Angular velocity and angular momentum of rod

  • Thread starter zhuyilun
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  • #1
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Homework Statement


A uniform rod of mass 2.90×10^−2 kg and length 0.410m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.160 kg, are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 4.90×10−2^2 on each side from the center of the rod, and the system is rotating at an angular velocity 28.0 rad/s. Without otherwise changing the system, the catches are released, and the rings slide outward along the rod and fly off at the ends. 1)find the angular velocity when the rings reach the end of the rod 2)fly off the rod


Homework Equations





The Attempt at a Solution


angular momentum= I*w
so i thought the initial angular momentum is (2.9*10^2*(0.41/2)^2+0.16*2*(4.9*10^-2)^2)*28
since angular momentum is conserved due to the absence of net torque, the final momentum should be equal to the initial which is (2.9*10^2*(0.41/2)^2+0.16*2*(0.41/2)^2)*w, w is what i need for the first part. but this is wrong
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
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Hi zhuyilun! :smile:

(have an omega: ω and try using the X2 tag just above the Reply box :wink:)
… so i thought the initial angular momentum is (2.9*10^2*(0.41/2)^2+0.16*2*(4.9*10^-2)^2)*28
No, you've used mL2 for the moment of inertia of a rod of length 2L about its centre …

look up a list of common moments of inertia (eg http://en.wikipedia.org/wiki/List_of_moments_of_inertia" [Broken]), and learn them! :wink:
 
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