Question

A uniform rod of mass M and length L can pivot freely at one end. Initially, the rod is oriented vertically above the pivot, in unstable equilibrium, and is released from rest. When the rod is again vertical, below the pivot, what is the speed of its center of mass (in terms of g and L)? The rotational inertia about the end of a uniform rod is 1 3 M L2 .

Answer 1

Answer:

The speed of its center of mass =$\sqrt{\frac{3}{2}gL}$

Explanation:

Consider the potential energy at the level of center of mass of rod below the pivot=0

Mass of uniform rod=M

Length of rod=L

The rotational inertia about the end of a uniform rod=$\frac{1}{3}ML^2$

Kinetic energy at the level of center of mass of rod below the pivot=$\frac{1}{2}I\omega^2$

Kinetic energy at the level of center of mass of rod above the pivot=0

Potential energy at the level of center of mass of rod above the pivot=mgh

We have to find the center of mass ( in terms of g and L).

According to conservation of law of energy

Initial P.E+Initial K.E=Final P.E+Final K.E

$mgh+0=0+\frac{1}{2} I\omega^2$

Where $K.E=\frac{1}{2} I\omega^2$

I=Moment of inertia

$\omega$=Angular velocity

Substitute the values then we get

$MgL=\frac{1}{2}\times \frac{1}{3}ML^2\omega^2$

$\omega^2=\frac{6g}{L}$

Now, we know that $\omega=\frac{v}{r}$, $r=\frac{L}{2}$

Substitute the values then we get

$\frac{v^2}{(\frac{L}{2})^2}=\frac{6g}{L}$

$\frac{v^24}{L^2}=\frac{6g}{L}$

$v^2=\frac{6g\times L^2}{4L}$

$v^2=\frac{3gL}{2}$

$v=\sqrt{\frac{3}{2}gL}$

Hence, the speed of its center of mass =$\sqrt{\frac{3}{2}gL}$