Question
The force experienced by a mass moving with a uniform speed v in a circular path of radius r experiences a force which depends on its mass, speed and radius. Prove that the relation is $\text{f} = \frac{\text{mv}^2} {\text{r}} $
✓
Answer
$\text{f }\infty\text{ m}^{a}\text{v}^{\text{b}}\text{r}^{\text{c}}$ $[\text{f}]=[\text{m}]^{\text{a}}[\text{v}]^{\text{b}}[\text{r}]^{\text{c}}$ $\text{MLT}^{-2}=\text{M}^{\text{a}}[\text{LT}^{-1}]^{\text{b}}\text{L}^{\text{c}}$ Equaiting the powers of $\text{M}:1=\text{a}$ $\text{L}:1=\text{b}+\text{c}$ $\text{T}:-2=-\text{b}$ Solve the equation to get a = 1, b = 2, c = -1, therefore, $\text{f}=\frac{\text{mv}^2}{\text{r}}$ is the correct relation.
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