Chemical Engineering - Apllied Mechanics

Exercise n1 on kinematics and dynamics

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APPLIED MECHANICS Exercise n.1 The mechanical system illustrated in Fig. 1 moves in a vertical plane. The disk of centre O is rolling, without any slippage, on an inclined plane (left side of Fig. 1). The disk has a mass m 1 and a mass moment of inertia J O. The bar OA is hinged at both ends, to the disk and a slider, respectively. The bar, whose barycentre is G 1, has a mass m 2 and a mass moment of inertia J G1. The slider, having a mass m 3 and a barycentre G 2, moves on an inclined plane (right side of Fig. 1) with velocity v and acceleration a. A force F is applied to the top of the slider. The kinetic friction coefficient between the slider and the supporting plane is f k. The rolling friction coefficient and the static friction coefficient between the disk and the plane are f r and f s respectively. All the geometrical parameters are assumed to be known . You are asked to describe the procedure to: 1 evaluate the angular velocity and angular acceleration of the disk; 2 evaluate the absolute velocity and acceleration of the barycentre G of the bar OA; 3 evaluate the drive torque M 1 applied to the disk; 4 evaluate the internal actions transmitted by the bar to the slider, at point A; 5 write the expression of the kinetic energy of the system; 6 describe the procedure to verify that no slippage occurs between disk and plane. Figure 1 System configuration Kinematic analysis Figures 2-a and 2-b show a kinematic model of the system. This model is based on position vectors that describe the absolute and relative position of some significant points of the considered system. The vector 3 33 ieϑ ρρ = =is able to model the absolute position of point A, fixed to the slider. The attitude angle= 3ϑ =is= constant while the magnitude= 3ρ =is a timeϑvarying quantity. We assume that the distance 3ρ , at the current= time, is known. The velocity= 3ρ =and acceleration= 3ρ =correspond to the kinematic parameters= Av =and Aa , respectively, which are input data of the problem.= = In accordance with the kinematic model shown in Figure=3=it is possible to write:= 12 3 12 3 ii iee eϑϑ ϑ ρρ ρ += = (1)= The two unknowns of this 2D vector equation are: 1ρ =and 2ϑ .= a) = = = = = b) c ) Figure 3 Kinematic model = By differentiating eq.(1) with respect to time we obtain:= ( ) 2 13 2 1 223 i iiee e ϑπ ϑϑ ρ ϑρ ρ + +=   = (2)= The two unknowns of this 2D vector equation are:= 1ρ =and 2ϑ . The vector= 1 11 ieϑ ρρ =  =is the absolute velocity of the centre O of the disk, while 2ϑ =is the=angular velocity of the bar OA. The angular velocity of the disk is given by: ( ) 11 OH ρ = −  ω . By differentiating eq.(2) with respect to time we obtain:= () () 22 13 22 2 1 222 23 ii ii e e ie e ϑπϑπ ϑϑ ρ ϑρϑρ ρ ++ ++ =   = (3)= Eq.(3) can be rewritten as:= ( ) ( ) 22 13 2 2 1 222 23 ii iiee e e ϑπϑ π ϑϑ ρ ϑρ ϑρ ρ ++ + +=   = (4)= The two unknowns of this 2D vector equation are: 1ρ =and 2ϑ .=The vector= 1 11 ieϑ ρρ =   =is the absolute acceleration=of the centre O of the disk, while 2ϑ =is the angular=acceleration=of the bar OA.= Now, let us consider the position vectors shown in Figure=3ϑc. The absolute position of the barycentre G 1 of the bar OA can be expressed as: 12 5 14 5 ii iee eϑϑ ϑ ρρ ρ += = (5)= The two unknowns of this 2D vector equation are:= 5ρ =and 5ϑ . By differentiating eq.(5) with respect to time we obtain:= () () 25 15 22 1 2 45 55 ii ii ee ee ϑπϑπ ϑϑ ρ ϑρ ρ ϑρ ++ +=+  = (6)= The two unknowns of this 2D vector equation are:= 5ρ =and 5ϑ. The sum of the two vectors on the right side of eq.(6) gives the absolute velocity of the barycentre G 1. By differentiating eq.