New formula for centripital force ? whats wrong

[ManishR]

new formula for centripital force ? what's wrong !

consider a circular motion with following variables with usual meanings,
\vec{r},\vec{F},\overrightarrow{\theta},t,v

v=r\frac{d\theta}{dt}

now

\frac{d\hat{r}}{dt}=(\frac{d\theta}{dt})\hat{\theta}

\Rightarrow\frac{d\hat{r}}{dt}=\frac{v}{r}\hat{\theta}

\Rightarrow\frac{d^{2}\hat{r}}{dt^{2}}=-\frac{v}{r}\hat{r}

now according to Newton's law

m\frac{d^{2}\overrightarrow{r}}{dt^{2}}=\overrightarrow{F}

\Rightarrow mr\frac{d^{2}\hat{r}}{dt^{2}}=\overrightarrow{F}

\Rightarrow-mv\hat{r}=\overrightarrow{F}

i am still not sure what actually this equation saying.
can someone recheck it please ? where i gone wrong ?

 

[K^2]

You've made a mistake taking \frac{d\hat{\theta}}{dt}.

\frac{d\hat{\theta}}{dt} = -\left(\frac{d\theta}{dt}\right)\hat{r}

So

\frac{d^2\hat{r}}{dt^2} = - \left(\frac{v}{r}\right)^2 \hat{r}

And then

\vec{F} = mr\frac{d^2\hat{r}}{dt^2} = -\frac{mv^2}{r}\hat{r}

Which is the correct formula for centripetal force.

 

[ManishR]

You've made a mistake taking \frac{d\hat{\theta}}{dt}.

\frac{d\hat{\theta}}{dt} = -\left(\frac{d\theta}{dt}\right)\hat{r}

So

\frac{d^2\hat{r}}{dt^2} = - \left(\frac{v}{r}\right)^2 \hat{r}

And then

\vec{F} = mr\frac{d^2\hat{r}}{dt^2} = -\frac{mv^2}{r}\hat{r}

Which is the correct formula for centripetal force.

thank u so much for ur help.

 

[litup]

You've made a mistake taking \frac{d\hat{\theta}}{dt}.

\frac{d\hat{\theta}}{dt} = -\left(\frac{d\theta}{dt}\right)\hat{r}

So

\frac{d^2\hat{r}}{dt^2} = - \left(\frac{v}{r}\right)^2 \hat{r}

And then

\vec{F} = mr\frac{d^2\hat{r}}{dt^2} = -\frac{mv^2}{r}\hat{r}

Which is the correct formula for centripetal force.

What is the difference between the two 'r''s? r by itself and r ^?

 

[Doc Al]

What is the difference between the two 'r''s? r by itself and r ^?

r is a magnitude; \hat{r} is a unit vector.