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Unformatted text preview: 19. Picture the Problem : The fish exerts a torque on the fishing reel and it rotates with constant angular acceleration. Strategy: Use Table 10-1 to determine the moment of inertia of the fishing reel assuming it is a uniform cylinder ( 2 1 2 MR ). Find the torque the fish exerts on the reel by using equation 11-1. Then apply Newtons Second Law for rotation (equation 11-4) to find the angular acceleration and equations 10-2 and 10-10 to find the amount of line pulled from the reel. Solution: 1. (a) Use Table 10-1 to find I : ( 29 ( 29 2 2 1 1 2 2 0.99 kg 0.055 m 0.0015 kg m I MR = = = 2. Apply equation 11-1 directly to find : ( 29 ( 29 0.055 m 2.2 N 0.121 N m r F = = = 3. Solve equation 11-14 for : 2 2 0.121 N m 81 rad/s 0.0015 kg m I = = = 4. (b) Apply equations 10-2 and 10-10: ( 29 ( 29 ( 29 ( 29 2 2 2 1 1 2 2 0.055 m 81 rad/s 0.25 s 0.14 m s r r t = = = = Insight: This must be a small fish because it is not pulling very hard; 2.2 N is about 0.49 lb or 7.9 ounces of force. Or maybe the fish is tired? 26. Picture the Problem : The person lies on a lightweight plank that rests on two scales as shown in the diagram at right. Strategy: Write Newtons Second Law in the vertical direction and Newtons Second Law for rotation to obtain two equations with two unknowns, m and cm x . Solve each to find m and cm x . Using the left side of the plank as the origin, there are two torques to consider: the positive torque due to the right hand scale and the negative torque due to the persons mass. Solution: 1. (a) Write Newtons Second Law in the vertical direction to find m : 2. (b) Write Newtons Second Law for rotation and solve for x cm : Insight: The equation in step 1 does not depend on the axis of rotation that we choose, but the equation in step 2 does. Nevertheless, we find exactly the same cm x if we choose the other scale,...
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- Spring '11