Answer to Question #141955 in Physics for Muhammad Kamran

Question

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 ].

Answer 1

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

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

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

$$R=L-r=1.25\text{ m}.$$

The centripetal acceleration of this end is

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