- #1
ManishR
- 88
- 0
new formula for centripital force ? what's wrong !
consider a circular motion with following variables with usual meanings,
[tex]\vec{r},\vec{F},\overrightarrow{\theta},t,v[/tex]
[tex]v=r\frac{d\theta}{dt}[/tex]
now
[tex]\frac{d\hat{r}}{dt}=(\frac{d\theta}{dt})\hat{\theta}[/tex]
[tex]\Rightarrow\frac{d\hat{r}}{dt}=\frac{v}{r}\hat{\theta}[/tex]
[tex]\Rightarrow\frac{d^{2}\hat{r}}{dt^{2}}=-\frac{v}{r}\hat{r}[/tex]
now according to Newton's law
[tex]m\frac{d^{2}\overrightarrow{r}}{dt^{2}}=\overrightarrow{F}[/tex]
[tex]\Rightarrow mr\frac{d^{2}\hat{r}}{dt^{2}}=\overrightarrow{F}[/tex]
[tex]\Rightarrow-mv\hat{r}=\overrightarrow{F}[/tex]
i am still not sure what actually this equation saying.
can someone recheck it please ? where i gone wrong ?
consider a circular motion with following variables with usual meanings,
[tex]\vec{r},\vec{F},\overrightarrow{\theta},t,v[/tex]
[tex]v=r\frac{d\theta}{dt}[/tex]
now
[tex]\frac{d\hat{r}}{dt}=(\frac{d\theta}{dt})\hat{\theta}[/tex]
[tex]\Rightarrow\frac{d\hat{r}}{dt}=\frac{v}{r}\hat{\theta}[/tex]
[tex]\Rightarrow\frac{d^{2}\hat{r}}{dt^{2}}=-\frac{v}{r}\hat{r}[/tex]
now according to Newton's law
[tex]m\frac{d^{2}\overrightarrow{r}}{dt^{2}}=\overrightarrow{F}[/tex]
[tex]\Rightarrow mr\frac{d^{2}\hat{r}}{dt^{2}}=\overrightarrow{F}[/tex]
[tex]\Rightarrow-mv\hat{r}=\overrightarrow{F}[/tex]
i am still not sure what actually this equation saying.
can someone recheck it please ? where i gone wrong ?