In uniform circular motion, acceleration is expressed by the equation a = v 2 r .

Question

In uniform circular motion, acceleration is expressed by the equation  a  =            v      2        r    .

Shyla Maldonado · Accepted Answer

That equation is misleading.You wrongly assume that a is dv/dt here, which it isn&#039;t.Let us first rewrite the centripetal acceleration equation properly:      |                                a                →              N        |    =                              |                                                    v                            →                                |                2            R        ⟺      a    N    =                    v        2            R      where       a    N   is the normal acceleration. The normal acceleration can be expressed as                     a            →        −                              a          →                      T   where       a    T   is the tangential compenent of acceleration a. For uniform circular motion,       a    T    =  0 so what we&#039;re left with is      |                            d                                                    v                →                                                              d          t                      |    =                              |                                                    v              →                                                            |                    2                    R        ⟺  a  =                    v        2            R      Note that a here is   a  =            a      x      2        +          a      y      2       in 2-D. Since                     v        →              =      (          v      x        ,          v      y        )  , we can say  a  =                    (                                            d                              v                x                                                    d              t                                      )            2        +                  (                                            d                              v                y                                                    d              t                                      )            2        .Putting this into our centripital acceleration equation, we can get,                    (*)                                                        (                                                                    d                                          v                      x                                                                            d                    t                                                              )                        2                    +                                    (                                                                    d                                          v                      y                                                                            d                    t                                                              )                        2                          =                                                            v                x                2                            +                              v                y                2                                      R                              This is the proper differential equation we are solving.Note that since       a    T    =  0, we have that             d              d        t                  |                            v          →                      |    =            d              d        t                        v      x      2        +          v      y      2        =  0, so       v    x        a    x    +      v    y        a    y    =                    v        →              ⋅                    a        →              =  0 , which serves as our second equation to solve (  ⋆).Now, since you know that the solution to uniform circular motion is                     r        →              =      (    R    cos    ⁡    (    ω    t    )    ,    R    sin    ⁡    (    ω    t    )    ), you can go ahead and verify that it indeed satisfies   ⋆

Why does solving the differential equation for circular motion lead to an illogical result?

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