Question

A particle of mass $m$ moves with constant speed $v$ on a circular path of radius $r$. Find the magnitude of the average force on it in half revolution.

Answer 1

Hint: In order to solve the question, we will first find the change in momentum and the time using the relation between distance, speed and time then we will find the average force substituting the value of change in momentum and time after then we will reach to the answer.

Formula used:
change in momentum = final momentum – initial momentum
${\text{Average force = }}\dfrac{{{\text{change in momentum}}}}{{{\text{time}}}}$
change in momentum = final momentum – initial momentum
${\text{time = }}\dfrac{{{\text{distance}}}}{{{\text{speed}}}}$

Complete step by step answer:
In the question we are given that a particle of mass $m$ is moving with a constant velocity of $v$ in the circular path which has a radius of $r$ and we have to find the average force which is taken in the half revolution.

Question states that’s,
Mass of particle = $m$
Constant speed of particle = $v$
Radius of path in which particle moves = $r$
According to question magnitude of velocity remains same but direction changes
Now we will find the momentum formula of momentum is,
change in momentum = final momentum – initial momentum
change in momentum = $mv - (- mv) = 2mv$
now we will find the average force using formula
${\text{Average force = }}\dfrac{{{\text{change in momentum}}}}{{{\text{time}}}}$
Substituting the value of change in momentum and time
${\text{Average force = }}\dfrac{{2mv}}{{{\text{time}}}}$

Now we will find time using the relation between time , velocity and distance that is
${\text{time = }}\dfrac{{{\text{distance}}}}{{{\text{speed}}}}$
Distance in half revolution=$\pi r$ (full revolution = 2$\pi r$)
So, after substituting the value of distance and speed we get
${\text{time = }}\dfrac{{\pi r}}{v}$
Now we will substitute the time in formula of average force
${\text{Average force = }}\dfrac{{2mv}}{{\dfrac{{\pi r}}{v}}}$
Therefore, taking v in numerator we get
${\text{Average force = }}\dfrac{{2m{v^2}}}{{\pi r}}$

Hence, the answer is ${\text{average force = }}\dfrac{{2m{v^2}}}{{\pi r}}$.

Note: Many of the students get confuse about how change in momentum is $2mv$ instead of zero this is because velocity and mass both are constant but they are changing direction which makes $-v$ and $+v$ this makes difference in initial momentum and final momentum hence answer will not be zero.