Same as .11, but use the following system: A bead of mass

Step 1 of 3

A bead of mass m threaded on a frictionless straight rod, which lies in a horizontal plane, is forced to spin with a constant angular velocity \(\omega\)  about a vertical axis, through the midpoint of the rod.

Therefore, the kinetic energy of the system is equal to the sum of the translational kinetic energy and rotational kinetic energy.

\(T=\frac{1}{2} m \dot{r^{2}}+\frac{1}{2} l \omega^{2}\)

Here \(\dot{r}\) is the velocity of rotation and r is the position of the bead along the rod.

Moment of inertia of the rod is equal to the product of mass, square of the separation between the axis of rotation and mass.

\(I=m r^{2}\)

Now, write the total kinetic energy as follows,

\(\begin{aligned}
T & =\frac{1}{2} m \dot{r^{2}}+\frac{1}{2} l \omega^{2} \\
& =\frac{1}{2} m \dot{r}^{2}+\frac{1}{2} m r^{2} \omega^{2} \\
& =\frac{1}{2} m\left(\dot{r^{2}}+r^{2} \omega^{2}\right)
\end{aligned}\)