Unformatted text preview: ω D =3 k ; v P =(3 k )x10 i = 30 j a P = α D k x10 iω 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 i8 j }/17 = 60 i32 j a A =8 α S i – ω S 2 (8 j )= 8 α S i32 j a C = a A + α R k x 15 iω R 2 ( 15 i) 60 i32 j = 8 α S i32 j +15 α R j60 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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 Spring '10
 richard
 Angular velocity, Angular Acceleration, Velocity, Physical quantities, constant speed

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