Physics question Ch x Three of the following X d5 all subred C The Drawing Show

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Physics question Ch x Three of the following X d5 all subred C The Drawing Show Wiley PLUS x M Your question was ar X e- C Secure https:// dug wiley plus.com/edugen/student/ma run Pem Assig IMESSAGE MY INSTRICTOR soREEN PRINTER VERSION BACK NEX ASSIGNMENT Chapter 09, Problem 50 Go RESOURCES Chapter 9 Incorrect. Homework Problems 03 Two thin rods of length L are rotating with the same angular speed (in rad/s) about axes that pass perpendicularly through one end. Rod A is massless but has a particle of mass 0.71 kg attached to its free end. Rod B has a mass of 0.71 kg, which ls distributed unlformly along its length. The length of each rod is 0.86 m, and the angular speed is 6.0 rad/s. Find the kinetic energies of rod A with its attached particle and of rod B. (a) KEA 2......................................... Chapter Problem Chapter 09, Problem (b) KE Click if you would like to show work for this question Open Show Work Chapter 09, Problem GO TUTORIAL E Chapter ng Prohlem LINK IO ILXI hapter Problem Question Attempts: 1 of 7 used SAVE FOR LATER SUBMIT ANSWER Chaner Problem powered by License Agreement l Prtvacy Pollcy I 2000 201Z lohn WLley & Sons, Inc. All Rights Reserved. A DIvlslon of ohn WLley & Sons, Inc. Version 4.22.3 308 PM ype here to search Desktop 5/7/2017

Explanation / Answer

(a)

Rod A is mass less but has a mass m = 0.71 kg attached to its end. So kinetic energy will be only due to the speed of mass attached to its end.

The particle is rotating on a circle of radius L. So the linear speed of the particle is v =  L

So kinetic energy of the rod is,

KEA = (1/2)mv2 = (1/2)m(L)2 = 0.5(0.71 kg)(6 rad/sX0.86 m)2

or, KEA = 9.45 J

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(b)

The rod is rotating about its end. The moment of inertia of the rod about its end is,

I = mL2/3 = (0.71 kg)(0.86 m)2/3

or, I = 0.175 kg-m2

So KEB  =(1/2)I2 = (1/2)(0.175 kg-m2)(6 rad/s)2

or, KEB = 3.15 J

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