PROBLEM: At the instant shown, the 12-in. crank on the small...

PROBLEM: At the instant shown, the 12-in. crank on the small aircraft landing gear actuation in Figure P7.41 rotates counterclockwise at 18 rpm and decelerates at 3.5 rad/s². Use the relative acceleration method to graphically and analytically determine the angular velocity and angular acceleration of the wheel assembly.

Transcript

00:01 In this question we are given ab is equal to 75 mm bd is equal to 200 mm omega ab is equal to 200 radiant per second and theta is equal to 45 degree so from here we find the value of n which is bd divided by ab so n is equals to 200 divided by 75 which is 2 .667.

00:34 So in a part of the question we have to find omega bd which is equal to omega a b b multiplied with cos theta divided by square root of n square minus sine square theta after putting the values we have omega bd is equal to 200 multiplied with cos 45 degree divided by square root of 2 .667 whole square minus sine square 45 degree.

01:08 So when we saw this, we have omega bd comes out to be 55 radian per second.

01:16 This is answer for a part of the question.

01:19 Moving towards the b part of the question, here we have to find angular acceleration of bd, that is alpha bd, which comes out to b .d, which comes out to b.

01:29 N square minus 1 omega square a b sine theta divided by n square minus sine square theta raised to the power 3 divided by 2 after putting all the values we have alpha bd is equal to 2 .667 square minus 1 here we have 200 square sine 45 degree divided by 2 .667 scale minus sine scale 45 degree and hold is to the power 3 divided by 2 so alpha bd comes out to be 10167 .27 radiant per second square this is answer for b part now moving towards c part of the question in c part we have to find velocity and acceleration of point d so velocity at d is equal to r at ab multiplied with omega at a b multiplied with sine theta plus sign of two theta divided by two n so after solve after putting the values v d is equal to 75 multiplied with 200 multiplied with sign of 45 degree plus sign of 90 degree and here two multiplied with 2 .667 so after solving v d comes out to be 1 .0...

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