Question

A uniform rod of mass $m$ and length $l$ is kept vertical with the lower end clamped. It is slightly pushed to let it fall down under gravity. Find its angular speed when the rod is passing through its lowest position. Neglect any friction at the clamp. What will be the linear speed of the free end at this instant?

Answer 1

R.E.F image

According to the question

We know moment of inertia of rod about

center of mass

Then, by parallel axis theorem the I about the

clamp will be

$I=I_{c}m+m(\frac{l}{2}{)}^2=\frac{1}{12}m{l}^2+\frac{m{l}^2}{4}=\frac{1}{3}m{l}^2$

Also, we know

Loss in potential energy = Gain in Kinetic energy

$mg\frac{l}{2}=\frac{1}{2}I{W}^2$

$mg\frac{l}{2}=\frac{1}{2}\left(\frac{1}{3}m{l}^2\right)W^2$

$W^2=\frac{3g}{l}$

$W=\sqrt{\frac{3g}{l}}$

Then, the linear speed of the free end at

this point

$V=W\times l$

$V=\sqrt{\frac{3g}{l}}\times l$

$V=\sqrt{3gl}$