Is Torque Determined by Centripetal or Tangential Acceleration?

[Haynes Kwon]

Since F=Δp/Δt...

Δp=mΔv+vΔm, Δm=0
Δp=mΔv

Therefore, Δp/Δt=mΔv/Δt=ma.
And if I multiply 'r' to both sides

rF=rma, which is τ=rma

My question is: is 'a' centripetal acceleration or is it tangential acceleration?
If 'a' is centripetal acceleration, which is v^2/r, τ=rmv^2/r=mv^2?

Also I don't really get what tangential acceleration is...

 

[A.T.]

Since F=Δp/Δt...

Δp=mΔv+vΔm, Δm=0
Δp=mΔv

Therefore, Δp/Δt=mΔv/Δt=ma.
And if I multiply 'r' to both sides

rF=rma, which is τ=rma

My question is: is 'a' centripetal acceleration or is it tangential acceleration?

Tangential

Also I don't really get what tangential acceleration is...

Perpendicular to centripetal (if the path is a circle).

 

[Haynes Kwon]

Thank you very much. But may I ask why?

 

[vanhees71]

The torque is an axial vector quantity, i.e., it is defined by
$$\vec{\tau}=\vec{r} \times \vec{F}.$$
Here, $\vec{r}$ is the radius vector from a fixed point (with respect to which you want to calculate the torque) and $\vec{F}$ the force acting on the particle. Obviously the torque is perpendicular to the plane spanned by $\vec{r}$ and $\vec{F}$, and it's 0 whenever $\vec{F} \propto \vec{r}$.

Defining the angular momentum as
$$\vec{L}=\vec{r} \times \vec{p}=m \vec{r} \times \dot{\vec{r}},$$
you get
$$\dot{\vec{L}}=m \dot{\vec{r}} \times \dot{\vec{r}}+m \vec{r} \times \ddot{\vec{r}}=m \vec{r} \times \ddot{\vec{r}}=\vec{r} \times \vec{F}=\vec{\tau}.$$

Particularly this shows that for central forces, i.e., for $\vec{F} \propto \vec{r}$ the angular momentum is conserved.

 

[Haynes Kwon]

Thank you so much.
So you mean that τ=mv^2 is correct?

 

[vanhees71]

No. I said $\vec{\tau}=\vec{r} \times \vec{F}$!

 

[Charles Link]

No. I said $\vec{\tau}=\vec{r} \times \vec{F}$!

Perhaps I can help clarify to @Haynes Kwon . When the symbol $\times$ is used between two vectors, it means vector cross product. If the two vectors are parallel or anti-parallel, the result is zero. The two vectors get the maximum cross product when they are perpendicular. In the case of an object moving in a circle, the centripetal force ($F=\frac{mv^2}{r}$) supplies zero torque because $\vec{r} \times \vec{F}=0$. The vectors $\vec{r}$ and $\vec{F}$ lie along the same line=they are anti-parallel, so their vector coss product is zero. Note: For a vector cross product, $| \vec{A} \times \vec{B} |=|\vec{A}||\vec{B}|sin(\theta)$ where $\theta$ is the angle between the two vectors.

 

[pasajerodeltoro]

Since F=Δp/Δt...

Δp=mΔv+vΔm, Δm=0
Δp=mΔv

Therefore, Δp/Δt=mΔv/Δt=ma.
And if I multiply 'r' to both sides

rF=rma, which is τ=rma

My question is: is 'a' centripetal acceleration or is it tangential acceleration?
If 'a' is centripetal acceleration, which is v^2/r, τ=rmv^2/r=mv^2?

Also I don't really get what tangential acceleration is...

[1st point]
We have:
\vec {Δp} =m( \vec Δv )+ ( \vec v )Δm, Δm=/=0, Δm="change in mass".
We can hypothesise that this "transfer of mass"Δm[x]+"change in internal mass distribution"Δm[r], in general.
So, we may have Δm=0 superficially, but Δm=Δm[x]+Δm[r] , in reality. (x='linear measure', r='radial measure')
To clarify, there may be cases Δm=Δm[x] [ii] Δm=Δm[r] [iii] Δm=0.

[2nd point]
Considering case [iii], (which makes sense for circular motion as the mass always returns to the same point):

We take the 'cross product' between the 'pair of vectors': \tilde P={ \vec r , \vec F } ;
such that: \vec r X \vec F = |r| |F| sin(θ) ( \hat n ) , where \hat n ="unit normal vector" to \tilde P .

So, for case [iii], we really have: \tilde τ = |r| |F| sin(θ) ( \hat n ) = |r| m|a| sin(θ) ( \hat n )
The acceleration resultant is comes from the vector addition of nutational (3d only), centripetal 'r' and tangential 'θ' components. We will consider only the latter 2 components, thus: \vec a = \vec a _r + \vec a _θ

From Newton's Principia, we have:
\vec a _r = \dot v _θ = \frac { \vec dv _θ } {dt} = ( \frac v ² , R ) \hat r ; with variable R=|r| , v=| \vec v _θ | = ωR
and \vec a_θ = \dot{ \vec v _r } = \frac { d \vec v _r } {dt}.

We could put \vec v _r = 0 for circular motion, or ; for simple harmonic motion r could vary sinusoidally wrt t,
so that: r= r ₀ sin ( ω _r t ) ,
and: vec v _r = \frac { \vec r} {dt} =\frac {d ( r ₀ sin ( ω _r t ) \hat r )} {dt}
⇒ vec v _r = ω _r r ₀ cos ( ω _r t ) \hat r +r ₀ sin ( ω _r t) \frac { d \hat r } {dt}
... substitute \hat r = \vec r / |r| and work out the derivative wrt t , then find \dot{ \vec v _r }

By letting |a|=a , we have: \tilde τ = R m a sin(θ) ( \hat n )
The centripetal component can be considered transversal as it varies 'up and down'.

We get a from the sum of the squares of the transversal and tangential components.