Assignment 36

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20:34:35 036. `query 36

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RESPONSE --> 8

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�X���`��o����n��I���i�E�� assignment #036 �[��޳���뺗��܃����ܝ��� Physics I 04-24-2006

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21:26:16 Query class notes #37 If we know the angular frequency `omega and the amplitude A of motion how do we obtain an equation of motion (i.e., the formula that gives us the position of the pendulum if we know the clock time t)? What are the corresponding velocity and acceleration functions?

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RESPONSE --> Position at clock time x = Acos(`omega* t) Velocity = -`omega *A*sin(`omega* t) Accel = -`omega * A * cos(`omega* t)

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21:26:20 ** Position at clock time is x = Acos(`omega* t) Velocity = -`omega *A*sin(`omega* t) Accel = -`omega * A * cos(`omega* t) University Physics students should note that velocity and acceleration are the first and second derivatives of the position function. **

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RESPONSE --> ok

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21:27:52 How is the acceleration of the pendulum related to the centripetal acceleration of the point on the reference circle?

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RESPONSE --> a = -`omega A sin(`omega *t) and aCent = v^2/r for SHM

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21:28:01 STUDENT ANSWER: a = -`omega A sin(`omega *t) and aCent = v^2/r for the circle modeling SHM INSTRUCTOR AMPLIFICATION: ** The centripetal acceleration of the point on the reference circle, which acts toward the center of the circle, has two components, one in the x direction and one in the y direction. The component of the centripetal acceleration in the direction of the motion of the oscillator is equal to the acceleration of the oscillator. If the oscillator is at position theta then the centripetal acceleration has direction -theta (back toward the center of the circle, opposite to the position vector). The centripetal acceleration is aCent = v^2 / r; so the x and y components are respectively ax = aCent * cos(-theta) = v^2 / r * cos(theta) and ay = aCent * sin(-theta) = -v^2 / r * sin(theta). **

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RESPONSE --> ok

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21:46:27 How is the kinetic energy of the pendulum related to its restoring force constant k, the amplitude of its motion, and its position x?

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RESPONSE --> The PE of the pendulum at displacement x is .5 k x^2. .5 m v^2 = .5 k A^2 - .5 k x^2 Solve for v v = +- sqrt( .5 k / m * (A^2 - x^2) )

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21:46:30 ** The PE of the pendulum at displacement x is .5 k x^2. By conservation of energy, if nonconservative forces are negligible, we find that the KE of the pendulum at position x is.5 k A^2 - .5 k x^2. This result is obtained from the fact that at max displacement A the KE is zero, and the KE change from displacement A to displacement x is the negative of the PE change between these points. Thus .5 m v^2 = .5 k A^2 - .5 k x^2. Solving for v we have v = +- sqrt( .5 k / m * (A^2 - x^2) ) . **

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RESPONSE --> ok

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21:49:26 How can we determine the maximum velocity of a pendulum using a washer and a rigid barrier?

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RESPONSE --> If a pendulum of length L is pulled back distance x, and stopped at the equilibrium point a washer on the pendulum will become a projectile to land at a distance determined by the horizontal velocity of the washer. That velocity is the same as the max velocity of the pendulum.

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21:49:31 GOOD STUDENT ANSWER: If we pullback a pendulum of length L a distance x (much smaller than L), and stop the motion at the equilibrium point (vertical limit of motion) a washer on the pendulum will become a projectile and project off the pendulum, to land at a distance from which we can determine the horizontal velocity of the washer. That velocity is the same as the max velocity of the pendulum, since the max velocity is that which is at the lowest point in its path.

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RESPONSE --> ok

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Your work looks good. Let me know if you have questions.