Question

A thin rod of length 1,5 m rotates at an angular velocity of 8 s^-1 about the axis, which is perpendicular to the rod and passes through one of its points. The linear velocity of one end of the rod is 2m/s. Determine the acceleration of the other end of the rod. Give your answer in m/s^2.

Answer 1

$$v_x=\omega x\\2=8x\\x=0.25\ m$$

$$l-x=1.5-0.25=1.25\ m$$

$$a(l-x)=\omega^2(l-x)=8^2(1.25)=80\frac{m}{s^2}$$