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A thin rod has a length of 0.288 m and rotates in a circle on a frictionless tabletop. The axis is perpendicular to the length of the rod at one of its ends. The rod has an angular velocity of 0.602 rad/s and a moment of inertia of 1.22 x 10-3 kg·m2. A bug standing on the axis decides to crawl out to the other end of the rod. When the bug (whose mass is 5 x 10-3 kg) gets where it's going, what is the change in the angular velocity of the rod?

1 Answer

4 votes

Answer:

0.152724283058 rad/s

Step-by-step explanation:


\omega_i=0.602\ rad/s

In this system the angular momentum is conserved


L_i=L_f\\\Rightarrow 1.22* 10^(-3)* 0.602=(1.22* 10^(-3)+5* 10^(-3)* 0.288)\omega_f\\\Rightarrow \omega_f=(1.22* 10^(-3)* 0.602)/((1.22* 10^(-3)+5* 10^(-3)* 0.288^2))\\\Rightarrow \omega_f=0.449275716942\ rad/s

Change in angular velocity is


\Delta \omega=0.449275716942-0.602=-0.152724283058\ rad/s

The change in angular velocity is 0.152724283058 rad/s

User Andrey Kon
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