friction, determine a system of two differential equations ofmotion for the particle in terms ofrand . tlmrKOP Figure P3-12 Solution to Question 3 12 KinematicsFirst, letFbe a fixed reference frame. Then, choose the following coordinatesystem fixed in reference frameF:Origin atOEx=AlongOPWhent=0Ez=Out of PageEy=Ez ExNext, letAbe a reference frame fixed to the arm. Then, choose the followingcoordinate system fixed in reference frameA:Origin atOex=AlongOPez=Out of Page(=Ez)ey=ez exFinally, letBbe a reference frame fixed to the direction along which the springlies ( , the directionPm).

2 Then, choose the following coordinate system fixed106 Chapter 3. Kinetics of Particlesin reference frameB:Origin atOur=AlongPmuz=Out of Page(=Ez=ez)u =uz urThe geometry of the bases{Ex,Ey,Ez},{ex,ey,ez}, and{ur,u ,uz}is shown inFig. 3-10. Using Fig. 3-10, we have the following relationship between the basise t turu eyuz,ez,EzExEyx Figure 3-10 Geometry of Bases{Ex,Ey,Ez},{ex,ey,ez}, and{ur,u ,uz}forQuestion 3 12 .{ex,ey,ez}and the basis{ur,u ,uz}:ex=cos( t)ur sin( t)u ey=sin( t)ur+cos( t)u ( )Next, observing that the basis{ex,ey,ez}rotates with angular rate relativeto the basis{Ex,Ey,Ez}, the angular velocity of reference frameAin referenceframeFis given asF A= Ez= ez( )Next, using Eq. ( ), we observe that the angle formed betweenthe basisvectorsurandex(and similarly betweenu andey) is t. Consequently,the angular velocity of reference frameBin reference frameAis given asA B=( )ez=( )uz( )Finally, observing that the basis{ur,u ,uz}rotates with angular rate relativeto the basis{Ex,Ey,Ez}, we obtain the angular velocity of reference frameBin107reference frameFasF B= uz( )The position of the particle can be written asr=rP+rm/P( )whererPis the position of pointPandrm/Pis the postion of the particle relativeto pointP.

3 In terms of the bases defined above, we have thatrP=Rexrm/P=rur( )Substituting the expressions from Eq. ( ) into Eq. ( ), we obtainr=Rex+rur( )Differentiating the expression for the position as given in Eq.( ) in referenceframeF, we have thatFv=Fdrdt=Fddt(rP)+Fddt rm/P =FvP+Fvm/P( )Now sincerPis expressed in the basis{ex,ey,ez}and{ex,ey,ez}is fixed inreference frameA, we can apply the rate of change transport theorem torPbetween reference framesAandFto giveFvP=Fddt(rP)=Addt(rP)+F A rP( )Now we have thatAddt(rP)=0F A rP= ez Rex=R ey( )Adding the two expressions in Eq. ( ), we obtainFvP=R ey( )Next, sincerm/Pis expressed in the basis{ur,u ,uz}and{ur,u ,uz}is fixed inreference frameB, we can apply the rate of change transport theorem torm/Pbetween reference framesBandFto giveFvm/P=Fddt rm/P =Bddt rm/P +F B rm/P( )108 Chapter 3. Kinetics of ParticlesNow we have thatBddt rm/P = rurF B rm/P= uz rur=r u ( )Adding the two expressions in Eq.

