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# 140 Note for CE 29800 with Professor Lyn at Purdue

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This 9 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Purdue University taught by a professor in Fall. Since its upload, it has received 33 views.

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Date Created: 02/06/15

Notes on Hmwk 8 solution 1540 This can be solved with various approaches We take the R approach emphasized in the lectures whereas the solution 60 manual takes a more vector 394 geometric approach VB VA VBA 20 V350 VA wABlAB V xdir gt VB cos 600 VA wABlAB sin 200 V VA a VA tan 6001AB ydir T VB sin60O 0 wABlAB cosZOO cos 600sin600tan 20quot AB 1 tan60quot1an20 cos20O Substituting VA 15 ms and AB 075 m we obtain the same solution as in the solution manual The other approach uses the instantaneous center of rotation We construct I perpendiculars to the velocity vectors VA and VB and find their intersection at O From the geometry we find the angles of the triangle OAB and use the law 700 W of sines to determine the lengths 10A and 103 39 O 39 O 39 O sm60 s1n50 3ZOAsin50 075 m066m 075 m 10A s1n60O 39 60O 39 70O 39 700 m m 03 S 075 m 081 m 075 m OB s1n60O With the lengths known we can now use the relations for the velocity magnitudes VA lOAwAB 3 wAB vAZOA 227s 3 VB lOBa2AB184 ms which is the same as before 1550 There are three rigid bodies involved the gear A which is in pure rotation about A the arm AB which also is in pure rotation also about A and the gear B which is in general plane motion ie both translation because B translates as well as rotation about the center B What is given are the angular velocities of the gear A wA and the arm AB MAB and what is asked for is the angular velocity of the gear B wB The key to the solution is that there is a kinematic constraint that where the gears mesh at point D the velocity at point D must be the same from the point of view of A VD rAwA from the point of view of B VD VB l wBrB wABlAB l wBrB Setting these two expressions equal we can solve for QB 1586 Because the solution manual takes the instantaneous center of rotation approach we use here the vector components approach We choose point B as reference and express the velocities ofother points in terms of the velocity of point B because VB 851 gt 4s0192 m gt 077 ms gt VD T VB gt wADlBD there we have assumed VD positive upwards and DAD positive clockwise Hence considering the separate components xdir gt 0 VB aBDlBD cos 600 V BD 1 B 077 ms W W 428s WAD VD aBDlBD sin60O 133 ms BD 39 ydir T vD0 aBDl3Dsin60O where the negative sign for VD indicates that the assumed mEDlAE lf ECU 4 upwards direction is incorrect and the direction of VD is in fact gt downwards note that the assumed clockwise direction of 035 is correct and that new now since this concerns the same rigid body The solution for VA is similarly obtained 24H mm VA d VB wADlAB W where the direction ova is not known 5Wquot 39 I so we need to fInd Its x and y components Zlfil mm maolmy xdir VAX vBwADlABsin30 128 ms ydir l VAy 0 cowlB cos30 089 ms l sothat VA1 Lv y 156 msand so VA156 ms 4318 z 15107 The solution manual uses a vector approach in terms of unit vectors but we will still use the vector approach empha sized in the lectures The solution manual also assumes the angular velocity to be zero but this is not necessary because we can show this from our equations Because information is given at both points A and B we express the acceleration at point C in terms of each of these points as ac aA l 011 l 76021 e a3 lalBC lea ch e where a single assumed clockwise angular acceleration or and a single angular velocity a applies because only a single rigid body is involved From the x direction equation we have 502 milB but this can only be true if a 0 From the ydirection equation we have aAiaB 7 1275 fts2 1 1 T 1071 ft ACT aAialACaBialBC 3 aAiaBalAcilBC 3 a 077s2 so after substituting back into the equation for ac we have ac 1A for 12 fts2 7077s2 10 ft 43 fts2 and ac 43 s2 l 15123 Note that AB undergoes pure rotation about A so 213 aABk xrBA in X row xrBA but because the angular frequency of rotation W is constant and hence or 0 such that 213 has magnitude 50131 and is directed towards the center of rotation A as obtained in the solution manual To determine the acceleration of D it will be necessary to determine WED which implies solving first the velocity problem For 6 90 application of the method of the instantaneous center of rotation leads to the conclusion that wBD 0 the perpendiculars are parallel and so intersect at infinity and so lengths such as 0D and DB will be infinite as seems implied in the solution manual Alternatively we can write the velocity vector equation as VD e VB 60le 3 ydirn l 0 0 wBDlBD cos 3 wBD 0 PROBLEM 1524 The rated speed of drum B of the belt sander shown is 2400 rpm When the power is turned off it is observed that the sander coasts from its rated speed to rest in 10 s Assuming uniformly decelerated motion determine the velocity and acceleration of Point C of the belt a immediately before the