2.003 Engineering Dynamics Problem Set 2 Solutions

Transcription

1 .003 Engineering Dynaics Proble Set Solutions This proble set is priarily eant to give the student practice in describing otion. This is the subject of kineatics. It is strongly recoended that you study the reading on kineatics provided with the course aterials. You should also read the handout on velocities and accelerations of oving bodies, which is referred to in this solution. Proble : A sall box of negligible size slides down a curved path defined by the parabola y = 0.4x. When it is at point A(x A =, y A =.6), the speed of the box is V box = 8/s and the tangential acceleration due to gravity is dv box /dt = 4/s. a. Deterine the noral coponent and the total agnitude of the acceleration of the box at this instant. b. Draw a free body diagra and deterine if the tangential acceleration is consistent with the forces acting on the body. In particular, can you deterine the friction coefficient which is consistent with the data given? a. The noral coponent of the acceleration is caused by the curvature of the path. The radius of curvature ay be obtained fro a well-known relationship fro analytic geoetry.

2 y 0.4x d y At (x,y)=(,.6) dy 0.8 x x 0.8(.0).6 dx x 0.8 dx The radius of curvature is given by: 3/ dy 3/ dx.6 = = 8.4 d y 0.8 dx The total acceleration on the box is the su of the noral and tangential coponents. a a u a u A/ t t n n a 4 / s a t n va / (8 / s) 7.63 / s 8.4 a a a / s A/ n t b. The free body diagra is on the left. In the direction tangent to the slope the ass ties the acceleration ust be equal to the su of the friction force and the tangential coponent of the gravitational force on the box. This is worked out in the equations below.

3 In the direction noral to the curve: va / Man M Fn N Mg cos( ) dy tan ( ) tan (.6) 58 dx va / () N M Mg cos( ) Tangential to the curve Newton's nd law provides that () Ft Mat N Mg sin( ) M (4 / s ) Substituting for N fro () into () leads to v A / Mat M Mg cos( ) Mg sin( ) M (4 / s ) M cancels out and the reaining quantities are known, allowing to be found. = Proble : The 0.5-lb ball is guided along a circular path, which lies in a vertical plane. The length of the ar P is equal to r P/ = rc cos θ. If the ar has an angular velocity θ = 0.4rad/s and an angular acceleration θ = 0.8rad/s at the instant θ = 30 o, deterine the force of the ar on the ball. Neglect friction and the size of the ball. Set rc = 0.4ft. Figure Solution: Fro the concept question we deduced that because friction is neglected, the force of the ar on the ball is perpendicular to the ar. We begin by describing the otion: one of the ost difficult steps in doing this proble is choosing a set of coordinates. There are any possibilities. In this case we seek 3

4 answers in ters of directions which are noral and parallel to the oving rod A and the angular rotation rate and angular acceleration are specified for the rod. This suggests that an appropriate set of coordinates would be polar coordinates with origin at, which describe the otion of the rod. The r vector is parallel to the rod, and the angle is defined as the angle between the rod and the horizontal, as shown in the figure. The ball follows the sei-circular track. The two unknowns in the proble are the force F n which is perpendicular to the curved track and the force F P which is perpendicular to the rod. First the kineatics: Fro trigonoetry it is known that r=r ccos( ). The other kineatic facts for the proble are: 0.4feet r c 30 degrees 0.4 radians/s 0.8 radians/sec r rr Expressions are needed for velocity and acceleration of the ass at point P. A set of polar coordinates, r and, are defined directly with respect to the inertial frae : dr V P / rr r () dt /r z dr a P / r r r r r () dt /r z The set of unit vectors r and define the direction of the r vector and the direction perpendicular to the r vector. is perpendicular to r, but it is not a unit vector which defines the direction of the angle. The angle is just a scalar coordinate which relates the position of the r vector in the reference frae in which it is defined. More on this confusing topic later on. Expressions are needed for r and r, which ay be obtained fro r=r cos( )r c. However these derivatives are tedious because of the tie derivatives of the unit vectors that are required. It is in fact easier to copute the velocity and acceleration of the ass in ters of polar coordinates based on the radius r c, which rotates about point B and use unit vectors r and. This is shown in the figure below. nce V P/ and a P/ are known in ters of the unit vectors r and it is then possible to convert the into expressions in ters of the unit vectors rand. This is done using the following relations. r z r cos( ) rsin( ) sin( ) r cos( ) (3) This approach is easier because the ass exhibits pure circular otion about point B and the velocity and acceleration expressions are well known. Because point B is a fixed point in the 4