(6) with respect to time= we obtain:= ( ) ( ) ( ) ( ) ( ) ( ) 22 1 555 5 5 2 2 1 242 4 222 2 5 5555555 5 ii i iii i i ee e eeee e ϑπϑ π ϑ ϑπϑπϑπϑ π ϑ ρ ϑρ ϑρ ρ ϑρ ϑρ ϑρ ϑ ρ ++ +++ + + += +++ +       = (7)= The two unknowns of this 2D vector equation are: 5ρ =and 5ϑ . The sum of the five vectors on the right= side of eq.(7) gives the absolute acceleration of the barycentre G 1.= = An alternative method can be used to solve the problem. Let us consider=a mobile reference frame with origin at the centre O of the disk (Figure=1). The=x-y axes of this frame do not rotate. Therefore, this is a translational reference frame. According to this reference frame it is possible to write: A rd= + v vv A rO= + v vv = (8)= That is,=the absolute velocity of point A (located on the slider) can be expressed as the sum of the relative velocity vector= rv =(relative with respect to the observer fixed to the translational reference frame) and the drag velocity vector dv . As the distance OA is constant, the relative trajectory of point A is a circle= with centre O and radius OA. The=drag velocity coincides with the absolute velocity of point O. The relative velocity vector rv =is tangent to the corresponding relative trajectory. That is, it is orthogonal to the segment OA.=The two unknowns of the 2D vector eq.(8) are=the magnitude of the=drag velocity vector dv =and the magnitude of the relative velocity vector= rv . The expression of this vector is given by:= () r2 AO =∧− vω = (9)= where 2ω =is the angular=velocity of the bar OA. According to=eq.(9) it is possible to evaluate the angular velocity= 2ω . The drag velocity dv =can be expressed as:= ( ) d1 OH =∧− v ω , where 1ω =is the angular= velocity of the disk. Therefore, the=angular velocity of the disk, 1ω ,=can be obtained from the previous expression.=Vector= rv =coincides with vector ( ) 2 2 2 22 ie ϑπ ρ ρϑ + =   =considered in the previous method used= to solve the kinematic problem, while vector= ≤v =coincides with vector 1 11 ieϑ ρρ =  .= Besides, in accordance with the same mobile reference frame it is possible to write:= A rd r O=+=+ a aa a a = (10)= That is, the absolute acceleration of point A can be expressed as:= ( ) ( ) A rt rn O 22O AO AO 2 = + + = ∧ − −ω − + a aa aa ω = (11)= Vector= rta =is the tangential component of the relative acceleration vector ra , while=vector rna =is the= normal=component of the relative acceleration vector ra .=Therefore, vector rta =is orthogonal to the segment= OA=while vector= rna =is parallel to the segment OA and it is directed from A to O (centripetal acceleration).= As the=drag trajectory is rectilinear (the trajectory described by point O), the normal component of vector da = is null.=The two unknowns of the 2D vector eq.(11) are the magnitude of the=drag acceleration vector dO= aa =and the magnitude of the tangential component of the relative acceleration vector rta . This vector= equation can be easily solved.= The drag acceleration da =can be expressed as:= ( ) d OH d =∧− a ω , where dω =is the angular= acceleration of the disk. Therefore, the angular acceleration of the disk,= 1 ω , can be obtained from the= previous expression.= Vectors rta =and rna =coincide=with vectors= ( ) 2 2 22 ie ϑπ ρϑ +  =and ( ) 2 2 22 ie ϑπ ρϑ +  , respectively, considered in the previous method used to solve the kinematic problem, while vector ≤a =coincides with= vector= 1 11 ieϑ ρρ =   .= Then, the angular acceleration 2ω , which coincides with the angular acceleration of the bar OA., can be obtained from the expression of the vector rta : ( ) rt 2 AO =∧− a ω (12) The same mobile reference frame can be used to evaluate the absolute velocity of the barycentre G 1 of the bar OA. That is: 1G rd = + v vv 1G rO = + v vv (13) The relative trajectory of point G 1 is a circle with centre O and radius OG 1. Then, the relative velocity rv is given by: ( ) r2 1 GO =∧− v ω (14) The two vectors on the right side of eq.