power is turned off b 9 5 later 25 mm SOLUTION 00 2400 rpm 2513 rads r 0025 m a VG ra 0025 m2513 rads vC 628 ms ac m2 0025 m2513 rads2 ac 1579 ms2 4 b Whent10 s 00 60 600 at 0 2513 rads a10 s a 2513 rad2 Whent9s aa0at cog 2513 rads 2513 radsz9 s 2513 rads VC Fab 0025 m2513 rads v9 0628 ms 4 ac Van 0025 m 2513 rad52 aCt 0628 ms2 aCn a 0025 m2513 rads2 810quot 1579 ms2 2 2 2 ac ac 510 0628 msz2 1579 n1522 ac 1580 ms2 4 PROPRIETARY MATERIAL 2009 The McGrawHill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw Hill for their individual course preparation If you are a student using this Manual you are using it without permission 984 PROBLEM 1529 The system shown is held at rest by the brakeand drum system A er the brake is partially released at t 0 it is observed that the cylinder moves 16 ft in 5 s Assuming uniformly accelerated motion determine a the angular acceleration of the drum b the angular velocity of the drum at t 4 s SOLUTION 1 2 Block A s y0t3at 16 0a5 s2 a 128 fts2 a 128 fts2 1 Drum hamsft A a a rat 2 128 ftsz 075 fta al707 rads2 xl707 rad52 4 b Uniforrnlv accelerated motion wo 0 when t O a 00 at When t 4 s a 0 1707 radsz4 s a 683 rads 0 683 rads D 4 PROPRIETARY MATERIAL 2009 The McGraw Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw Hill for their individual course preparation If you are a student using this Manual you are using it without permission 991 A PROBLEM 1540 Small wheels have been attached to the ends of rod AB and roll freely along the surfaces shown Knowing that wheel A moves to the left with a constant velocity of 15 ms determine a the angular velocity of the rod b the velocity of end B of the rod 20 750 mm SOLUTION 4 15 an ifquot R 3 00 quot3 VB VA VBA v3 3 60 15 ms lt vBA 4 70 Law of sines vB vBM 15 ms sin 70 sin 60 sin 50 b v3 1840 ms 60 4 VBA 1696 ms 4 70 VBA ABwAB 1696 ms 075 maAB a cow 2261 rads DAB 226 rads 4 PROPRIETARY MATERIAL 2009 The McGraw Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGrawHillfor their individual course preparation If you are a student using this Manual you are using it without permission 1003 PROBLEM 1586 Knowing that at the instant shown the angular velocity of rod BE is 4 rads counterclockwise determine a the angular velocity of rod AD b the velocity of collar D c the velocity of Point A 240 mm 360 mm SOLUTION Rod AD VB rBEwBE 0192X4 0768 ms gt a Instantaneous center C is located by noting that CD is perpendicular to VD and CB is perpendicular to V3 quotBc 0360 sin 30 0180 m v3 0768 a 42667 AD rm 0180 DAD 427 rads 4 b Velocig ofD rm 0360 cos30 031177 m vD rDCa 03117742667 vD1330 ms14 c Veloci of A M 0240 cos 30 020785 In CE 0600 sin 30 0300 m 020785 0 tan 0300 347 CA 0207852 03002 036497 m VA CAwAD 03649742667 1557 ms VA 1557 ms 4 347 4 PROPRIETARY MATERIAL 2009 The McGraw Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw Hill for their individual course preparation If you are a student using this Manual you are using it without permission 1061 PROBLEM 15107 A lO steel beam is lowered by means of two cables unwinding at the same speed from overhead cranes As the beam approaches the ground the crane operators apply brakes to slow down the unwinding motion At the instant considered the deceleration of the cable attached at A is 12 fts while that of the cable attached at B is 5 W5 Determine a the angular acceleration of the beam b the acceleration of Point C SOLUTION I 9 99A aA12 sl a aB5 sl W Assume a 0 a Angular acceleration on wk 3 as aA aBA 5j12jakgtlt9i12j9aj k 5 12 k a 0778 90 x 0778k a 0778 rads b Acceleration of Point C 12j0778kx10i 12j 7778j 4222j ac 422 mszi 4 PROPRIETARY MATERIAL 2009 The McGrawHill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw Hill for their individual course preparation If you are a student using this Manual you are using it without permission 1088 PROBLEM 15123 Arm AB has a constant angular velocity of 16 rads counterclockwise At the instant when 6 90 determine the acceleration a of collar D b of the midpoint G of bar BD SOLUTION Rod AB a3 ABij 3 inl6 rads2 a3 768 inszl Rod BD instantaneous center is at CD 130 0 sin 6 3 in10 in 03 6 1746 Acceleration 39 3 8 43nd a quotC Plane motion Trans with B Rotation about B a 3D as 6103 B aDB 2 aDB n 51 1BDa 13 5 1ng in 768 ins2 t 10 in al B 10 in02 4C 6 aD H768insi10ina j a Vector diagram lt39 7 3amp6quot 3D 768 insztan1746 wadquot 941294quot 24162 ins2 aD 242 ins2 4 10 in0t 768 ins2 cosl746 10 ina 80508 ins2 a 805 rads PROPRIETARY MATERIAL 2009 The McGraw Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 1113 PROBLEM 15123 Continued 17 3G 33 363 as 3619 aGB n aB l Ban W Email 413 768 ins2 L 5 in805 rad52 J 6 BGa2 a6 768ins2 l 4025 ins2 J 1746 2 components aGy 4025inszsin1746 6 21G 12077 ins2 lt Y a v 1 components aG y 768 ins2 4025 insz cos 1746 6 309 arkquot t 768 ins2 384ins2 I 7 Q 36 y 384 inszl aG 403 ins27 725 4 PROPRIETARY MATERIAL 2009 The McGrawHill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by MeGrawHillfor their individual course preparation If you are a student using this Manual you are using it without permission 1114

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