Solution () urpose and Requirement Solution Kinetics of rigid bod(newton s Second Law) In rob, kinematics stud regarding acceleration of mass center should be done before Newton s second law is used to setup 3 scalar equations for a bod in plane motion In rob, rinciple of ngular Momentum about a fied point is applied Students should know how to calculate I b using equation (445) 3 rob 3 is a basic problem of planar motion It should be emphasized that the mass center must be taken as the reference point to set up the dnamic governing equations 4 In roblem 4, the individual masses of blocks could be taken into the sstem and a equivalent moment of inertia can be defined such that the whole sstem can be modeled as a single disk rotating about a fied point 5 roblem 5 is a tpical problem of constrained motion of a sstem including two bodies Kinematics analsis is crucial, students must know how to use the condition rolling without slipping More time ma be used in this problem to help students full understand the logic of analsis n alternative solution is also given for it The problem 6 is similar to this problem Translation of a rigid slab (roblem 6 modified on page 80) The 7 kg uniform slender rod connected to a disk centered at and to a crank Knowing that the disk is made to rotate at the constant angular velocit of 80 rmin (rotations per minute), determine for the position shown (a) the resultant force of all the eternal forces acting on, (b) the resultant moment about mass center of all the eternal forces acting on, *(c) constraint forces at and 30 750 mm 30 00 mm Solutions Rotation of the disk: 80 rmin =885 rads, and =0 Rod is in translation (eplain wh) 0 (a) a a r60 = 706 0 (ms ) Resultant force acting on a Kinematics diagram ma = 7 (7060) = 4974 0 (N) () (b) or rod = 0, so M = 0 () (c) *ssume the constraint forces at and to be i j i j which include 4 ubknowns Using () we have i j i j mgj = 4974 0 mg ree bod diagram
Solution () in i direction: 487 (N) in j direction: 36 (N) Using () we have = 8 (N)(ans) We cannot solve for,, because for one rigid bod in planar motion there are onl three independent scalar governing equations One additional condition must be provided for solving 4 unknowns (or eample, if rod is assumed to be massless and with no friction torque acting on it, then we can have one additional equation What is it?) L a N Rotation of a rigid slab about a fied point (roblem 678 on page 88) uniform slender rod, of length L = 750 mm and mass m = kg and I ml, hangs freel from a hinge at If a force of magnitude N is applied at horizontall to the left (h = L), determine (a) the angular acceleration of the rod, (b) the components of the constraint force at Solution: Rod rotates about a fied point, so we have mg N iagram = ma Kinetic iagram I I mr ml m L ml = 0375 (kg m ) 3 (a) Use rinciple of ng Momentum about fied point : M I )(075) = 0375 rads) (ans) (b) a r 4k ( 0375 ) j 9i Newton s second law gives gj i i j ( 9i) 6 (N) 9 6 (N) (ans) (lternativel, we ma use (=K), and take moment about )
Solution (3) eneral planar motion of a rigid slab 3 (roblem 660 on page 806) disk of radius r = 5 mm, mass m = 7 kg, and I mr, rests on a frictionless horizontal surface force of magnitude 4 N is applied to a tape wrapped around the disk as shown etermine (a) the angular acceleration of the disk, (b) the acceleration of the center of the disk (c) the length of tape unwound after s 50 mm a rob 3 Solution: raw the kinematics agram and ree bod diagram (a) I (b) mr M 05 ( 7)( 05 ) (kg m ) I 4(05) = 05(7)(05 ) rads ma = 4N ma K 4 = 7 (a ) a =0574 (ms ) (c) a r = 943(05) i = 49 i (ms ) s =05a t = 05(49)(4) = 858 (m) (ans)
