# B Τ=mv^2 ? (Torque question)

1. Mar 23, 2017

### 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...

2. Mar 23, 2017

### A.T.

Tangential

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

3. Mar 23, 2017

### Haynes Kwon

Thank you very much. But may I ask why?

4. Mar 23, 2017

### 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.

5. Mar 23, 2017

### Haynes Kwon

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

6. Mar 23, 2017

### vanhees71

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

7. Mar 23, 2017

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.

8. Mar 23, 2017

### Haynes Kwon

You precisely pointed out where I was wrong. I know what cross product is, but I didn't even think of it. My bad!

Thanks a lot. I feel as if I had been hit by a hammer.
Thank you again. @vanhees71 @Charles Link

It's five am already. I should get a sleep.