4 ( ), we have thatFvm/P= rur+r u ( )Then, adding Eq. ( ) and Eq. ( ), we obtain the velocity of the particle inreference frameFasFv=FvP+Fvm/P=R ey+ rur+r u ( )Now the acceleration of the particle in reference frameFis given asFa=Fddt Fv =Fddt FvP +Fddt Fvm/P =FaP+Fam/P( )Observing that the expression forFvPas given in Eq. ( ) is expressed in thebasis{ex,ey,ez}and{ex,ey,ez}is fixed in reference frameA, we can applythe rate of change transport theorem toFvPbetween reference framesAandFto giveFaP=Fddt FvP =Addt FvP +A F FvP( )Now sinceRand are constant, we have thatAddt FvP =0F A FvP= ez R ex= R 2ey( )Adding the two expressions in Eq. ( ), we obtain the acceleration of pointPin reference frameFasFaP= R 2ex( )Next, sinceFvm/Pis expressed in the basis{ur,u ,uz}and{ur,u ,uz}is fixedin reference frameB, the acceleration of the particle relative to pointPin refer-ence frameFcan be obtained by applying the rate of change transport theoremtoFvm/Pbetween reference framesBandFasFam/P=Fddt Fvm/P =Bddt Fvm/P +F B Fvm/P( )Now we have thatBddt Fvm/P = rur+( r +r )u F B Fvm/P= uz ( rur+r u )= r 2ur+ r u ( )109 Adding the two expressions in Eq.

5 ( ), we obtain the acceleration of theparticle relative to pointPin reference frameFasFam/P=( r r 2)ur+(2 r +r )u ( )Then, adding Eq. ( ) and Eq. ( ), we obtain the accelerationof the parti-cle in reference frameFasFa= R 2ex+( r r 2)ur+(2 r +r )u ( )Finally, using the expression forexin terms of{ur,u }from Eq. ( ), theacceleration of the particle in reference frameFcan be written in terms of thebasis{ur,u ,uz}asFa= R 2[cos( t)ur sin( t)u ]+( r r 2)ur+(2 r +r )u ( )Simplifying Eq. ( ), we obtainFa=[ r r 2 R 2cos( t)]ur+[2 r +r +R 2sin( t)]u ( )Kinetics and Differential Equations of MotionIn order to obtain the two differential equation of motion for the particle, weneed to apply Newton s 2ndlaw, ,F=mFa. The free body diagram of theparticle is shown in Fig. 3-11. It can be seen that the only force acting on theFsFigure 3-11 Free Body Diagram of Particle for Question 3 12 .particle is due to the linear spring,Fs.

6 Consequently, we have thatFs= K 0 us( )Now we are given that the unstretched length of the spring isrOwhich impliesthat 0=r0. Furthermore, the attachment point of the spring isrA=rP. Conse-quently, the stretched length of the spring is given as = kr rAk = kr rPk( )Using the expression forrfrom Eq. ( ) and the expression forrPfromEq. ( ), we obtain = krur+Rex Rexk = krurk =r( )110 Chapter 3. Kinetics of ParticlesFinally, we have thatus=r rAkr rAk=r rPkr rPk=rurr=ur( )The spring force is then given asFs= K(r r0)ur( )The resultant force acting on the particle is then given asF=Fs= K(r r0)ur( )Then, settingFin Eq. ( ) equal tomFawhereFais obtained from Eq. ( ),we obtain K(r r0)ur=m[ r r 2 R 2cos( t)]ur+m[2 r +r +R 2sin( t)]u ( )Equating components in Eq. ( ), we obtain the following two scalar equa-tions:m[ r r 2 R 2cos( t)]= K(r r0)( )m[2 r +r +R 2sin( t)]=0( )It can be seen that neither Eq. ( ) nor Eq.

7 ( ) contains any unknown re-actions forces. Consequently, Eq. ( ) and Eq. ( ) are the two differentialequations of motion for the particle. We can re-write Eq. ( ) and Eq. ( )in a slightly different form to givem[ r r 2 R 2cos( t)]+K(r r0)=0( )m[2 r +r +R 2sin( t)]=0( )111 Question 3 13A particle of massmslides without friction along a surface in form of a paraboloidas shown in Fig. P3-13. The equation for the paraboloid isz=r22 Rwherezis the height of the particle above the horizontal plane,ris the distancefromOtoQwhereQis the projection ofPonto the horizontal plane, andRisa constant. Knowing that is the angle formed by the directionOQwith thex-axis and that gravity acts downward, determine a system of Figure P3-13 Solution to Question 3 13 KinematicsFirst, letFbe a reference frame fixed to the paraboloid. Then, choose thefollowing coordinate system fixed in reference frameF:Origin atOEx=AlongOxEy=AlongOyEz=Ex Ey112 Chapter 3.