(13) are known while the unknown of this vector equation is the vector 1Gv . Then, the absolute acceleration of the barycentre G 1 of the bar OA can be expressed as: ( ) ( ) 1G rt rn O 2 12 1O GO GO 2 = + + = ∧ − −ω − + a aa aa ω (15) All the vectors on the right side of eq.(15) are known while the unknown of this vector equation is the vector 1Ga . Both magnitude and attitude angle of this vector are un known. A numerical solution of this problem has been carried out considering the following input data: Table 1 Input data INPUT DATA KA = 3ρ = 0.5478 m β = 3ϑ = 50� 3ρ = 0.15 m/s KG 1 = 4ρ = 0.2860 m 1 α πϑ = − = 15� 3ρ = 0.08 m/s 2 Figure 4 shows the map of the position vectors of the system evaluated for the given distance KA. Figure 5 shows the map of the velocity vectors of eq.(8). Map of position vectors O A K Map of velocity vectors vO vA vd Figure 4 Position vectors Figure 5 Map of the velocity vectors Figure 6 shows the map of the acceleration vectors of eq.(15). Figure 7 shows the map of the position vectors that define the absolute position of the barycentre G 1 of the bar OA. Ma p of ac c ele r at io n vec t or s aO aA adt adn Map of position vectors O G1 K Figure 6 Map of acceleration vectors Figure 7 Map of the position vectors Figure 8 shows the map of the velocity vectors that define the absolute velocity of the barycentre G 1 of the bar OA. Map of velocity vectors vO vG1 vd Figure 8 Evaluation of the absolute velocity of the barycentre G 1: map of the velocity vectors Figure 9 shows the map of the acceleration vectors that define the absolute acceleration of the barycentre G 1 of the bar OA. Ma p of ac c ele r at io n vec t or s aO aG1 adt adn Figure 9 Evaluation of the absolute acceleration of the barycentre G 1: map of the acceleration vectors Dynamic analysis After solving the kinematics of the system it is possible to perform the dynamic analysis. Now, all the kinematic parameters that must be used to define the inertia actions are known. In order to evaluate the drive torque M 1 applied to the disk it is possible to isolate the global system by removing all the external constraints. Obviously, all the corresponding reaction forces and moments must be considered. Figure 10 shows the isolated system. The force N 2 and the moment M 2 are the reactions transmitted to the slider by the inclined supporting plane (note that the slider cannot rotate with respect to the supporting plane: this causes the presence of the moment M 2). The force T 2 is caused by the kinetic friction between the slider and the supporting plane. This force is opposite to the relative velocity of the slider. The disk is rolling on the inclined supporting plane, without any slippage, that is the relative velocity – at the contact point H – between the disk and the plane is null. The tangential force T 1 is caused by the rolling friction. In this case, in order to take into account the energy dissipated by the rolling friction it is possible to shift the normal reaction force N 1 by a very short distance u (Figure 10). The distance u is given by Rrd uf= , where rf=is the rolling friction coefficient between the disk and the supporting plane, while= Rd= is the disk radius. Note that the moment of the force N 1, evaluated with respect to the pole H, is opposite to the angular velocity 1ω =of the disk (which is clockwise).=The inertia=actions exerted on the disk are the force O 1O i m = − Fa =and the moment= O O1 i J = − C ω .=The inertia actions exerted on the bar OA are the force G11 2G i m = − Fa =and the moment= G11 G2 i J = − C ω . In the end, the inertia actions exerted on the disk is only= given by the force= G22 3G i m = − Fa .