Question

A particle of mass m is executing uniform circular motion on a path of radius $r$. If \[p\] is the magnitude of its linear momentum, the radial force acting on the particle is _____.
A. $pmr$
B. $\dfrac{rm}{p}$
C. \[\dfrac{m{{p}^{2}}}{r}\]
D. $\dfrac{{{p}^{2}}}{mr}$

Answer 1

Hint: Uniform circular motion is defined as the motion in which the object covers equal distance in equal interval of time. The radial force is given by \[F=\dfrac{m{{v}^{2}}}{r}\] and linear momentum is given by$p=mv$.

Formula used: \[F=\dfrac{m{{v}^{2}}}{r}\], where $m$ is the mass of the particle, \[v\]is the linear velocity and $r$is the radius of the path.
Linear momentum of the object is given by$p=mv$, where $m$ is the mass of the particle, \[v\]is the linear velocity.

Complete Step by Step Solution:
Uniform circular motion is defined as the motion in which the object covers equal distance in equal interval of time. The radial force is given by \[F=\dfrac{m{{v}^{2}}}{r}\] and the liner momentum is given $p=mv$.
Therefore,
$p=mv$

Taking $m$on the left side and deriving the value of$v$, we get:
$v=\dfrac{p}{m}$……………….. $(i)$

The other equation is given by:
\[F=\dfrac{m{{v}^{2}}}{r}\]…………… $(ii)$
Putting the value of $v=\dfrac{p}{m}$ from equation $(i)$ to $(ii)$.
$\Rightarrow $\[F=\dfrac{m{{(\dfrac{p}{m})}^{2}}}{r}\]
$\Rightarrow $$F=\dfrac{m}{r}(\dfrac{{{p}^{2}}}{{{m}^{2}}})$, Now solving it further we can see that
$\Rightarrow $$F=\dfrac{{{p}^{2}}}{mr}$
Hence, the correct option is D. $\dfrac{{{p}^{2}}}{mr}$

Note: In order to solve this question you must understand uniform circular motion. To proceed further, we must first recall certain fundamental concepts. Then, in order to solve the problem, we will make use of the relation between the linear momentum and radial force then using suitable substitutions we will derive the formula for radial force in terms of linear momentum, mass of particle and radius of the path.