Solved Problems Chapter 3: Mechanical Systems
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1 ME 43: Sytem Dynamic and Contro Probem A-3-8- Soved Probem Chapter 3: Mechanica Sytem In Figure 3-3, the impe penduum hown conit of a phere of ma m upended by a tring of negigibe ma. Negecting the eongation of the tring, find a mathematica mode of the penduum. In addition, find the natura frequency of the ytem when i ma. Aume no friction. Soution Aumption:. Ma of the rope i negected compared to the ma of the bob.. Motion occur in one pane ony. 3. Friction in the pivot i negected. 4. Ange i ma. Uing Newton econd aw co h in T m mg Figure 3-3 Simpe Penduum Taking CCW moment about the pivot. The ony fce that i producing moment about the pivot i the weight mg. Thi fce i producing a CW moment. Therefe, the moment produced by mgi mgin. M J && mg in J && Where J m. Therefe, m && + mg in F ma, in, and the above equation of motion impifie to && g + which i of the fm && + ω From the above, the natura frequency of the penduum i n ω n g Chapter 3: Mechanica Sytem
2 ME 43: Sytem Dynamic and Contro Uing The Conervation of Energy Method The potentia energy i V mgh mg ( co The kinetic energy i T mv m& m ( & T + V m ( & + mg ( co d ( T + V m &&& + mg ( & in dt d ( T + V ( m && + mg in & dt Since &, then the term in bracket mut be equa to zero, Therefe m && + gin Or ( && g + Which i imiar to the differentia equation obtained befe. Probem A-3- F the pring-ma-puey ytem of Figure 3-7, the moment of inertia of the puey about the axi of rotation i J and the radiu i R. Aume that the ytem i initiay at equiibrium. The gravitationa fce of ma m caue a tatic defection of the pring uch that k δ t mg. Auming that the dipacement x of ma m i meaured from the equiibrium poition, obtain a mathematica mode of the ytem. In addition, find the natura frequency of the ytem. Soution Aumption:. Lever i rigid and mae.. Reaction at ever P are negected. 3. Dipacement x i ma. 4. The wire i inextenibe. x T T m mg R k δ Uing Newton econd aw Figure 3-7 Chapter 3: Mechanica Sytem
3 ME 43: Sytem Dynamic and Contro Appying Newton econd aw f the ma m F mx&& T mx&& ( Where T i the tenion in the wire (Notice that ince x i meaured from the tatic equiibrium poition the term mg doe not enter into the equation. Appying Newton econd aw f the puey T J&& Eiminating T from equation ( and ( give TR kxr J&& ( && (3 mxr kxr J&& Noting that x R, one can impify the above equation ( && (4 J + mr + + J + mr && (5 which repreent the mathematica mode f the ytem. The natura frequency of the ytem i given by ω n J + mr (6 Uing The Energy Method The kinetic energy of the ytem i T mx& + J& 3 3 Due to Tranation Due to Rotation (7 The potentia energy of the ytem i V kx (8 { Strain energy of the pring Chapter 3: Mechanica Sytem 3
4 ME 43: Sytem Dynamic and Contro Noting that x R, one can impify the above equation T + V mx& + J & + kx T + V mr & + J & + T + V ( mr + J & + (9 d ( T + V ( mr + J &&& + & ( dt Or ( mr + J && + ( which i imiar to the one obtained befe Probem A-3-3 In the mechanica ytem of Figure 3-8, one end of the ever i connected to a pring and a damper, and a fce f ( t i appied to the other end of the ever. Derive a mathematica mode of the ytem. Aume that the dipacement x i ma and the ever i rigid and mae. ( ft P k x b Soution Aumption:. Lever i rigid and mae.. Reaction at ever P are negected. 3. Dipacement x i ma. Figure 3-8 Chapter 3: Mechanica Sytem 4
5 ME 43: Sytem Dynamic and Contro Free Body Diagram (FBD: The FBD i hown in the Figure beow ft ( P F k k x x Figure 3-8 F b b x& Equation of motion: Appying Newton econd aw, f a ytem in rotation about point P give M P f ( t bx & kx DID YOU ASK YOURSELF WHY THE RHS of the above equation i ZERO? Rearranging, one can write: ( ( & f t bx kx + bx& + kx f ( t which i the mathematica mode of the ytem. It i cear that thi differentia equation repreent i a firt der ytem ince it i repreented by a firt der dinary differentia equation with contant coefficient. The RHS of the above differentia equation i not zero. Sytem Repone: In the above ytem, the input i the fce F whie the output i the dipacement x. We wi try to find a reationhip between the input and the output. Taking Lapace tranfm of both ide of the above equation provided we have zero initia condition give Chapter 3: Mechanica Sytem 5
6 ME 43: Sytem Dynamic and Contro where X ( and ( ( b + k X ( F ( F are Lapace tranfm of x and f ( t, repectivey. The above equation can be written a Output X ( C G ( Input F b k b k where C ( where C * ( ( + ( +. The above reation can be written in the tandard fm a: Output X ( Cb C* G ( Input F ( k + + b C b and b k ( F 443 Input C + * ( Tranfer Function ( X 4443 Output x t wi depend on the input fce f ( t. If f ( t i a tep input of magnitude : In thi cae F ( / and C * C * α β X ( F ( The repone ( where α Therefe, and hence C * + C * and β X ( C + + C * + C * Chapter 3: Mechanica Sytem 6
7 ME 43: Sytem Dynamic and Contro ( * x t C e t Chapter 3: Mechanica Sytem 7
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