SolutionsSet07

SolutionsSet07 - Homework Solutions Set 07 Physics 121...

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Homework Solutions Physics 121 Set 07 Spring 2008 - 1 - WeBWorK Problem 1 The blocks move with a constant acceleration. Starting from rest, they move a distance d during the first t seconds after being released. The distance d and the time t are related in the following manner: d = 1 2 at 2 Based on the known values of d and t we can determine a : a = 2 d t 2 = 1.11 m/s 2 The net force on mass M 1 can be expressed in terms of the acceleration and in terms of the force acting on it: ! M 1 a = ! M 1 g + T 1 The tension T 1 is thus equal T 1 = M 1 g ! M 1 a = 3.57 N The tension T 2 can be found in a similar manner: T 2 = M 2 g + M 2 a = 2.29 N Since the ropes are not slipping, the angular acceleration of the pulley is directly related to the linear acceleration a of the blocks: ! = a R = 17.3 rad/s 2 The magnitude of the torque on the pulley is equal to ! = T 1 R " T 2 R = M 1 g " M 1 a ( ) " M 2 g + M 2 a ( ) ( ) R = M 1 " M 2 ( ) g " M 1 + M 2 ( ) a ( ) R The torque is also equal to the product of the moment of inertia and the angular acceleration: = I " = I a R
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Homework Solutions Physics 121 Set 07 Spring 2008 - 2 - Combining the last two equations we can determine the moment of inertia of the pulley: I = R a ! = M 1 " M 2 ( ) g a " M 1 + M 2 ( ) # $ % & ( R 2 = 0.00471 kg m 2
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Homework Solutions Physics 121 Set 07 Spring 2008 - 3 - WeBWorK Problem 2 This problem is most easily solved by using the parallel axis theorem, and the known moment of inertia for a rectangular plate of length a and width b . According to Figure 10.20 in our textbook, the moment of inertia of such a plate with respect to a rotation axis perpendicular to the plate and going through its center of mass is equal to I cm = 1 12 M ( a 2 + b 2 ) In this problem the rotation axis is parallel to the rotation axis through the center of mass, and the distance d between the axes is equal to d = 1 2 a ! " # $ % & 2 + 1 2 b ! " # $ % & 2 = 1 4 a 2 + b 2 ( ) The parallel axis theorem can now be used to determine the moment of inertia with respect to this axis: I = I cm + Md 2 = 1 12 M ( a 2 + b 2 ) + 1 4 M a 2 + b 2 ( ) = 1 3 M a 2 + b 2 ( ) = 0.384 kg m 2
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Homework Solutions Physics 121 Set 07 Spring 2008 - 4 - WeBWorK Problem 3 In order to calculate the angular acceleration of the cylinder we first determine the torque exerted on it by the four forces. Given our sign convention, force F 1 exerts a positive toques, force F 2 and F 3 exert a negative torque, and force F 4 exerts no torque. The net torque is equal to ! = F 1 R 1 " F 2 R 1 " F 3 R 3 Since the torque is equal to the product of the moment of inertia of the cylinders and its angular acceleration, we need to determine the moment of inertia in order to be able to calculate the angular acceleration. The moment of inertia of the cylinder is equal to I = 1 2 MR 1 2 (see Figure 10.21 in our textbook). The angular acceleration of the cylinder is thus equal to = " I = F 1 R 1 # F 2 R 1 # F 3 R 3 1 2 MR 1 2 = # 4.12 rad/s 2
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Homework Solutions Physics 121 Set 07 Spring 2008 - 5 - WeBWork Problem 4 In this problem we use conservation of mechanical energy.
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This note was uploaded on 10/20/2010 for the course PHY PHY 121 taught by Professor Wolfs during the Spring '08 term at Rochester.

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SolutionsSet07 - Homework Solutions Set 07 Physics 121...

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