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Coriolis Forces and centrifugal forces-Explanation with Problem & Solution

Whenever anybody rotates uniformly about any axis and any other small body translates on a rotating body, then the translating body experiences Coriolis forces and centrifugal forces.

I am going to discuss theory and problem based on Coriolis Forces and centrifugal forces. Dear Student, I hope you will like it and this will solve your problem related to Coriolis Forces and Centrifugal Forces.

What are Coriolis and centrifugal forces? When do they come into play?

            The rotating frame of reference: Coriolis Forces and Centrifugal forces – We shall now consider a rotating (non-inertial) reference frame and the fictitious forces it gives rise to.

Rotating frame of reference

Let S\left ( X,Y,Z \right )be an inertial frame, and S^\prime (X^\prime,Y^\prime,Z^\prime) a non-inertial frame rotating with a uniform angular velocity  \vec{\omega } relative toS(figure 1). We assume, for simplicity, thatSand S^\prime have a common origin O and that the axis of rotating of S^\prime about S passes through O.

Coriolis Forces described with image


Let us consider a particle P whose position vector \vec{r} is the same in both the frames (because the origins are coincident), though the components of \vec{r} along the three axes are different. If the particle is at rest with respect toS, it would appear moving with linear velocity -\vec{\omega \times }\vec{r} to an observer in the rotating frame S^\prime.
Thus, if it has a linear velocity \vec{v} in the frame S, its velocity \vec{v} in the frame S, its velocity \vec{v^\prime} as observed from S^\prime will be given by

\vec{v^\prime}=\vec{v}-\vec{\omega }\times \vec{r}

Or,  \vec{v}=\vec{v^\prime}+\vec{\omega }\times \vec{r}

Or,  \left ( \frac{d\vec{r}}{dt} \right )_{s}=\left ( \frac{d\vec{r}}{dt} \right)_{s^\prime}+\vec{\omega }\times \vec{r}            …(i)

This equation, which relates the time-derivative of a vector \vec{r} in the frames S and S^\prime, holds for all vectors. Hence it may be written in the form of an operator equation.

\left ( \frac{d}{dt} \right )_{s}=\left ( \frac{d}{dt} \right )_{s^\prime}+\vec{\omega }\times

We apply this to the velocity vector \vec{v} when we get

\left ( \frac{d\vec{v}}{dt} \right )_{s}=\left ( \frac{d\vec{v}}{dt} \right )_{s^\prime}+\vec{\omega }\times\vec{v}

Substituting the value of \vec{v} from eq. (i) in the right-hand side of this equation, we get

\; \; \; \left ( \frac{d\vec{v}}{dt} \right )_{s}= \left [ \frac{d}{dt}\left ( \vec{v^\prime}+\vec{\omega }\times \vec{r} \right ) \right ]_{s^\prime}+\vec{\omega }\times \left ( \vec{v^\prime}+\vec{\omega }\times \vec{r} \right )

or, \left ( \frac{d\vec{v}}{dt} \right )_{s}= \left ( \frac{d\vec{v^\prime}}{dt} \right )_{s^\prime}+\frac{d\vec{\omega }}{dt}\times \vec{r}+\vec{\omega }\times \left ( \frac{d\vec{r}}{dt} \right )_{s^\prime}+\vec{\omega }\times \vec{v^\prime}+\vec{\omega }\times \left ( \vec{\omega }\times \vec{r} \right )

now,\left ( \frac{d\vec{v}}{dt} \right )_{s}= \vec{a}(acceleration of the particle as observed in S)

and\left ( \frac{d\vec{v^\prime}}{dt} \right )_{s^\prime}= \vec{a^\prime} (acceleration as observed inS). Also, \left ( \frac{d\vec{r}}{dt} \right )_{s^\prime}= \vec{v^\prime}.

Therefore \vec{a}=\vec{a^\prime}+\frac{d\vec{\omega }}{dt}\times \vec{r}+2\vec{\omega }\times \vec{v^\prime}+\vec{\omega }\times \left ( \vec{\omega }\times \vec{r} \right ).


Since the rotation is uniform ( \vec{\omega } is constant), \frac{d\vec{\omega }}{dt} is zero.


