Question

Check the correctness of the equation $F{\text{ }} = {\text{ }}\dfrac{{m{v^2}}}{r}$ where F is a force, m is mass, v is velocity and r is a radius.

Answer 1

To solve this question, i.e., to check the correctness of the given equation. We will take standard formula of force, which is $F{\text{ }} = {\text{ }}m \times a.$ Then we will connect both the equations of force and the given equation, after converting those into their respective dimension formula, we will check whether both sides of the equation are equal or not, if it is equal, then the equation is correct, if not, then the equation will be incorrect.

Complete step by step answer:
We need to check the correctness of the equation given to us, which is, $F{\text{ }} = {\text{ }}\dfrac{{m{v^2}}}{r}$
where, F $=$ force, m $=$ mass, v $=$ velocity and r $=$ radius.
We know that, Force $=$ mass $\times$ acceleration, i.e., $F{\text{ }} = {\text{ }}m \times a........eq.(1)$
And the given equation of force is, $F{\text{ }} = \dfrac{{m{v^2}}}{r}........eq.(2)$
Now if we equate both the equations of force, we get
$\Rightarrow ma = \dfrac{{m{v^2}}}{r}$
Now, if we convert the above equation into dimension, we get

$$\Rightarrow {20}{l}{ML{T^{ - 2}} = M{L^2}{T^{ - 2}}{L^{ - 1}}} \\\ \Rightarrow {ML{T^{ - 2}} = ML{T^{ - 2}}}$$

Since, the dimension formula of Force and $\dfrac{{m{v^2}}}{r}$ are equal, therefore, the equation is dimensionally correct.

Note: In the solutions, we have changed the formula of force and also the given formula into its dimension formula. Let us understand about it in detail. So, the expression which shows the powers to which the fundamental units are to be raised to obtain one unit of a derived quantity is called the dimensional formula of that quantity. And for the dimension formula here, we have used MLT, where M stands for mass, L stands for length and T stands for Time.