SolutionsSet06

SolutionsSet06 - Homework Solutions Set 06 Physics 121...

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Homework Solutions Physics 121 Set 06 Spring 2008 - 1 - Problem 1 Consider the equations of motion for a system that rotates with a constant angular acceleration α . The angular velocity ω as function of time is equal to ! ( t ) = 0 + " t At the current time, the period of rotation is equal to 0.19 s, which corresponds to an angular velocity of 2 π / T = 33.1 rad/s. The period of the rotation at time t is equal to T ( t ) = 2 ( t ) = 2 0 + # t The period of rotation at a time t + Δ t is equal to T ( t + ! t ) = 2 ( t + ! t ) = 2 0 + $ ( t + ! t ) The difference in the period of rotation is thus equal to ! T = T ( t + ! t ) " T ( t ) = 2 0 + % ( t + ! t ) " 2 0 + t = 2 ( 0 + t ) " ( 0 + ( t + ! t )) ( 0 + ( t + ! t ))( 0 + t ) & ( ) * + , " T 2 ! t 2 The angular acceleration is thus equal to = " 2 T 2 $ T $ t = " 5.28 % 10 -11 rad/s 2 The pulsar stops rotating when its angular velocity is 0 rad/s. This happens at a time t when ( t ) = 0 + t = 0 or t = ! 0 = 19,833 yr The pulsar originated in a super-nova explosion in 1054 or 950 years ago. The time dependence of the angular velocity is given by ( t ) = 0 + t
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Homework Solutions Physics 121 Set 06 Spring 2008 - 2 - where ω 0 is the angular velocity at the time of the super-nova explosion. At time t = 950 yr, the angular velocity is 33.1 rad/s. Based on this information we can determine 0 : ! 0
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SolutionsSet06 - Homework Solutions Set 06 Physics 121...

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