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CORIOLIS COMPONENT ACCELERATIONPr OCSlider - BA on CDDRDraw crank and slotted mechanism as per given configuration.OA = 60 mm, NOA = 200 RPM clockwise, CD = 300 mm, DR = 400 mm, Angle BOC = 120, OC = 160 mm, distance between horizontal line from R & point O = 120 mm. Find out velocity & acceleration of ram R, acceleration of block A along slotted bar CD.

Pr OCSlider - BA on CDDRo , cbFind = (2N) / 60, substituting value of N=200 rpm, = 20.952 rad/s.Now velocity of slider b with respect to o, Vbo = OB = 20.952 0.06 = 1.257 m/s.Take two fixed reference points, o & c.Draw vector Vbo from o perpendicular to crank OB.

Pr OCSlider - BA on CDDRo , cbVelocity of point A on CD w. r. t. slider B will be parallel to link CD.Hence from b draw a vector parallel to link CD. Value is unknown so draw vector taking any arbitrary length.

Pr OCSlider - BA on CDDRo , cbaVelocity of point A (on link CD) with respect to point C is perpendicular to link CD.Hence from c draw a vector perpendicular to CD.Intersection of two vectors will give point a in velocity diagram.

Pr OCSlider - BA on CDDRo , cbaNow, points C, A and D are on a single link.So in velocity diagram points c, a and d will be co-linear.To draw velocity of point d, extend the vector ca taking ratio ca/cd = CA/CD Take dimension from velocity diagram in case of small letters.Take dimension from mechanism in case of capital letters.Values of ca from velocity diagram & CA, CD from mechanism can be found.The value of cd for velocity diagram can be found. Draw vector as per the derived length.d

Pr OCSlider - BA on CDDRo , cbadFrom d draw velocity of r with respect to point d (Vrd), perpendicular to RD.

Pr OCSlider - BA on CDDRo , cbadrNow, velocity of slider r with respect to fixed point o or c will be in horizontal line. So, from o or c draw horizontal line.Intersection of two vectors will be point r.

Pr OCSlider - BA on CDDRo , cbadr 1f cOB = Vob2 / OB = 1.2572 / 0.06 = 26.33 m/sParallel to OB O2f cAC = Vac2 / AC = 0.89342 / 0.19698 = 4.052 m/sParallel to AC C3f tAC = unknown to AC-4f cAB = unknownParallel to AB-5f crAB = 2Vab CD = 2 x 0.8842 x 4.54 = 8.021 m/s to CD-6f cRD = Vrd2 / RD = 0.3592 / 0.4 = 0.322 m/sParallel to DR D7f tRD = unknown to DR-8f t R = unknownParallel to Vro-Derive all components for acceleration analysis.

Pr OCSlider - BA on CDDRo , cbadro1 , c1b1From acceleration table draw 1st acceleration vector.Centripetal acceleration of slider B with respect of O, will be parallel to OB & it will be toward centre of rotation of link OB, i.e. O.So from O1 draw vector parallel to OB & head of vector towards O.Magnitude of vector will be same as the value which we have found. 1f cOB = Vob2 / OB = 1.2572 / 0.06 = 26.33 m/sParallel to OB O

Pr OCSlider - BA on CDDRo , cbadro1 , c1b1oa22f cAC = Vac2 / AC = 0.89342 / 0.19698 = 4.052 m/sParallel to AC CNow centripetal acceleration of A with respect to C, it will be parallel to AC & towards the centre of rotation of link AC, i.e. towards C.So from C1 draw vector parallel to CD & magnitude of vector will be as per the value derived.

Pr OCSlider - BA on CDDRo , cbadro1 , c1b1oa3f tAC = unknown to AC-Now tangential acceleration of A with respect to C, it will be perpendicular to AC.So from Oa draw vector perpendicular to CD & magnitude of vector is unknown.

Pr OCSlider - BA on CDDRo , cbadro1 , c1b1baoa5f crAB = 2Vab CD = 2 x 0.8842 x 4.54 = 8.021 m/s to CD-Here we are interested in finding out acceleration of slider B with respect to C. It is addition of acceleration of B with respect to A & acceleration of A with respect to C.Here coriolis component will come into picture.It can be found our by the method shown in red figure. Blue vector is coriolis component of acceleration.Pick coriolis component & put its head at b1.

Pr OCSlider - BA on CDDRo , cbadro1 , c1b1baoaa14f cAB = unknownParallel to AB-From ba draw centripetal acceleration of B with respect to A.Draw a line parallel to CD from ba.Intersection of two vectors will be point a1.

Pr OCSlider - BA on CDDRo , cbadro1 , c1b1baoaa1d1a1b1 = Total acceleration of B with respect to A.c1a1 = Total acceleration of A with respect to C.Now links C, A & D are co-linear. Hence in acceleration diagram these three points must be co-linear.So taking ratio, c1a1/c1d1 = CA / CD.Capital letter indicates measurements from mechanism drawn.Small letter indicates measurements from acceleration diagram.

Pr OCSlider - BA on CDDRo , cbadro1 , c1b1baoaa1d1

Pr OCSlider - BA on CDDRo , cbadro1 , c1b1baoaa1d1rd6f cRD = Vrd2 / RD = 0.3592 / 0.4 = 0.322 m/sParallel to DR DNow draw centripetal component of point R with respect to D. It is parallel to DR & it is towards centre of rotation of link DR, i.e. towards D.Do from rd draw a line parallel to DR.Magnitude is same as the derived one.d1rd

Pr OCSlider - BA on CDDRo , cbadro1 , c1b1baoaa1d1rd7f tRD = unknown to DR-Now tangential acceleration of point R with respect to D.Value is unknown.So from rd draw a line perpendicular to DR.

Pr OCSlider - BA on CDDRo , cbadro1 , c1b1baoaa1d1r1rd8f t R = unknownParallel to Vro-Now draw tangential acceleration of slider R.Value is unknown.From c1 draw a horizontal line.Intersection of two points will be point r1.

Pr OCSlider - BA on CDDRo , cbadro1 , c1b1baoaa1d1r1rdConfiguration DiagramVelocityDiagramAccelerationDiagram