Answer to Question #141955 in Physics for Muhammad Kamran

Question #141955
A thin rod of length 1, 5 m rotates at an angular velocity 8 s- about the axis, which is perpendicular to the rod and passes through
one of its points. Linear velocity of one end of the rod is 2 m/s. Determine the acceleration of the other end of the rod. Give your
answer In [m/s ].
1
Expert's answer
2020-11-03T10:32:05-0500

Find the distance from the end with 2 m/s to the axis of rotation:


r=vω=28=0.25 m.r=\frac{v}{\omega}=\frac28=0.25\text{ m}.

Thus, the distance from the axis of rotation to the other end is


R=Lr=1.25 m.R=L-r=1.25\text{ m}.

The centripetal acceleration of this end is


a=ω2R=821.25=80 m/s2.a=\omega^2R=8^21.25=80\text{ m/s}^2.


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