Question

85. A thin rod of length and mass$\mathit{M}$ is held vertically with its lower end resting on a frictionless, horizontal surface. The rod is then released to fall freely. (a) Determine the speed of its center of mass just before it hits the horizontal surface. (b) What If? Now suppose the rod has a fixed pivot at its lower end. Determine the speed of the rod’s center of mass just before it hits the surface.

Short answer

The solution is

a) The speed of the center of mass$v_{CM}=\sqrt{\left(\frac{3gh}{4}\right)}$

b) This comes down to$v_{CM}=\sqrt{\left(\frac{3gh}{4}\right)}$

Step-by-step solution

Given information

It is given that

Length of the thin rod$=h$

Mass of the rod$=M$

The speed of its center of mass just before it hits the horizontal surface

a) We can observe that there are no horizontal forces acting on the rod in the presented issue, hence the center of mass will not move horizontally.

The rod rotates around the center of mass as it falls, and the center of mass drops straight down a distance of$\frac{h}{2}$ .

We obtain from energy saving.

$\begin{array}{l}{K}_f+U_{gf}=K_i+U_{gi}\\ \frac{1}{2}M{v}_{CM}^2+\frac{1}{2}I{\omega }^2+0=0+Mg\left(\frac{h}{2}\right)\end{array}$

$\frac{1}{2}M{v}_{CM}^2+\frac{1}{2}\left(\frac{1}{12}M{h}^2\right){\left(\frac{v_{CM}}{\left(\frac{h}{2}\right)}\right)}^2=Mg\left(\frac{h}{2}\right)$

We get the speed of the center of mass as by solving,

$v_{CM}=\sqrt{\left(\frac{3gh}{4}\right)}$

The speed of the rod’s center of mass if the rod has a fixed pivot at its lower end

b) In this section, the motion is purely rotational, with the center of mass traveling in a circular path of radius $\frac{h}{2}$around a fixed pivot point, which is the lower end of the rod.

We obtain from energy saving

$\begin{array}{l}{K}_f+U_{gf}=K_i+U_{gi}\\ \frac{1}{2}I{\omega }^2+0=0+Mg\left(\frac{h}{2}\right)\end{array}$

Or,

$\frac{1}{2}\left(\frac{1}{3}M{h}^2\right){\left(\frac{v_{CM}}{\frac{h}{2}}\right)}^2=Mg\left(\frac{h}{2}\right)$

As a result, this comes down to$v_{CM}=\sqrt{\left(\frac{3gh}{4}\right)}$