= = = Figure 10 Isolated system= = Three equations of motion can be written for the system shown in Figure=10. For instance, the balance of= the horizontal forces, the balance of the vertical forces and the balance of the all the moments about a pole, e.g. point O. Let us consider a 2D Cartesian frame=x-y, with axes x and y that coincide with the horizontal and vertical axes. All the forces can be projected on these to axes. Figure 11 Isolated subϑsystem composed of the= slider only = Figure 12 isolated subϑsystem composed of the bar= OA only = The projections of a generic force jF =can be expressed as:= Fjx=and Fjy. The moment of the same force, applied at point P j, evaluated about the pole O, can be expressed as: ( ) O PO jj j = −∧ MF .= Therefore we can write:= OG 1G2 112 2N T F +F N T +F 0 x x ix i x x x i x++ + + = = OG 1G2 1 12 212 3N T F +F N T +F P P P F 0 y y iy i y y y i y+ + + + + + + += = ( ) 12 11 PO0 nn jj j jj==    −∧ + =      ∑∑ FM = (16)= This system composed of three equations contains the following six unknowns: = 1 M , 1N , 2 N , 1T , 2T , 2 M . In order to solve the problem it is necessary to write further equations.= Therefore, we can isolate the slider from the remaining part of the system (Figure=11). Vectors= AH =and AV =express the internal=actions transmitted to the slider, at point A, by the remaining part of the system. Three equations of motion can be written for the system shown in Figure=11. For instance, the balance of the= horizontal forces, the balance of the vertical forces and the balance of the all the moments about a pole, e.g.= point A. Let us consider a 2D Cartesian frame x-y, with axes x and y that coincide with the horizontal and vertical axes. All the forces can be projected on these to axes. Therefore we can write: G2 A2 2H N T +F 0 x x ix ++ = = G2 A2 23V N T +F P F 0 y y iy + + + += = () 12 11 PA0 nn jj j jj==     −∧ + =       ∑∑ FM = (17)= This system composed of three equations contains the following unknowns: = AH , AV , 2 N , 2T , 2 M . Therefore, we have written=six=equations of motions that contains the following the=eight=unknowns: 1 M , 1N , 2 N , 1T ,= 2T , 2 M , AH , AV . In order to solve the problem it is necessary to write further equations.= Now, we can isolate the bar OA (Figure 12). = Vectors OH =and OV =express the internal actions transmitted to the bar, at point O, by the disk. Three equations of motion can be written for the system shown in Figure=12. For instance, the balance of the= horizontal forces, the balance of the vertical forces and the balance of the all the moments about a pole, e.g.= point O. Let us consider a 2D Cartesian frame x-y, with axes x and y that coincide with the horizontal and vertical axes. All the forces can be projected on these to axes. Therefore we can write: G2 AOH HF 0 iy ++ = = G2 OA2V V +F P 0 iy + += = () 12 11 PO0 nn jj j jj==     −∧ + =       ∑∑ FM = (18)= This system composed of three equations contains the following unknowns: AH , AV , OH , OV ,. Therefore, we have written nine=equations of motions that contains the following the=ten=unknowns: 1 M , 1N , 2 N , 1T , 2T , 2 M , AH , AV , OH , OV . The friction force is given by:= 22TN kf = = (19)= where kf=is the kinetic friction coefficient between the slider and the supporting plane.=This last equation is the tenϑth equation that allows one to evaluate the ten unknowns above listed. These ten equations are= uncouple≤ =each other, therefore they are able to solve the dynamic problem.= This way, we can evaluate the drive torque= 1 M =applied to the disk, all the reaction forces and moments= and the internal actions= AH , AV , OH , OV .= = The kinetic energy of the global system is given by:= { } 1112 2 1O O O d d 2 G G G 1 1 3G G 1 T= 2 mJ m J m ×+ ×+ × + ×+ × vvv vv v ωωωω = (20)= = In order to avoid any slippage between the disk and the supporting plane it is necessary that the following expression is fulfilled:= 11TN sf ≤ = (21)= where sf=is the static friction coefficient between the disk an≤=the supporting plane.= =