Question

A particle of mass $m$ rotating along a circular path of the radius $r$ with uniform speed. Its angular momentum about the axis of rotation is $L$, the centripetal force acting on the particle is:
A) $\dfrac{{{L}^{2}}}{m{{r}^{3}}}$
B) $\dfrac{{{L}^{2}}}{mr}$
C) $\dfrac{L}{m{{r}^{2}}}$
D) $\dfrac{{{L}^{2}}m}{r}$
E) $\dfrac{Lm}{{{r}^{2}}}$

Answer 1

Angular momentum in rotational motion is equivalent to linear momentum in translation motion. Angular momentum is a vector quantity that is a measure of the rotational momentum of a rotating body. The angular momentum is directed along the rotation axis.

Formula Used:
$L=I\omega$, $I=m{{r}^{2}}$, $\omega =\dfrac{v}{r}$ and $F=\dfrac{m{{v}^{2}}}{r}$

Complete step by step solution:
As stated in the hint, angular momentum is equivalent to linear momentum. We know that linear momentum is equal to the product of the mass of the body and its linear velocity. The equivalent of mass in rotational motion is the Moment of Inertia and the equivalent of linear velocity in rotational motion is angular velocity.
Mathematically, we can say that $L=I\omega$ where $L$ refers to the angular momentum of the body, $I$ is the Moment of Inertia of the body and $\omega$ is the angular velocity.
The Moment of Inertia $I$ for a particle of mass $m$ is given as $I=m{{r}^{2}}$ and the angular velocity $\omega$ of the body is given as $\omega =\dfrac{v}{r}$ where $v$ is the uniform speed of motion of the particle and $r$ is the radius of the circular path.
Substituting the values of the moment of inertia and the angular velocity in the expression for angular momentum, we get

& L=I\times \omega =m{{r}^{2}}\times \dfrac{v}{r} \\\ & \Rightarrow L=mvr \\\ & \Rightarrow v=\dfrac{L}{mr} \\\ \end{aligned}$$ Now centripetal force $$F$$ acting on a body in a circular motion is given as $$F=\dfrac{m{{v}^{2}}}{r}$$ Substituting the value of velocity obtained in terms of angular momentum, we get $$\begin{aligned} & F=\dfrac{m{{v}^{2}}}{r} \\\ & \Rightarrow F=\dfrac{m}{r}\times {{v}^{2}}=\dfrac{m}{r}\times {{\left( \dfrac{L}{mr} \right)}^{2}} \\\ & \Rightarrow F=\dfrac{m{{L}^{2}}}{r\times {{m}^{2}}\times {{r}^{2}}} \\\ & \Rightarrow F=\dfrac{{{L}^{2}}}{m{{r}^{3}}} \\\ \end{aligned}$$ **Hence option (A) is the correct option.** **Note:** An alternative method of approaching this problem is as follows. Angular momentum is also equivalent to the torque of linear momentum or the moment of momentum. Now the torque of linear momentum will be the product of linear momentum and the distance from the axis of rotation, that is $$L=mvr$$. Hence we can skip the angular velocity part.