Question

A uniform rod of mass 1 kg and length 2 m is free to rotate about one end. If the rod is released from rest at an angle of 60° with respect to the horizontal, what is the speed of the tip of the rod as it passes the horizontal position?

Answer 1

We will have the following:

First, we remember that by the conservation of energy the following is true:

$KE_f+PE_f=KE_i+PE_i$

Now, this in terms of movement with respect to the angle given is:

$\frac{1}{2}I\omega^2_f+m\cdot g\cdot(0)=\frac{1}{2}I(0rad/s)^2_i+m\cdot g\cdot h$

So:

$\frac{1}{2}I\omega^2_f=m\cdot g\cdot h$

From this and the information provided we can see that:

$h=(2m)\sin (60)\Rightarrow h=\sqrt[]{3}m$

And from the inertia of a rod we have that:

$I=\frac{1}{3}(1kg)(2m)^2\Rightarrow I=\frac{4}{3}kg\cdot m^2$

Now, we plug in this information and solve for the angular velocity:

$\frac{1}{2}(\frac{4}{3}kg\cdot m^2)\omega^2_f=(1kg)(9.8m/s^2)(\sqrt[]{3}m)\Rightarrow\omega^2_f=\frac{9.8\cdot\sqrt[]{3}}{(2/3)}$$\Rightarrow\omega_f=\sqrt[]{\frac{9.8\sqrt[]{3}}{(2/3)}}\Rightarrow\omega_f=5.04590397\ldots$$\Rightarrow\omega_f\approx5$

So, the velocity of the tip of the rod as it passes the horizontal position is approximately 5 radians/s.