Hw09 - Problem 10.40 Since the net external torque is zero,...

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Problem 10.40 Since the net external torque is zero, the angular momentum is conserved 0 f LL = . That is, 00 f f II ω = . Since f is given (0.40 rev/s), we only need to calculate 0 I and f I , by considering that the total moment of inertia I is the sum of I hands and arms and I remainder of the body. Note I remainderis is 0.40 . 2 kgm Initially, the hands and arms can be considered a slender rod, I hands and arms(initial) = 22 11 12 12 8 1.8 2.16 mL kgm =×× = 2 2 0 2.16 0.40 2.56 –> I kgm =+= Finally, the hands and arms can be considered a hollow cylinder, I hands and arms(initial) = mR –> 8 0.25 0.50 kgm = 2 2 0.50 0.40 0.90 f I kgm Thus, / 2.56 0.40/ 0.90 1.14 ff == × = rev/s. Problem 10.62 Part A: T In horizontal direction: 0 xx x FNTm a =−= = (1) N In vertical direction: 0 yy s y F T mg F f ma =− −−= = (2) The paper roll is slipping down the wall, f s mg F sk f N µ = (3) From (1), we get (4) sin30 / 2 o x NT T T = From (3) and (4), we get 0.25 /8 s f NT (5) Substitute (5) into (2), we have 3 2 16 9.8 40 0 TT ×−− = Æ TN 266 = Part B: From A, 33.3 s f , () net s f FR I τ α = −= 18/100/ 0.260
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This homework help was uploaded on 04/02/2008 for the course PHYSICS 7A taught by Professor Lanzara during the Spring '08 term at Berkeley.

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Hw09 - Problem 10.40 Since the net external torque is zero,...

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