Question

Derive the formula mv²/r
(Class XI - Kinemetics- Circular Motion)

Answer 1

 h(2R-h) = x2 
but 2R>> h therefore 2h = x2 and so h = x2/2R (equation 1) 
now x = AK which is almost the arcAB = vt (equation 2) 
 
Combining 1 and 2, h = (vt)2/2R (equation 3) 
h is the vertical fall and so using s = 1/2 at2 = h (equation 4) 
 
Then from (3) and (4) 1/2at2 = (vt)2/2R leading to a = v2/R 
Using F = ma then F = mv2/R 

Answer 2

Answer -

It can be derived easily by the following way.....

Explanation -

This is Newton’s method.

The circle represents the orbit of a satellite of radius R, moving with speed v . The satellite moves from A to B in a time t. Without a force the satellite would have moved to K at constant speed.

From the crossed chords property, h(2R-h) = x ²

but 2R>h therefore 2v = x ² and so h =

$\frac{ {x}^{2} }{2r}$(equation 1)

now x = AK which is almost the arc AB = vt (equation 2)

Combining 1 and 2, h =

$\frac{(vt^{2}) }{2r}$

(equation 3)

h is the vertical fall and so using s =

$\frac{1}{2}a {t}^{2}$= h (equation 4)

Then from (equation 3) and (equation 4) -

$\frac{1}{2}a {t}^{2} \: = \: \frac{(vt^{2}) }{2r}$

Thus leading to -

$a \: = \: \frac{ {v}^{2} }{r}$

Hence, By the Equation -

F = ma

$f \: = \: \frac{ {mv}^{2} }{r}$

Thus , Proved .

My answer is absolutely CORRECT .... MY GUARANTEE....AM I RIGHT or AM I RIGHT....