341_PartUniversity Physics Solution

341_PartUniversity Physics Solution - 10-36 Chapter 10 (c)...

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10-36 Chapter 10 (c) The final kinetic energy is () ()( ) 2 2 34 , M vM a t = so the change in kinetic energy is 2 22 0 k0 0 k 31 1 . 434 6 R KMg M R M R g ω μ ωω ⎛⎞ Δ= = ⎜⎟ ⎝⎠ EVALUATE: The fraction of the initial kinetic energy that is removed by friction work is 2 1 0 6 2 1 0 4 2 3 MR MR = . This fraction is independent of the initial angular speed 0 . Figure 10.101 10.102. IDENTIFY: The vertical forces must sum to zero. Apply Eq.(10.33). SET UP: Denote the upward forces that the hands exert as and L R FF . ( ) LR r τ =− , where 0.200 m r = . EXECUTE: The conditions that and L R must satisfy are FFw + = and I r −= Ω , where the second equation is , L divided by r . These two equations can be solved for the forces by first adding and then subtracting, yielding 1 2 L I Fw r =+ Ω and 1 . 2 R I r Ω Using the values 2 (8.00 kg)(9.80 m s ) 78.4 N and wm g == = 2 (8.00 kg)(0.325 m) (5.00 rev s 2 rad rev) 132.7 kg m s (0.200 m) I r ωπ × gives 39.2 N (66.4 N s), 39.2 N (66.4 N s). Ω⋅ (a) 0, 39.2 N Ω= = = . (b) 0.05 rev s 0.314 rad s, 60.0 N, 18.4 N. = = = (c) 0.3 rev s 1.89 rad s, 165 N, 86.2 N = = , with the minus sign indicating a downward force. (e) 39.2 N 0 gives 0.575 rad s, which is 0.0916 rev s. 66.4 N s R F = = EVALUATE: The larger the precession rate Ω , the greater the torque on the wheel and the greater the difference between the forces exerted by the two hands. 10.103. IDENTIFY: The answer to part (a) can be taken from the solution to Problem 10.92. The work-energy theorem says WK . SET UP: Problem 10.92 uses conservation of angular momentum to show that 11 2 2 rv = . EXECUTE: (a) 3 . Tm v rr = (b) and d Tr ! ! are always antiparallel. dT d r ⋅= ! ! . 21 12 2 2 1 1 32 2 . 2 dr mv WT d r m v r r r = = ∫∫ (c) 1 2 , s o vv r r = 2 2 1 1 2 1 ( / ) 1 mv Km v v r r Δ = , which is the same as the work found in part (b). EVALUATE: The work done by T is positive, since T ! is toward the hole in the surface and the block moves toward the hole. Positive work means the kinetic energy of the object increases.
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11-1 E QUILIBRIUM AND E LASTICITY 11.1. IDENTIFY: Use Eq.(11.3) to calculate cm x . The center of gravity of the bar is at its center and it can be treated as a point mass at that point. SET UP: Use coordinates with the origin at the left end of the bar and the x + axis along the bar. 1 2.40 kg, m = 2 1.10 kg, m = 3 2.20 kg. m = EXECUTE: 11 2 2 33 cm 123 (2.40 kg)(0.250 m) 0 (2.20 kg)(0.500 m) 0.298 m 2.40 kg 1.10 kg 2.20 kg mx x mmm ++ + + == = + + . The fulcrum should be placed 29.8 cm to the right of the left-hand end. EVALUATE: The mass at the right-hand end is greater than the mass at the left-hand end. So the center of gravity is to the right of the center of the bar.
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341_PartUniversity Physics Solution - 10-36 Chapter 10 (c)...

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