5 r z inertial frae, then a non-rotating frae at B is also an inertial frae. Velocities and accelerations easured with respect to a frae at B are the sae as would be observed fro the frae r z. Therefore V P/ and a P/ easured with respect to B are the sae as with respect to r z. Using polar coordinates with origin at point B, V P/ and a P/ are easily obtained. r r r c Let, and dr V r r r (4) P / dt /xyz 0 r r c c This is intuitively satisfying because it is siply circular otion about point B. Taking another tie derivative of V P / in equation (4) with respect to the fixed inertial frae leads to: a P / r r r r r a P/ rc rc r rc rc a P/ 0 rc r rc 0 rc r rc As expected of pure circular otion, there is a centripetal acceleration ter and an Eulerian acceleration ter. These can be expressed in ters of r and unit vectors by the transforation given above in equation (3). r cos( ) rsin( ) sin( ) r cos( ) which leads to the following expression in ters of r and unit vectors. a P/ rc cos( ) r sin( ) rc sin( ) r cos( ) (5) ft ft a P/ r cos( c ) r c sin( ) r r c sin( ) r c cos( ) r (6) s s 5

6 The above expression is not intuitively obvious. However it is now in ters of coordinates aligned with the rod P, which akes it uch easier to evaluate the noral force the rod exerts on the ass. This copletes the kineatics portion of the proble. Next the appropriate physical laws ust be applied. In this case Newton s nd law is the priary one that is needed. By suing forces fro the free body diagra two equations ay be obtained relating force to acceleration of the ass in directions noral to and parallel to the rod. The proble has two unknown forces, F P and F N, as shown in the figure above. F N is the force that the rod exerts on the ass. It is perpendicular to the rod because there is no friction. F P is the force on the ass exerted by the circular curve that the ass ust follow. F P is noral to the surface of the circular curve, again because there is no friction. The only other force acting on the ass is due to gravity. All three forces ust be reduced to coponents noral to and parallel to the rod. These are in the r and directions. This leads to two equations in the two unknown forces. r F Mg sin F cos( ) Ma (7) P r, P/ F Mg cos F F sin( ) Ma (8) N P, P/ where a and a are the r and coponents of the acceleration a r, P/, P/ P/ Equation (7) ay be solved for F P and substituted into equation (8). This final equation ay be solved for F N, the noral force of the rod on the ass. This is ade draatically sipler by first substituting in for all of the known angles, lengths, asses, velocities and accelerations in equations 5 through 8. FP Mg sin( ) Mar, P/ / cos( ) 0.79 lbs F F sin( ) Mg cos Ma [(0.055)0.463] 0.30 lbs N P, P / The forces are ostly due to the Mg ter. 6

7 Proble 3 Solution The blades of a helicopter rotor are of radius 5. and rotate at a constant angular velocity of 5rev/s. The helicopter is flying horizontally in a straight line at a speed of 5.0 /s and an acceleration of 0.5/s, as sketched in the figure. Deterine the velocity and acceleration of point A, which is the outerost tip of one of the blades, (with respect to ground) when θ is 90 o. z B A y x AB / constant z. VB/ 5.0, a B/ = 0.5 s s r 5. r y x rev rad rad rad s rev s s Figure 3 The reference fraes defined in the figure are as follows: xyz is fixed to ground; B xyz is fixed to the translating helicopter. This is a proble which asks only for a solution for velocity and acceleration. This a pure kineatics question. No physical laws need be applied. Just describe the otion and do the ath. In this solution we will begin with the full 3D vector equations expressing velocity and acceleration. Since this proble involves otion in only D, and the rotational aspect of the proble is siple circular otion, it ay be siplified considerably, using polar coordinates, rather than using a rotating frae of reference, which is fixed to the rotating body. In polar coordinates the r vector is allowed to rotate through an angle, which is defined with respect to a fixed reference frae, in this case B xyz. For siple circular otion, such as the blades of the helicopter, the r vector ay be defined so as to rotate with the rigid body. In this proble we will require the r vector to rotate as if fixed to blade A. Fro the notes provided on velocities and accelerations we have expressions in cylindrical coordinates, given below, for translating and rotating rigid bodies. Since this is otion confined to a plane the linear velocity and acceleration in the z direction are set to zero, resulting in forulas in siple polar coordinate for: V V rr r zk A/ B/ 7

