311_PartUniversity Physics Solution

# 311_PartUniversity Physics Solution - 10-6 Chapter 10 (d) F...

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10-6 Chapter 10 (d) y y Fm a = applied to the cylinder gives 0 nT M g −= and 2 42.0 N (12.0 kg)(9.80 m/s ) 160 N nTm g =+ = + = . EVALUATE: The tension in the rope is less than the weight of the bucket, because the bucket has a downward acceleration. If the rope were cut, so the bucket would be in free-fall, the bucket would strike the water in 2 2(10.0 m) 1.43 s 9.80 m/s t == and would have a final speed of 14.0 m/s. The presence of the cylinder slows the fall of the bucket. Figure 10.14 10.15. IDENTIFY: Apply m = F a ! ! to each book and apply z z I τ α = to the pulley. Use a constant acceleration equation to find the common acceleration of the books. SET UP: 1 2.00 kg m = , 2 3.00 kg m = . Let 1 T be the tension in the part of the cord attached to 1 m and 2 T be the tension in the part of the cord attached to 2 m . Let the -direction x + be in the direction of the acceleration of each book. aR = . EXECUTE: (a) 2 1 00 2 x x x xv t a t −= + gives 2 0 22 2( ) 2(1.20 m) 3.75 m/s (0.800 s) x xx a t = . 2 1 3.75 m/s a = so 11 1 7.50 N Tm a and () 1 18.2 N Tmga =− = . (b) The torque on the pulley is 21 0.803 N m, TTR and the angular acceleration is 1 50 rad/s , so 0.016 kg m . I ατ = = EVALUATE: The tensions in the two parts of the cord must be different, so there will be a net torque on the pulley. 10.16. IDENTIFY: Apply m F =a ! ! to each box and z z I = to the pulley. The magnitude a of the acceleration of each box is related to the magnitude of the angular acceleration of the pulley by = . SET UP: The free-body diagrams for each object are shown in Figure 10.16a-c. For the pulley, 0.250 m R = and 2 1 2 IM R = . 1 T and 2 T are the tensions in the wire on either side of the pulley. 1 12.0 kg m = , 2 5.00 kg m = and 2.00 kg M = . F ! is the force that the axle exerts on the pulley. For the pulley, let clockwise rotation be positive. EXECUTE: (a) x x a = for the 12.0 kg box gives a = . y y a = for the 5.00 kg weight gives 222 mg T ma . z z I = for the pulley gives 2 1 2 TTR M R . = and 1 2 TT M a . Adding these three equations gives 1 2 2 mg m m M a =+ + and 2 1 12 2 5.00 kg (9.80 m/s ) 2.72 m/s 12.0 kg 5.00 kg 1.00 kg m ag mm M ⎛⎞ = ⎜⎟ ++ + + ⎝⎠ . Then 2 (12.0 kg)(2.72 m/s ) 32.6 N a = . gives ( ) (5.00 kg)(9.80 m/s 2.72 m/s ) 35.4 N ga = = . The tension to the left of the pulley is 32.6 N and below the pulley it is 35.4 N. (b) 2 2.72 m/s a = (c) For the pulley, x x a = gives 1 32.6 N x FT and y y a = gives 2 2 (2.00 kg)(9.80 m/s ) 35.4 N 55.0 N y FM gT = + = .

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Dynamics of Rotational Motion 10-7 EVALUATE: The equation 1 21 2 2 () mg m m M a =+ + says that the external force 2 must accelerate all three objects. Figure 10.16 10.17. IDENTIFY: Apply z z I τ α = to the post and m F =a !
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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311_PartUniversity Physics Solution - 10-6 Chapter 10 (d) F...

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