Whakaoti i te F=mg+mv^2/R

$$F = m g + \frac { m v ^ { 2 } } { R }$$

Whakaoti mō R

$$\left\{\begin{matrix}R=-\frac{mv^{2}}{gm-F}\text{, }&v\neq 0\text{ and }m\neq 0\text{ and }F\neq gm\\R\neq 0\text{, }&\left(F=gm\text{ and }v=0\right)\text{ or }\left(v\neq 0\text{ and }F=0\text{ and }m=0\right)\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tē taea kia ōrite te tāupe R ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te R.

$$FR=mgR+mv^{2}$$

Tangohia te mgR mai i ngā taha e rua.

$$FR-mgR=mv^{2}$$

Whakaraupapatia anō ngā kīanga tau.

$$-Rgm+FR=mv^{2}$$

Pahekotia ngā kīanga tau katoa e whai ana i te R.

$$\left(-gm+F\right)R=mv^{2}$$

He hanga arowhānui tō te whārite.

$$\left(F-gm\right)R=mv^{2}$$

Whakawehea ngā taha e rua ki te F-mg.

$$\frac{\left(F-gm\right)R}{F-gm}=\frac{mv^{2}}{F-gm}$$

Mā te whakawehe ki te F-mg ka wetekia te whakareanga ki te F-mg.

$$R=\frac{mv^{2}}{F-gm}$$

Tē taea kia ōrite te tāupe R ki 0.

$$R=\frac{mv^{2}}{F-gm}\text{, }R\neq 0$$

Whakaoti mō F

$$F=\frac{m\left(v^{2}+Rg\right)}{R}$$

$$R\neq 0$$