8 a a zk ( r r ) r ( r r ) A/ B/ For the helicopter proble any ters are zero: r r z z 0, because the point A is fixed to the blade and there is no otion in the z direction. The angular acceleration is also given as zero. The angular velocity has to be converted to radians per second. Hence (5 rev/s)=0 radians/second. When all quantities are substituted in the velocity becoes: rad V A / 5.0 i 5. (0 ) 5.0 i 5 s s s s The answer is now in ters of unit vectors fro two different coordinate systes. We need to convert to the unit vectors of the inertial reference frae xyz because that was what was specified in the proble. The unit vectors in the B xyz and the xyz fraes align. Hence, i i, j j, k k. The conversion of unit vectors fro polar to Cartesian coordinates is given by: r i cos( ) j sin( ) i sin( ) j cos( ) Substitution of these quantities leads to: V A / 5.0 i i 5 sin( ) i 5 cos( ) j (5 5 ) i s s s V A / 68.4 i at = / s The contribution to this velocity which coes fro the forward velocity of the helicopter will increase with tie due to the translational acceleration of the helicopter. This has been ignored in coputing this answer. The acceleration at A reduces to : a A/ ab/ zk ( r r ) r ( r r ) 0.5 i r r s for = / aa / 0.5 i 5. (0 rad/s) r [ 0.5i 53 j] s s The agnitude of the 8

9 acceleration is ore than 500g s!! The forces on the rotor hub are very large. A final note on polar coordinates: In this proble the coordinates rz,,and are defined in the non-rotating B xyz frae which is attached to the body of the helicopter. They are not a rotating reference frae. In siple circular otion probles the rotating r vector, which ay have coponents in the r and k directions, is adequate to describe the velocity and acceleration of a point on the rotating body. In such cases the need for an additional reference frae attached to the rotating body is avoided. It is necessary to use a rotating frae of reference attached to the body when velocities and accelerations ust be described which cannot be accoplished with an r vector. An exaple would be to describe the velocity of a stone which has been struck by the helicopter blade. The r vector can account for the oving blade but not the velocity of the stone with respect to the blade. In such a case a rotating frae attached to the blade would be required. Proble 4: An auseent park ride: A ride at an auseent park consists of a rotating horizontal platfor with an ar connected at point B to the outside edge. The ar rotates at 0 rev/in with respect to the platfor. The platfor rotates at 6 rev/in with respect to the ground. The axes of rotation for both are in the z direction, out of the plane of the page as shown in the drawing. Gravity acts in the z direction. A seat is located at C at the end of ar BC, which is long. The radius fro A to B is 6 long. The ass of the passenger is 75kg. Find the velocity and acceleration of the passenger. The reference fraes defined in Figure 4 are as follows: xyz is fixed to ground; A xyz is fixed to the platfor at the axis of rotation. B x y z is fixed to the ar BC at B and rotates with the ar. Proble 4 Solution: For this solution we will practice using the full rotating coordinate syste forulation. Let s begin by writing the general expression for the velocity of point C with respect to the inertial frae xyz, the vector su of the velocities at B and the velocity of C with respect to B. 9

10 V C / but, r V C / drc / drb / drc / B d( ra / rb / A) drc / B dt dt dt dt dt A / / xyz / xyz / xyz / xyz / xyz 0, and therefore drb / A drc / B dt dt / xyz / xyz () For ephasis the right hand side has been written as the tie derivative of the position vectors r B/ and r C/B. Because the origins of fraes at and A are coincident, then r B/ =r A/ +r B/A = 0+ r B/A because r A/ =0. The tie derivatives in the above expression ust be taken in the inertial frae xyz as shown in the above equation. Since both of these position vectors are rotating, then the velocity expressions for both will have coponents of the for r. These will be expanded in coplete detail in this early proble set solution so as to ephasize that the derivatives of rotating vectors always have a coponent due to the rotation of the vector and a coponent associated with the change in the agnitude of the vector. Taking the one at a tie: dr r V r V r 0 k R i R j B/ A B/ A B/ A / B/ A B/ A / B/ A A dt x yz / t xyz / Ax yz drc / B rc / B 0 k dt t B x y z V r V r k C/ B / C/ B C/ B / C / B / xyz / Bx y z V k R j R i C/ B where it ust be used that k=k k R i The notation syste here uses the / slash sybol to ean with respect to. Therefore the rcb / expression VCB / 0 eans the partial tie derivative of r C/B with t / B x y z / Bx y z respect to the rotating frae B xyz. It is shown as a partial derivative to ephasize that it accounts for the part of the total tie derivative in the inertial frae that coes fro the velocity of the point C as seen fro within the rotating frae B xyz. Since the frae is attached to the rotating body any velocity coponent due to the rotation of the body will not be observable. What is observable is any oveent of point C relative to the body. Since point C is fixed in this case to the body this relative velocity is zero. Expressed as a relative velocity, CB / / Bx y z V 0. Siilarly the velocity coponent of point B with respect to A ay be written as rb/ A VB/ A 0 / A It is also equal to zero, because B is fixed to the platfor and xyz t / Ax yz does not ove with respect to the rotating a ayz frae. () 0

11 Therefore the velocities we are attepting to copute coe only fro the V V R j B/ B/ A r ters. V C/ B Adding the two coponents together yields an expression for the V : V R j V V R j R j C/ B/ A C/ B C / (3) An iportant subtlety is the use of the correct rotation rate. Note that in equation () the rotation rate vectors in each case are specified as being the rotation rate with respect to xyz, the inertial frae. The platfor rotates at / k. This is with respect to the inertial frae. The ar BC rotates at k. This rotation rate is specified as the relative rotation rate / Ax yz between the ar and the platfor. In evaluating this r ter one ust use the rotation rate relative to the inertial frae, which in this case is the su of the two rotation rates; / k k. In this proble the z, z and z axes are all parallel to one another. Therefore they have the sae unit vector. That is to say, k k k. The final expression for the velocity of the passenger at C involves unit vectors fro two different reference fraes: V C R j R j / In order to go any further with the proble the actual angular position of the two reference fraes ust be evaluated and the unit vectors transfored to one coon set of unit vectors, for exaple the unit vectors of the inertial frae. For exaple: Let the x axis be at an angle of +30 degrees with respect to the x axis. Let the ar BC be aligned with the radius AB as shown in the following figure. It iediately follows that the unit vectors are related in the following way: 3 j j j i cos(30 ) j sin(30) i Fro this we can express V C/ in ters of the units vectors in the inertial frae. / 3 j V CR j R j [( R R ) R ]( i )

12 y, j C x, i y z A z y x 30 x B Evaluation of the acceleration of point C, a C/. With the above discussion to infor us, the evaluation of the acceleration proceeds in uch the sae way. There are just ore ters to evaluate. Fro the reading on expressing velocities and accelerations in translating and rotating fraes, coes the following expression for acceleration in full 3-diensional vector for. a a ( a ) r ( r ) ( V ) a B/ A/ B/ A / Ax yz B/ A / / B/ A / B/ A / Ax yz B / (3) (4) 5 Translational acceleration of frae A with respect to the inertial frae. xyz Relative acceleration of the point B as seen fro within the oving frae. (3) Eulerian acceleration due to the angular acceleration of the rotating body. (4) Centripetal acceleration ter due to the body's rotational velocity. 5 Coriolis acceleration due to translatlational velocity of an bject at B relative to the rotating rigid body. xyz (4) This solution will be done in such way as to show you how to accuulate the ters when there is a cascade of rotating fraes, each one connected to the next. We begin by coputing a B/. B is s fixed point on the platfor and is a fixed point in the frae A xyz which rotates with the platfor. This is a straight forward application of the forula in (4) above. Begin by identifying all the ters known to be zero. a 0, because the A frae rotates about point. A/ xyz (a ) 0, because B is a fixed point in the A frae. B/ A / Ax yz xyz 0, no rotational acceleration of the platfor. B/ A / A x yz xyz (V ) 0, because B is a fixed point in the A frae. Therefore a ( ) 0 B k k R i R i / Next we set out to find the acceleration of point C with respect to point B so that we can

