# When does the Newton's second law for progressive motion apply to a rigid-body and why? [closed]

Disclaimer: I may seem to be using strange terms because although I know English rather well, it is not the language I learn physics in.

Consider a cylinder (with mass $$m$$ and radius $$r$$) on an inclined plane (incline is $$\alpha$$, an angle in radians). As far as I know this is a valid way to calculate its acceleration ($$a$$): $$\epsilon$$ - angular acceleration, $$\tau$$ - moment of force, $$I$$ - moment of inertia, $$g$$ - gravitational acceleration

$$\epsilon=\tau/I,\tau =F_f *r, a/r=\epsilon \rightarrow a/r=F_f*r/I$$

$$F_f=F-am$$ (this is the part I am asking about)

$$Ia=(F-am)r^2$$

$$1/2mr^2*a=(F-am)r^2$$

$$1/2am+am=F$$

$$3/2am=mg\sin\alpha$$

$$a=2/3g\sin\alpha$$

I am asking when does the Newton's second law for progressive movement ($$F=am$$) apply to rigid-bodies (with torque) and why is that the case? Does it only apply when the movement is without sliding or always etc.? I am uncertain because it seems to me that the force of friction does not affect the entire body evenly. I would appreciate an answer with an explanation why can this law be applied.

As another example consider that the cylinder would have a string attached to it. Its movement would differ were the string be attached to its center of mass or its bottom, so it seems that the point where the force is applied does matter.

To be clear I asking when and why $$F=am$$ applies for rigid-bodies (especially ones that are rotating). I know that the angular laws such as $$\tau=\epsilon*I$$ apply.

• What is “Newton's second law for progressive movement”? What are $\epsilon$, $\tau$, $I$, and $r$? Jun 24, 2020 at 19:54
• @G.Smith I meant the law where Force=mass*acceleration, but for movement where the movement is without rotation, it is progressive, forward. I don't know how it is called in english, that's the name I found (in contrast to the law where Moment of Force = Angular acceleration * Moment of Inertia). epsilon is angular acceleration, tau is moment of force, I is moment of inertia and r is the radius. I thought these were standard symbols across the world. Jun 24, 2020 at 20:05
• I don't think so @enthusiastic_3d_graphics_pr... , in fact, the symbol which is commonly used (atleast, in most cases) for angular acceleration is $\alpha$. We usually use the $\epsilon$ to denote the permittivity. And, in most cases, it isn't generally called the 'Newton's Second Law of Progressive Motion', as the motion can not only be progressive, it can be any sort of motion. It's usually just called the 'Newton's Second Law of Motion'. Jun 24, 2020 at 20:15
• @Firefox1921 Ok, thanks for pointing that out. I used the symbols from my country and I mistakenly thought that they were uniform across the world. Jun 24, 2020 at 20:17
• Could you please clarify what do you mean by $F_f$ in your question? Is it friction? Jun 24, 2020 at 20:20

I am asking when does the Newton's second law for progressive movement ($$F=am$$) apply to rigid-bodies (with torque) and why is that the case? Does it only apply when the movement is without sliding or always etc.?

No, $$N2L$$ applies regardless of whether the object is rolling without sliding, sliding without rolling or sliding only.

In order to have rolling (with or without sliding) we need torque $$\tau$$ about the CoG of the object, provided by the friction force $$F_f$$:

$$\tau=R F_f=\mu mgR\cos\alpha$$

This causes angular acceleration $$\alpha$$:

$$\tau=I\alpha$$

where $$I$$ is the inertial moment of the object about the axis running through the CoG.

So we have:

$$\alpha=\frac{\mu mgR\cos\alpha}{I}$$

'In the mean time', the object also undergoes translational acceleration $$a$$, with $$N2L$$:

$$F-F_f=ma$$

If the object rolls without any sliding, then:

$$a=\alpha R$$

So that for rolling without slipping:

$$\frac{F-F_f}{m}=\frac{RF_f}{I}$$

$$F_f=\frac{FI}{I+mR}$$

$$\mu mg\cos\alpha=\frac{FI}{I+mR}$$

So for strict rolling without sliding:

$$\mu \geq \frac{FI}{mg\cos\alpha (I+mR)}$$

For pure sliding, there can be no torque, so:

$$\mu=0$$

And for anything in between, that is rolling with some sliding:

$$0 \geq \mu \geq \frac{FI}{mg\cos\alpha (I+mR)}$$

To answer your question, $$ma$$ is just the the acceleration of the mass multiplied by the mass itself. It is equivalent to the net force applied on a system or a singular body(The acceleration changes accordingly) i.e. $$\vec F_{net}$$. It doesn't really matter whether the force is actually producing torque or not. There is simply no requirement of the need for a force to produce torque before it is included in the Second Law Equation.

There's also no such condition for the Newton's Second Law to be applied, whether the body is sliding or not sliding, rolling or not rolling or any random condition you can think of - it doesn't matter. It simply has to be in motion. This is strictly with regards to Newtonian Mechanics though, in Quantum Mechanics, this principle doesn't hold, as the mass also varies as the speed of the object approaches the speed of light ($$c$$)

As another example consider that the cylinder would have a string attached to it. Its movement would differ were the string be attached to its center of mass or its bottom, so it seems that the point where the force is applied does matter.

Again, the string, will produce a tension on the rigid body, which can be added as an additional parameter in the the equation.

Regarding the why aspect, I suggest you read this thread: Why are Newton's Laws valid for rigid bodies. There are some very good explanations to it.

• Why did you delete your answer, @Gert ? It was a great answer and you had even include the bit on friction. IMO, there was absolutely no need to. There's room for two on this wooden panel :P! Jun 24, 2020 at 21:10