Chapter_11_solved_problems

Chapter_11_solved_problems - Chapter 11 solved problem. 2....

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Chapter 11 solved problem. 2. Picture the Problem : The weed is pulled by exerting a downward force on the end of the tool handle. Strategy: Set the torque on the tool equal to the force exerted by the weed times the moment arm and solve for the force. Solution: Solve equation 11-1 for F : weed weed weed weed 1.23 N m 31 N 0.040 m F r F r τ = × = = = Insight: The torque must be the same everywhere on the tool. Therefore, the hand must exert a 1.23 N m 0.22 m 5.6 N × = force to produce a 31-N force at the weed. The force is multiplied by a factor of 22 4 5.5. = 12. Picture the Problem : The CD rotates about its axis, increasing its angular speed at a constant rate. Strategy: Determine the angular acceleration of the CD using equation 10-11 and its moment of inertia (treat it as a disk) from table 10-1. Then use equation 11-4 to find the torque exerted on the CD. Solution: 1. Solve equation 10- 11 for α : ( 29 ( 29 2 2 2 2 0 2 450 rev/min 2 rad rev 1 min 60 s 0 2 2 3.0 rev 2 rad rev 59 rad/s π ϖ θ × × - - = = × = 2. Use table 10-1 to find 2 1 2 I MR = : ( 29 ( 29 2 2 5 2 1 1 2 2 0.017 kg 0.060 m 3.1 10 kg m I MR - = = = × × 3. Apply equation 11-4 directly: ( 29 ( 29 5 2 2 3.1 10 kg m 59 rad/s 0.0018 N m I - = = × × = × Insight: The CD takes 0.80 s to accelerate to its final angular velocity, accounting for some of the delay between when you press “play” and when you first hear the music. 13. Picture the Problem : The ladder rotates about its center of mass, increasing its angular speed at a constant rate. Strategy: Use table 10-1 to find the moment of inertia of a uniform rod of mass M and length L that is rotated about its center of mass: 2 1 12 I M L = . Then use equation 11-4 to find the required torque to produce the acceleration. Solution: 1. Use table 10-1 to find 2 1 12 I M L = : ( 29 ( 29 2 2 2 1 1 12 12 8.22 kg 3.15 m 6.80 kg m I M L = = = × 2. Apply equation 11-4 directly: ( 29 ( 29 2 2 6.80 kg m 0.302 rad/s 2.05 N m I = = × = × Insight: This torque is about 1.51 ft·lb, so if the person has the ladder on his shoulder and exerts the torque with his hand at arm’s length (3.00 ft), he need only exert 0.505 lb of force to produce the angular acceleration of 0.302 rad/s 2 .
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15. Picture the Problem : The object consists of three masses that can be rotated about any of the x , y , or z axes, as shown in the figure at right. Strategy:
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This note was uploaded on 01/05/2012 for the course CHEM 3321 taught by Professor Bean during the Spring '11 term at American Jewish University.

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Chapter_11_solved_problems - Chapter 11 solved problem. 2....

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