\overset{\vec{a}}{\binom{acceleration\, in }{inertial\, frame}}=\overset{\vec{a^\prime}}{\binom{observed \, acceleration }{in\, rotating \, frame}}+\overset{2\vec{\omega }\times \vec{v^\prime}}{\binom{coriolis }{acceleration}}+\overset{\vec{\omega }\times \left ( \vec{\omega }\times \vec{r} \right )}{\binom{centripetal }{acceleration}}

If m is the mass of the particle, then the force acting on the particle in the rotating frame is given by

m\vec{a^\prime}=m\vec{a}-2m\vec{\omega }\times \vec{v^\prime}-m\vec{\omega }\times \left ( \vec{\omega } \times \vec{r}\right )

here, m\vec{a^\prime}is the observed force \vec{F^\prime} on the particle in the rotating frame S’ and \vec{ma^\prime} is the actual force \vec{F^\prime} acting in the inertial frame S.

Thus,\vec{F^\prime}= \vec{F}-2m\vec{\omega }\times \vec{v^\prime}-m\vec{\omega }\times \left ( \vec{\omega }\times \vec{r} \right )

We know that the force \vec{F^\prime}  acting on a particle in a non-inertial frame is given by

\vec{F^\prime}= \vec{F} (actual force)+fictitious force.

\therefore fictitious force = -2m\vec{\omega }\times \vec{v^\prime}-m\vec{\omega }\times \left ( \vec{\omega }\times \vec{r} \right ).

Here, -2m\vec{\omega }\times \vec{v^\prime} is the ‘Coriolis force’ and -m\vec{\omega }\times \left ( \vec{\omega }\times \vec{r} \right ) is the ‘centrifugal force’.

Coriolis Forces

The Coriolis force -2m\vec{\omega }\times \vec{v^\prime}is a fictitious force experienced by a particle moving (with velocity \vec{v}) relative to a rotating (non-inertial)frame (with angular velocity \vec{\omega }).

Since it is the vector-2m \vec{\omega } produce of and \vec{v}, it acts in a direction perpendicular to the particle’s path as seen by an observer moving with the rotating frame. The negative sign indicates that the Coriolis forces is in the direction opposite to that given by the right-hand screw rule.

Centrifugal Force

The centrifugal force m\vec{\omega }\times \left (\vec{\omega }\times \vec{r} \right )is the only fictitious force experienced by a particle at rest \left (\vec{v^\prime} = 0 \right )in the rotating frame. Being responsible for keeping the particle at rest in the rotating frame.

Centrifugal Forces explain with diagram


The related problem on this topic is given below:

Q. A frame of reference that is accelerated with respect to an inertial frame of reference is called a non – inertial frame of reference.

A co-ordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity \omega is an example of non-inertial frame of reference. The relationship between the force \vec{F}_{rot} experienced by a particle of mass m moving on the rotating disc and the force\vec{F}_{in} experienced by the particle in an inertial frame of reference is

\vec{F}_{rot}= \vec{F}_{in}+2m\left ( \vec{v }_{rot}\times \vec{\omega } \right )+m\left ( \vec{\omega }\times \vec{r} \right )\times \vec{\omega }

Where \vec{v}_{rot} is the velocity of the particle in the rotating frame of reference and \vec{r} is the position vector of the particle with respect to the center of disc.

Now consider a smooth slot along a diameter of a disc of radius R rotating counter-clockwise with a constant angular speed \omega about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the x-axis is along with the slot, y-axis perpendicular to the slot and z-axis along the rotation axis \vec{\omega }= \omega \hat{k}. A small block of mass m is gently placed in the slot at\vec{r}= \frac{R}{2}\hat{i} at t= 0 and is constrained to move only along the slot

Coriolis Forces Images problem based diagram

  1. The distance r of the block at time t is
\left ( a \right )\frac{R}{4}\left ( e^{\omega t}+e^{-\omega t} \right )\\


\left ( b \right )\frac{R}{2}cos\omega t\\


\left ( c \right )\frac{R}{4}\left ( e^{2\omega t}+e^{-2\omega t} \right )\\


\left ( d \right )\frac{R}{2}cos2\omega t\\


Ans:- Force acting on the particle due to which body is moving away from the centre of the disc.

