Question

Consider the simplest rigid body, consisting of two point objects each of mass m, connected by a massless stick of length 2f. Suppose that the motion of this baton is to spin about an axis we'll choose to call , i.e. w = w2. We choose our origin at the center of mass, which is stationary. The angle between the AB axis and i is B.

Answer 1

The whole situation is equivalent to that of a top precessing about an axis passing through its center of mass.

The relation for the torque in the latter case is given by:

$\vec{\tau }=\vec{\omega }*\vec{L}$.

We have $\vec{\tau }=\vec{r}*\vec{F}$ where $\vec{F}$ is the force acting on the body and $\vec{r}$ is the distance from the origin to the point at which the force is applied.

The force $\vec{F}$ is the tension in the stick T and r is the radius of the circle which the top part of the stick makes as it rotates about the z-axis.

Now $\vec{\tau }=\vec{r}*\vec{F}=lsin\beta *Tsin(90-\beta )=lTsin\beta cos\beta$ .

Also $\vec{\tau }=\vec{\omega }*\vec{L}=\omega Lsin(90-\beta )=\omega Lcos\beta$ .   

Equating the two expressions we have$lTsin\beta cos\beta =\omega Lcos\beta$ i.e.$\omega L=lTsin\beta$ or $L=lTsin\beta/\omega$. eq 1)

Now we know that the radial component of the tension in the stick provides the necessary centripetal force responsible for its rotational motion i.e $Tcos(90-\beta )=Mv^{2}/r=m\omega ^{2}r=m\omega ^{2}lsin\beta$ since $r=lsin\beta$.

Therefore we have $T=m\omega ^{2}l$ .

Substituting this expression for T in eq 1) we obtain:
$L=m\omega l^{2}sin\beta$ .