Question

3.) Given: The rod AB is confined to move along the inclined planes at A and B. At the instant shown the rod becomes horizontal and the point A has an acceleration of 3 m/s and a velocity of 2 m/s, both directed down the plane. At the same time on the other side of the rod, the velocity of point B is 2 m/s directed up the plane. In addition, the angular velocity of the rod is AB = 0.283.k.rad/s. Please, refer to the Acceleration-Difference Equation for planar rigid bodies below: à rà, +ä'ba +ä"BA a = absolute acceleration of point B à, -absolute acceleration of point A a B = tangent, relative acceleration of B with respect to A ab/1 = normal, relative acceleration of B with respect to A Note: For planar problems ā" B/4 =-WBPA Find: (a) Write the position vector of B with respect to A when the rod is horizontal (EW). (b) Use the Acceleration-Difference Equation to calculate the angular acceleration of the rod (QAR). (c) Use the Acceleration-Difference Equation to calculate the linear acceleration of point B in polar notation - 10 m = 2 m/s = 3 m/s2 a

Answer 1

DATE PAGE NO. lom 450 W450 AVAZ 2m/s A =3 m/s2 145 45° Velocity of A, VA ²2 m/s AY acceleration of A,a 3 m lom S2 A B velocity of B, VB ² 2 m/s ly1 asor 7.071 m. now, from the fig AC= CB = 5m A1 = BI= 5 = Cos 45° so, Angular velocity WAB = W- VA = 2 = 0.282 rad AT 7.071 sec Now, from the fig (a) Position of B with respect to A B/A = 108 m

DATE PAGE NO. Now, By accelerationDifference Equation a an + at eta + BIA pod also, cap a ap los459 + AB Sin 45 a - 3 Los450 - 3 sin45 07 a = 2.1219 - 2.121 7 al so, Angulare acceleration, the = XÃ - Coo (8) © of AB Now From Equation a = a + Table of Com BLA molte 13ltbmet BSAZAR aga hat a XMBIA - W2MBIA aplos450 i + aglin 450. = 2.1218 - 2.121 / + & R x 10 - (0.282 x 107 a los 4500 + assints = 2.121 - 2.121 + 1007 - .0.79522 Now, compaining ? and of components i components: ap Cos 45 = 2.121 - 0.7952 28% AB = = 21.87 mis 2.121 - 0.7952 Cos450 A 2 1.87 m/s²

DATE PAGE NO.. - Component : aplin 45°- - 2.12 1 8 + 100 1.87X Sin 45 = -2-121 + 100 x = 0.344 read / 52 (b) Angular Acceleration of rod, LAB E = = & 0.344 Ř read IS2 of acceleration point ß , ab = a (os45° it as Sin 45° AB = 1.87 Cos 450+ 1.87 sin 45 a = 1.322 i + 1.322 m/s2