From here:Why does torque lead to rotation? I have got good understanding of rotational motion. But no one has answered the question: why both the ends of rods move in opposite direction?
If the all the molecules and atoms of a rod are bind with the intermolecular forces, both the end should move in same direction.
Pls do explain.
Torque is a vector quantity and it's defined as the moment of force. That can be mathematically written as:
$τ=\vec{r}×\vec{F}$
Torque will have only two directions perpendicular to the plane of rotation. Now according to your question let's assume that both ends moves in the same direction. So now the position vector $\vec{r}$ must be measured form the axis of rotation and if we do it for the one end then the torque will come in some fixed direction let's assume the object is rotating in the x-y plane.
So the direction for torque is let's say outwards in the positive $z$ direction and now as the assumption taken above that another particle at the end of the rod moves in the same direction as the particle at the other side then the torque might get a same magnitude but the different directions and the angular acceleration too! How this can be possible for a single body to get the angular acceleration in both the directions if only one force is acting on the body which is either rotating clockwise or anticlockwise? So it should move opposite to the direction of the first particle so that the angular acceleration should only be in the one direction and hence the torque. (If there is only one force acting on the system and the system is not in the Rotational equilibrium)
Edit:- If there is some net angular acceleration then the angular velocity will generate anyway and Angular velocity is also a vector quantity so it also has to be in only one direction either outwards or inwards a particle can neither have more than one translational velocity or more than one angular velocity (I'm not talking about the components of the velocity I'm talking about the net velocity), otherwise that particle would displace in the two different directions in the same time and that is unphysical.
After the force is applied, the molecule which was pushed doesn't just move in the direction of the force, it also rotates slightly around it's neighboring molecules, because IMF's try to keep all the molecules separated by a certain distance. This force doesn't just act on the molecule that the original force was applied to, it also applied to it's neighboring molecules, thereby rotating those neighboring particles slightly. This continues on until all of the molecules are rotated. However, a significant portion of that original force is still applied in the original direction, and so the rod is rotated and it's center of mass is moved.
A rigid body can be modelled as a system of N particles where the distance between any pair of them doesn't change. By definition, the center of mass is such that if it is the origin of the coordinate system: $\sum m_k\mathbf r_k = 0$. If we differentiate the equation with respect to time: $\sum m_k\mathbf v_k = 0 \implies \sum \mathbf p_k = 0$. The sum of the linear momentum is always zero in the frame of the COM. But there is a lot of freedom of movement of the individual particles compatible with the requirement that $\sum \mathbf p_k = 0$. That is the natural state of movement of a rigid body free of external forces and can be quite complex.
If we are initially in the frame of the COM, and an arbitrary net force is applied over the body for a small time $\delta t$, the new momentum configuration is: $\sum \mathbf p_k' = \sum \mathbf p_k + \sum \boldsymbol {\delta p_k} = \sum \boldsymbol {\delta p_k}$. As the sum of the linear momentum is always zero in the frame of the COM, that means: the COM has now in our frame a linear momentum $\boldsymbol{\delta p} = \sum \boldsymbol {\delta p_k}$.
In the particular case of all $\boldsymbol {\delta p_k}$ are equal, the state of movement of the body doesn't change except for the translation of the COM. But if it is not the case (an imbalanced external force), there are many possibilities. If all $\mathbf p_k$ were zero initially, (the OP question) it is perfectly possible that for some $k$, $\boldsymbol {\delta p_k}$ is momentarilly in a direction opposed to $\sum \boldsymbol {\delta p_k}$, the translational movement of the COM.
The motion would be translational if both ends of the rod were to move in the same direction.
