Circular Motion
4.0 Rigid body rotating in a vertical circle
4.0 Rigid body rotating in a vertical circle
Suppose a rod of mass $m$ and length $L$ is rotated about one of its end in a vertical circle.
$l$ is the distance of center of mass of rod from the hinge point $O$ such that $\left( {l = \frac{L}{2}} \right)$, as shown in the figure.
Properties of a rod
- It is a rigid body
- All the forces acts at the centre of mass
- Tension can be away from the rod or towards the rod which means that the tension vector can be both positive or negative
Note: Tension is a vector quantity. So, positive or negative tension gives us idea about the direction.
$u$ is the initial velocity imparted to the rod in horizontal direction at point $A$. Rod rotates in a circular path about the hinge point $O$.
$h$: is the rise in height of the center of mass of rod
$L$: total length of the rod
$l$: distance between centre of mass and hinge point $O$
At point $B$, the rod makes an angle $\theta $ with the vertical and has tangential velocity $v$.
We know, $${v^2} = {u^2} - 2gh$$
From the figure we get, $$h = + \frac{L}{2}\left( {1 - \cos \theta } \right)$$
Note: $h$ is positive as the centre of mass moves upward.
From FBD at point $B$ we get, $${T_B} - mg\cos \theta = \frac{{m{v^2}}}{l}$$
The above equation is a basic equation for rod in circular motion.
