# 121&amp;122 - TAM 212 Spring 2004 1.121 Given r = k1t =...

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TAM 212 Spring, 2004 1.121 Given: t k r 1 = , 2 2 t k = θ , 2 2 / R Hr z = , and k 1 , k 2 , H , and R are known constants Problem: Find the expressions for r , v , and a when the point reaches the top edge, i.e., z = H. Solution: k e r z r r + = ; k e e v z r r r & & & + θ + = θ ; ( 29 ( 29 k e e a z r r r r r & & & & & & & & & + θ + θ + θ - = θ 2 While at the top, r = R, since in general, t k r 1 = , 1 k r = & , and 0 = r & & ; Since 2 2 t k = θ : t k 2 2 = q & ; and 2 2 k = q & & = = 2 2 2 1 2 2 / / R t Hk R Hr z 2 2 1 / 2 R t Hk z = & ; 2 2 1 / 2 R Hk z = & & When H R Hr z = = 2 2 / , solving for r gives R r = . Further, solving R t k r = = 1 for t gives 1 / k R t = Plugging these into the equations for r , v , and a k e e v 2 2 1 2 1 / 2 2 R t Hk t k R k r + + = q ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 k e e a 2 2 1 2 2 1 2 2 / 2 2 2 2 R Hk k R t k k t k R r - + + - = q ( 29 ( 29 ( 29 ( 29 ( 29 k e e a 2 2 1 2 2 2 1 2 / 2 2 4 / 2 R Hk k R R k k R k R r - + + - = q 1.122a Given: z = -p θ /2 π ; r = R Problem: Write equations for r , v , and a in cylindrical coordinates. Solution: k e r z r r + = k e e v z r r r & & & - θ + = θ ( 29 ( 29 k e e a z r r r r r & & & & & & & & & - θ + θ + θ - = θ 2 1.122b Given: z = -p θ /2 π ; r = R=0.3m; p=0.2m, and t 6 . 0 = θ & rad/s Problem: Find and sketch velocity and acceleration vectors. Solution: We can use the equations we derived above k e v ) 2 / ( π θ
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