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TAM212.SAExam3soln

# TAM212.SAExam3soln - ω D =-3 k v P =-3 k)x10 i =-30 j a P...

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TAM 212 Sample Exam no. 3 Solution Oct 25, 2004 1] Disk D with attached pin P located 10 inches from the center C of the disk rotates at a constant speed of 3 radians/second in the clockwise direction. Pin C is fixed. Triangularly shaped slotted body T is pinned at O; the pin P slides in the slot. Find the angular velocity and angular acceleration of T in the diagrammed position (in which the slot lies at an 45 degree angle to the horizontal.) Remember to indicate the sense of these quantities. 2] The two links S and R are pinned at A. (See figure below) Point C moves at a constant speed of 34 m/sec towards the upper right in the circular track of radius 17m. Find the angular velocity ω and angular acceleration α of rod R . Point O is fixed. Remember to indicate the sense of ω and α . 3] The balls in the ball-bearing assembly roll without slipping on both the shaft and the housing. Find the velocity and acceleration of the point C at the center of the top Ball. The housing is at rest, and the shaft is rotating at constant rate ω =3rad/sec clockwise; R=3 cm; r = 1cm.
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Unformatted text preview: ω D =-3 k ; v P =(-3 k )x10 i = -30 j a P = α D k x10 i-ω D 2 10 i = -90 i v P = v O + ω T k x(-15 i ) + v rel { i + j }/ √ 2 =-15 ω T j +v rel i / √ 2 + v rel j / √ 2=-30 j a P = α T k x(-15 i )- ω T 2 (-15 i) + 2 ω T k x v rel + a rel { i + j }/ √ 2 =-90 i = -15 α T j +60 i+ a rel i / √ 2 + a rel j / √ 2 v rel =0; ω T =2 a rel =-150 √ 2; α T =-10 v C = 16 i +30 j ; v A =-8 ω S i ; v C = v A + ω R k x 15 i or 16 i + 30 j = -8 ω S i+15 ω R j => ω R =2 ; ω S =-2 a C = (34 2 /17){15 i-8 j }/17 = 60 i-32 j a A =-8 α S i – ω S 2 (8 j )= -8 α S i-32 j a C = a A + α R k x 15 i-ω R 2 ( 15 i) 60 i-32 j = -8 α S i-32 j +15 α R j-60 i => α S =-15 ; α R =0 v Q = ω shaft k x R j =(-3 k x 3 j ) = 9 i cm/sec v P = 0 (P is ICR of Ball) => v Q = v P + ω Ball k x (–2r j ) or 9 i = 2r ω Ball i => ω Ball = 4.5 => v C = v P + ω Ball k x (–r j ) = 4.5 i cm/sec Speed of C is constant; C’s path is circular, with radius ρ = R + r = 4cm => a C = e n v C 2 / ρ = (4.5) 2 /4cm = -5.06 j cm/sec 2...
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