Hi there, how do we get the formula a = (L/2)*alpha...(I circled it in red), ALSO, what acceleration is that(the acceler
Posted: Sun Apr 10, 2022 6:30 pm
Hi there, how do we get the formula a = (L/2)*alpha...(I circled
it in red), ALSO, what acceleration is that(the acceleration I
indicated with a green arrow) referring to? Why don't we indicate
normal and tangential acceleration at end B?? Please let me know if
my questions make sense...
PROBLEM 16.79 A uniform rod of length L and mass m is supported as shown. If the cable attached at B suddenly breaks, determine (a) the acceleration of end B, (b) the reaction at the pin support. B SOLUTION 0 = 0, a = -C A ID G B 182 A = Wem G B & m NIT NI NIN wž = Ta + ma 2 EMA = (MA)er: w L 1 -mla + m m 15 2 12 L mg mla, 2 , mg O 2 a =- 2 L L + EF, = E(F): A - mg = -mā =-ma = eff bo L 4-mg = -ml 2 2 L A-mg =--mg 4 A=--mg A-mg - 4 a; = a, +aBA = 0 + La =a 4 ag = 2 11 3 g 21 1 B g
it in red), ALSO, what acceleration is that(the acceleration I
indicated with a green arrow) referring to? Why don't we indicate
normal and tangential acceleration at end B?? Please let me know if
my questions make sense...
PROBLEM 16.79 A uniform rod of length L and mass m is supported as shown. If the cable attached at B suddenly breaks, determine (a) the acceleration of end B, (b) the reaction at the pin support. B SOLUTION 0 = 0, a = -C A ID G B 182 A = Wem G B & m NIT NI NIN wž = Ta + ma 2 EMA = (MA)er: w L 1 -mla + m m 15 2 12 L mg mla, 2 , mg O 2 a =- 2 L L + EF, = E(F): A - mg = -mā =-ma = eff bo L 4-mg = -ml 2 2 L A-mg =--mg 4 A=--mg A-mg - 4 a; = a, +aBA = 0 + La =a 4 ag = 2 11 3 g 21 1 B g