Solution (4) 4 (J 646) Each of the double pulles shown has an moment of inertia I = 0 kgm, and is initiall at rest The outside radius is 400 mm and the inner radius is 00 mm etermine (a) the angular acceleration of each pulle, (b) the angular velocit of each pulle at t = 3 s, (c) the angular velocit of each pulle after point on the cord has moved 3 m Neglect friction Solution: onsider sstem = disk + blockes 400 mm 00 mm O, O O O 00 kg 300 kg 00 kg 98 N 98 N 398 N 98 N 0598 N () () (3) (4) I O = I = 0 (kgm ) I O = 0 + 00 (0 ) I O = 0 + 500 (0 ) I O = 0 + 50 (04 ) (a) M I O 98( 0 ) 98( 0 ) 98( 0 ) 50( 98 )( 0 4) 9 6 4 04 654 09 0 4 30 8 (b) t (3)=5886 (rad) (3)= 4043(3)=96 (3)= 37 (c) U s 05I O s 05I O 98 ( 3 ) =46 98 ( 3 ) =05 98 ( 3 ) =40 05 I 0 5I O 0 5I O 50( 98 )( 3) =79 05 I O Note: This problem can be easil solved using standard method b separating blocks and disk
Solution (5) 5 (H 7-7 p395) pipe (thin ring) has a mass of 500 kg and radius of 05 m and rolls without slipping down a 300 kg ramp as shown If the ramp is free to move horizontall (frictionless), determine the acceleration of the ramp Neglect the masses of the wheels at and Solution: Step Kinematics analsis 30 r=05 m enote the mass centers of pipe and ramp b and, respectivel onsider is the mass center of the ramp, is the contact but on the ramp Since the ramp is in translation a a raw the kinematics diagram, and find the relation: a a a a a ue to the constraints, directions of a a, and a known, so we ma epress them b where a a i a a e t e t 30, e n 0 are e n e t a a 30 kinematics diagram 500g a The angular acceleration of pipe a r Step ree bod diagrams of ipe and Ramp: ipe: Ne e 500g j 500 ( a i a e ) n t t e n e t 30 N (ot e n ) N 500g ( 0866) 500a ( 05 ) () (ot e t ) 500g ( 05) 500a ( 0 866) 500a () ipe r I I a r ( I 500 r ) N 500 a (3) Ramp: Ne e 300g j Rj 300 a i n t e n e t 300g (ot i ) N( 05 ) ( 0866 ) 300a (4) Using (3) in (4) gives N 866a 600a (5) Ramp R
Solution (6) s 0 Substituting (3) and (5) into () and () ields Solving (6) and (7) gives 866a 850a 433g (6) 000a 433a 50g (7) a 303 (m s ), a 7336 (m s ) (ans) lternative solution Step The sstem is a conservative sstem (wh?) ssume the pipe and ramp are released from rest uring the histor when pipe travels s relative to the ramp, the work done b the applied force is U0 500g( 05 s) 50gs The kinetic energ of the sstem at state is v 30 v T v v I v ( 500) ( 300) where v v v v i v e t v v v v v v cos 30 Using angular velocit of pipe v, and I r r 500, we can epress the kinetic energ at state as T 50 [ v v v v ( 0866 ) ] ( 500) v 50v Then using U 0 50gs T 0, we have 50gs = 50 [ v v v v ( 0866 ) ] ( 500) v 50v () Step onsider ipe +Ramp, the linear Momentum of the sstem should be conserved in i direction: ( 500 v 300v ) i 0
Solution (7) ie, 500( v i v e ) i 300v i 0 t 800v 500( 0866 ) v 0 v 054 v () Using () in () gives 50gs = 388 v (3) Since the relative motion of with respect to the ramp is rectilinear motion with constant acceleration a, we have a s v (4) Using (4) in (3) we have a 5g 303 (ms ) 388 rom () we also have a 054 a = 7336 (ms ) (ans) 6 (H 7-5 modified on page 395) The disk of weight 50 N, radius r = 05 m, I = (mr ), and block of weight 0 N are released from rest as shown The friction at contact point between the disk and the surface is large enough that the disk rolls without slipping Neglect the masses of pules and cable etermine (a) the acceleration of block at the instant, (b) the minimum value of the static friction coefficient s between the disk and the surface 05 m Solution: 0 Step Kinematics analsis constant 05 m 0 0, 0 r 05 0 (r) 03, (wh? ecause is the instant center) 05 05
Solution (8) Step ree od iagram & Kinetics analsis lock : 0 T = 0 g = 0 05 g ( ) () T = isk : 0N 50 (N) m I T m 05 (m) N T = m I 0N 50 (N) N T m ( i ) T + = 50 g = 50 05 g ( ) () ( j ) N = 50 (N) (3) ( k ) ( T) r = I ( T )( ) ( ) 50 05 05 g Solve() () and (4) for Tand : = 377 (rads ) = 635 (N) namic equilibrium diagram T = (ill in b ourself) Since = s N, we have the minimum friction coefficient s = N 635 = 00563 (ans) 50 ood students are encouraged to tr lembert rinciple in this problem appling inertia force and inertia moment at, we combine and K as one dnamic equilibrium diagram as shown