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B Τ=mv^2 ? (Torque question)

  1. Mar 23, 2017 #1

    Haynes Kwon

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    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. jcsd
  3. Mar 23, 2017 #2

    A.T.

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    Tangential

    Perpendicular to centripetal (if the path is a circle).
     
  4. Mar 23, 2017 #3

    Haynes Kwon

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    Thank you very much. But may I ask why?
     
  5. Mar 23, 2017 #4

    vanhees71

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    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.
     
  6. Mar 23, 2017 #5

    Haynes Kwon

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    Thank you so much.
    So you mean that τ=mv^2 is correct?
     
  7. Mar 23, 2017 #6

    vanhees71

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    No. I said ##\vec{\tau}=\vec{r} \times \vec{F}##!
     
  8. Mar 23, 2017 #7

    Charles Link

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    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.
     
  9. Mar 23, 2017 #8

    Haynes Kwon

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    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.
     
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