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Question

A horizontal smooth rod AB rotates with a constant angular velocity ω=2.00rad/s about a vertical axis passing through its end A. A freely sliding sleeve of mass m=0.50kg moves along the rod from the point A with the initial velocity v0=1.00m/s. Find the Coriolis force acting on the sleeve (in the reference frame fixed to the rotating rod) at the moment when the sleeve is located at the distance r=50cm from the rotation axis in Newton. (Round off to the nearest integer.)

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Solution

The sleeve is free to slide along the rod AB. Thus only the centrifugal force acts on it. The equation is,
m˙v=mω2r where v=drdt.
But ˙v=vdvdr=ddr(12v2)
so, 12v2=12ω2r2+constant
or, v2=v20+ω2r2
v0 being the initial velocity when r=0. The Corialis force is then,
2mωv20+ω2r2=2mω2r1+(v0ωr)2
=2.83N on putting the values.

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