Are Newton's Laws applicable to torque too?

Question

Since Torque is $Fd$ where $F$ is force, $d$ is perpendicular distance between line of action of force and axis of rotation, will Newtons Laws be applicable to torque too?

For e.g., if a force od $5N$ is acting on a body fixed $2m$ away from the point of action of force, producing a constant torque of $10Nm$, if we stop applying the force of $5N$, or we apply a force of $-5N$ to make the net force $0$, will the object still rotate due to Newtons first Law, or stop? I am asking this as torque depends on force directly.

Answer 1

You can still have a net torque with the net force being $0$. You do this wherever you turn a door knob, for example. The knob turns about its center, but the center of mass of the knob doesn't accelerate (assuming you don't move the door itself). Such an instance of zero net force and non-zero net torque is called a couple.

There are analogs of Newton's laws for torques and rotations though. For example, for an object rotating about a principle axis we have the analog of $F=ma$ as $\tau=I\alpha$, where $\tau$ is the torque $I$ is the moment of Inertia about the principle axis and $\alpha$ is the angular acceleration (all about the specified principle axis).

A more general form of this comes from the linear momentum $\mathbf p=m\mathbf v$ in the form of $\mathbf L=\hat I\boldsymbol\omega$ where $\mathbf L$ is the angular momentum vector, $\boldsymbol\omega$ is the angular velocity vector, and $\hat I$ is the moment of inertia tensor. Then, just like how $\dot{\mathbf p}=m\dot{\mathbf v}\to\mathbf F=m\mathbf a$, we have $\dot{\mathbf L}=\hat I\dot{\boldsymbol\omega}\to\boldsymbol\tau=\hat I\boldsymbol\alpha$

Answer 2

I am unsure what you are asking because torque is by definition $r \times F$ so of course it depends on the force directly. However, for your question:

if a force of $5N$ is acting on a body fixed $2m$ away from the point of action of force, producing a constant torque of $10Nm$, if we stop applying the force of $5N$, or we apply a force of $-5N$ to make the net force $0$, will the object still rotate due to Newtons first Law, or stop?

The answer is the object will continue to rotate. Rotational motion has its own set of kinematics equations; without a torque there is no change in angular velocity and the object continues to rotate.

Answer 3

Yes, Newton's laws are applicable. Torque is a convenient measure for rotational motion. For a system of particles (not a point body) it would be possible, but very difficult, to describe the rotational motion using forces only; but using torque the motion can be described much more easily. For a rigid body rotating about an axis in a plane we have simply $\tau = I \alpha$ where $\tau$ is the total external torque, $I$ is the moment of inertia, and $\alpha$ is the angular acceleration. This is the rotational analog of $F = ma$ for translational motion of the center of mass in one dimension. For general translation/rotation of a rigid body, you can use the Euler equations which use torque and principal axes, and inertia is a tensor (see a physics mechanics text such as Goldstein, Classical Mechanics).

In general, $\vec \tau = {d \vec L \over dt}$ where $\vec L$ is the angular momentum; this is the rotational analog of $\vec F = {d \vec p \over dt}$ where $\vec p$ is the linear momentum.

Your question is what happens if we start with a net force that produces a torque on a point mass, then change the net force to zero therefore also changing the net torque to zero. At the time the force/torque is reduced to zero, the point mass has a certain linear momentum $\vec p$ and a certain angular momentum $\vec L$. When the net force and net torque are reduced to zero, as you can see from the relationships in the second paragraph above, there is no change in linear or angular momentum, so the point mass retains the linear and angular momenta present when the net force/torque were reduced to zero.

Suggest you also look at Why is the concept of torque necessary? on this exchange.

Hope this helps.