Question 526: Centre of Mass and Rotational Motion

Question

A particle of mass ' m ' is rotating in a circular path of radius ' $r$ '. Its angular momentum is ' $L$ ' The centripetal force acting on it is ' $F$ '. The relation between ' $F$ ', ' $L$ ', ' $r$ ' and ' $m$ ' is

Accepted Answer

The correct relationship between the centripetal force $ F $, angular momentum $ L $, radius $ r $, and mass $ m $ can be derived using the concepts of circular motion and angular momentum.
Angular momentum $ L $ is given by:
$$ L = mvr $$
where $ v $ is the tangential velocity. For an object in circular motion, the centripetal force $ F $ is also defined as:
$$ F = \frac{mv^2}{r} $$
By rearranging the first equation for velocity, we have:
$$ v = \frac{L}{mr} $$
Substitute $ v $ into the expression for $ F $:
$$ F = \frac{m \left(\frac{L}{mr}\right)^2}{r} = \frac{m \cdot \frac{L^2}{m^2r^2}}{r} $$
Simplifying further:
$$ F = \frac{L^2}{mr^3} $$
Thus, the correct relationship is:
$$ \frac{L^2}{m} = Fr^3 $$
Therefore, the option that matches this expression is
Option C
:
$$ \frac{L^2}{m} = Fr^3 $$

Answer

$\mathrm{F}=\frac{\mathrm{L}}{\mathrm{mr}^2}$

Answer

$\mathrm{L}=\mathrm{m}^2 \mathrm{Fr}^2$

Answer

$\frac{\mathrm{L}^2}{\mathrm{~m}}=\mathrm{Fr}^3$

Answer

$\frac{\mathrm{F}}{\mathrm{L}^3}=\mathrm{mr}^2$

Centre of Mass and Rotational Motion

A particle of mass ' m ' is rotating in a circular path of radius ' r '. Its angular momentum is ' L ' The centripetal force acting on it is ' F '. The relation between ' F ', ' L ', ' r ' and ' m ' is

Choose the correct answer: