Question

A thin, uniform rod of length L and mass M is placed vertically on a horizontal table. If tilted ever so slightly, the rod will fall over.
a. What is the speed of the center of mass just as the rod hits the table if there’s so much friction that the bottom tip of the rod does not slide?
b. What is the speed of the center of mass just as the rod hits the table if the table is frictionless?

Short answer

a) Speed of the center of mass just as the rod hits the table is $v=\sqrt{\frac{3gl}{4}}$

b) Speed of the center of mass just as the rod hits the for frictionless table is $v=\sqrt{gl}$

Step-by-step solution

Part(a) Step1 : Given information

Length= L

Mass = M

Part(a) Step 2: Explanation

Moment of inertia of a rod is given by

$I=\frac{1}{3}M{L}^2..............................................\left(1\right)$

Total energy , when rod is standing on table is

$E=\frac{MgL}{2}.......................\left(2\right)$

Kinetic energy of the rod when it hits the table

$KE=\frac{1}{2}I{\omega }^2$

Substitute the value of inertia, we get

$KE=\frac{1}{2}\times \left(\frac{M{L}^2}{3}\right)\times {\omega }^2..........................\left(3\right)$

From the law of conservation of energy

$$\begin{gathered} E=KE \\ \frac{1}{2}MgL=\frac{1}{2}\times \frac{1}{3}M{L}^2\times {\omega }^2 \\ \omega =\sqrt{\frac{3g}{L}}\dots \dots \dots \dots \dots \dots .\left(4\right) \end{gathered}$$

Velocity is $v=\frac{L}{2}\omega$

Substitute in equation (4), we get

$$\begin{gathered} v=\frac{L}{2}\times \sqrt{\frac{3g}{L}} \\ v=\sqrt{\frac{3gL}{4}} \end{gathered}$$

Part(b) Step 1 : Given information

Length= L

Mass = M

Part(b) Step2 : Explanation

Total energy of the standing rod is

$E=mg\frac{L}{2}$( as center of mass is at middle of rod)

From the law of energy conservation

K=KE

As there is no friction so there will be no radial kinetic energy. so

$$\begin{gathered} \frac{1}{2}MgL=\frac{1}{2}M{v}^2 \\ v=\sqrt{gL} \end{gathered}$$