13 coplete the coputation of a C/. a a a C/ B/ C/ B a ( a ) r ( r ) ( V ) C/ B C/ B / Bx yz / C/ B / / C/ B / C/ B / Bx y z where ( a ) 0, 0, and ( V ) 0 C/ B / Bx yz / C/ B / Bx y z k k ( ) k and therefore a / / Bx y z / xyz C/ B 0 0 ( ) k (( ) k R i) 0 ( ) R i Putting it all together a a a R i ( ) R i C/ B/ C/ B This ethod can be generalized to copute the velocity or acceleration of a point which is attached to a cascade of rigid bodies: each one rotating with respect to the one it is connected to. By starting at the inertial frae and working ones way out, the velocity of each velocity or acceleration that one coputes can becoe the reference point for the kinetics of the next point further out the chain. As was found for the velocity coputation, this acceleration is expressed in the unit vectors of the rotating fraes. To reduce the to the inertial frae the transforations fro oving unit vectors to the inertial ones ust be invoked. This is done here for the exaple in which the two oving fraes align with one another and both ake an angle with respect to the inertial frae of 30 degrees. Then we can express a C/ as: a R i ( ) R i ( R ( ) R ) i C / 3 j ac / [ R ( ) R ][cos( ) i sin( ) j] [ R ( ) R ][ i ] Proble 5: Cart and roller syste: A block of ass M b is constrained to horizontal otion by rollers. Let the acceleration of the block in the inertial frae xyz be designated as a xi which is assued to be known. Point A is fixed to the block. A assless rod rotates A / about point A with an angular rate of. A rotating reference frae, A xyz is attached to the rod, with origin A at the axis of rotation of the rod. The unit vectors in the x and y are i and j. A point ass,, is attached to the end of the rotating rod at point B. The distance fro A to B is e, which is known in the rotating achinery world as the eccentricity. In vector ters r A/B =ei. a. Find an expression for the total acceleration of the ass,, in the fixed inertial frae xyz. That is find a B/. Express the result in ters of x and y coponents in the xyz frae. In other words use the unit vectors associated with the xyz frae. Reeber to include the contribution fro a A / xi in your answer. b. What is the agnitude and direction of the axial force that the rod applies to the ass,? What is the force and direction that the rod 3

14 places on the pivot point at A? y e A x B y M b z x Proble 5 a. Solution: We begin by recalling the forula for the acceleration of an object using translating and rotating reference fraes. a a ( a ) r ( r ) ( V ) B/ A/ B/ A / Ax yz B/ A / / B/ A / B/ A / Ax yz Those parts known to be zero are: ( a ),, and ( V ) B/ / / B/ A B/ A / Ax yz B/ A / Ax yz which leads to a xi ( r ) x i k ( k ei ) xi e i B / applying the transforation of coordinates: i icos( ) jsin( ) a xi e [ i cos( ) j sin( )] The horizontal and vertical coponents of the acceleration are given by: a [ x e cos( )] i e sin( ) j B / b. To obtain the force that the assless rod puts on the rotating ass requires that Newton s nd law be satisfied. This requires a free body diagra for the ass. 4

15 Frod g z y x y A x c. Newton s nd law requires that F a [( x e cos( )) i e sin( ) j ] F gj B/ rod F rod gj x e i e j Therefore: [( cos( )) sin( ) ] Frod [ x e cos( )] i [ g e sin( )] j Newton s 3 rd law tells us that if the rod puts a force on the sall ass then it ust put an equal and opposite force on the pivot at A. F A F Assue for a oent that the ass M b is not allowed to ove in the x direction. It is a achine fixed in place. Assue also that we can ignore gravity(this would be the case if the axis of rotation were vertical, for exaple). Then the force the rotor puts on the ain ass M b is siply given by: [cos( )) sin( ) ] [cos( )) sin( ) ] FM a b B/ e i j e i j This is an iportant result for all echanical achines which have rotating parts. If the axis of rotation does not pass through the center of ass of the rotating coponent, then the rotating coponent places a force on the bearings given by F e [cos( )) i sin( ) j ] M b Where e is the distance fro the axis of rotation to the center of ass of the rotating coponent. Great effort is taken to balance rotating achinery. Everything fro car tires to jet engines receive attention to eliinate probles associated with unbalanced rotating parts. When the unbalance is large, the achine will eventually beat itself to death, as in this video of the washing achine. 5

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