8 Kinetics of ParticlesNext, letAbe a reference frame fixed to the plane formed by the vectorsEzandOQ. Then, choose the following coordinate system fixed in reference frameA:Origin atOer=AlongOQEz=Upe =Ez erThe position of the particle is then given asr=rer+r22 REz( )Furthermore, the angular velocity of reference frameAin reference frameFisgiven asF A= Ez( )The velocity of the particle in reference frameFis then obtained from the rateof change transport theorem asFv=Fdrdt=Adrdt+F A r( )Now we note thatAdrdt= rer+r rREzF A r= Ez rer+r22 REz!=r e ( )Adding the two expressions in Eq. ( ), we obtain the velocity of the particlein reference frameFasFv= rer+r e +r rREz( )Next, applying the rate of change transport theorem toFv, we obtain the accel-eration of the particle in reference frameFasFa=Fddt Fv =Addt Fv +F A Fv( )Now we have thatAddt Fv = rer+( r +r )e + r2+r rREzF A Fv= Ez rer+r e +r rREz = r e r 2er( )Adding the expressions in Eq.

9 ( ), we obtain the accelerationof the particlein reference frameFasFa=( r r 2)er+(r +2 r )e + r2+r rREz( )113 KineticsWe need to apply Newton s 2ndLaw, The free body diagram ofthe particle is shown in Fig. 3-12. We note from Fig. 3-12 thatNis the reactionNmgFigure 3-12 Free Body Diagram of Particle for Question 3 of the paraboloid on the particle and thatmgis the force due to further note thatNmust lienormalto the surface at the point of we note that the vector that is normal to a surface is in the direction ofthe gradient of the function that defines the surface. In order to compute thegradient, we re-write the equation for the paraboloid in the following form:f (r , , z)=z r22R=0( )Then, from calculus, the gradient in obtained in cylindrical coordinates as f= f rer+1r f e + f zEz( )Computing the gradient, we obtain f= rRer+Ez( )The unit vector in the direction of the gradient is then given asen= fk fk= rRer+Ezs1+ rR 2( )which implies that the normal force isN=Nen=N rRer+Ezs1+ rR 2 ( )Next, the force of gravity ismg= mgEz( )114 Chapter 3.

10 Kinetics of ParticlesThe resultant force on the particle is then given asF=N+mg=N rRer+Ezs1+ rR 2 mgEz( )which can be re-written asF= NrRs1+ rR 2er+ N1s1+ rR 2 mg Ez( )SettingF=mFausingFafrom part (a), we obtain NrRs1+ rR 2er+ N1s1+ rR 2 mg Ez=m( r r 2)er+m(r +2 r )e +m r2+r rREz( )We then obtain the following three scalar equations: NrRs1+ rR 2=m( r r 2)0=m(r +2 r ) N1s1+ rR 2 mg =m r2+r rR( )A system of two differential equations can then be obtained as follows. The firstdifferential equation is simply the second equation in Eq. ( ), (r +2 r )=0( )Droppingm, we obtainr +2 r =0( )Next, rearranging the third equation in Eq. ( ) by addingmgto both sides,we obtainN1s1+ rR 2=mg+m r2+r rR( )115 Then, dividing the first equation in Eq. ( ) by this last result, we obtain rR= r r 2 r2+r rR+g( )Rearranging and simplifying this last equation, we obtain the second differentialequation as"1+ rR 2# r+rR2 r2 r 2+grR=0( )The system of two differential equations is then given asr +2 r =0"1+ rR 2# r+rR2 r2 r 2+grR=0( )Conservation of EnergyTwo forces act on the particle:Nandmg.