\vec{F}= m\left ( \vec{\omega }\times \vec{r} \right )\times \vec{\omega }

F= m\omega ^{2}r

or,    \not{m}a= \not{m}\omega ^{2}r

\frac{d\upsilon }{dt}= \omega ^{2}r

\frac{dv }{dr}\cdot \frac{dr}{dt}= \omega ^{2}r

Integrating both sides,

\int_{o}^{\upsilon }\upsilon \left ( d\upsilon \right )= \int_{\frac{R}{2}}^{r}\omega ^{2}r\left ( dr \right )

\frac{v ^{2}}{2}= \frac{\omega ^{2}}{2}\left [r ^{2} \right ]_{\frac{R}{2}}^{r}

v = \omega \sqrt{r^{2}-\left ( \frac{R}{2} \right )^{2}}

Using ,    v = \frac{dr}{dt}.  

\therefore \frac{dr}{dt}=\omega \sqrt{r^{2}-\left ( \frac{R}{2} \right )^{2}}

  \int_{\frac{R}{2}}^{r}\frac{dr}{\sqrt{r^{2}-\left ( \frac{R}{2} \right )^{2}}}= \omega \int_{o}^{t}dt

As we know,

\int \frac{dx}{\sqrt{x^{2}-a^{2}}}= ln\mid x+\sqrt{x^{2}-a^{2}}\mid+c

\therefore \left [ ln\mid r+\sqrt{r^{2}-\left ( \frac{R}{2} \right )^{2}\mid } \right ]_{\frac{R}{2}}^{r}= \omega \left [ t \right ]_{o}^{t}

ln\mid r+\sqrt{r^{2}-\left ( \frac{R}{2} \right )^{2}}\mid -ln\mid \frac{R}{2}\mid = \omega t

ln \frac{r+\sqrt{r^{2}-\left ( \frac{R}{2} \right )^{2}}}{\frac{R}{2}}= \omega t

r+\sqrt{r^{2}-\frac{R^{2}}{4}}=\frac{R}{2}e^{\omega t}

\sqrt{r^{2}-\frac{R^{2}}{4}}=\left ( \frac{R}{2}e^{\omega t}-r \right )

Squarring both sides;

r^{2}-\frac{R^{2}}{4}=\frac{R^{2}}{4}e^{2\omega t}+r^{2}-\frac{2R}{2}e^{\omega t}.r

re^{\omega t}=\frac{R}{4}\left ( 1+e^{2\omega t} \right )

r= \frac{R}{4}\left ( e^{-\omega t}+e^{\omega t} \right )

Answer -(a)

(2)  The net reaction of the disc on the block is

(a) \frac{1}{2}m\omega ^{^{2}}R\left ( e^{2\omega t}-e^{-2\omega t} \right )\hat{j}+mg\hat{k}

(b) \frac{1}{2}m\omega ^{^{2}}R\left ( e^{\omega t}-e^{-\omega t} \right )\hat{j}+mg\hat{k}  

(c) -m\omega ^{2}R cos\omega t\hat{j}-mg\hat{k}.

(d) m\omega ^{2}R sin\omega t\hat{j}-mg\hat{k}.


The net reaction of the disc on the block can be written as

\sum \vec{F}_{reaction}=-2m\left ( \overrightarrow{v _{rot}}\times \overrightarrow{\omega } \right )+mg\hat{k}.

here,-2m\left ( \vec{v}_{rot}\times \vec{\omega } \right ) is the reaction force exerted by disc on the block (against the direction of Coriolis force experienced by the block) and mg\hat{k} is normal reaction force exerted by the lower contact surface of the disc on to the block.

\sum \vec{F}_{reaction} = -2m\left ( \frac{dr}{dt}\hat{i}\times \omega \hat{k} \right )+mg\hat{k}

\sum \vec{F}_{reaction}= 2m\omega \frac{d}{dt}\left \{ \frac{R}{4}\left ( e^{-\omega t}+e^{\omega t} \right ) \right \}\hat{j}+mg\hat{k}

\sum \vec{F}_{reaction} = \frac{m\omega r}{2}\left \{ e^{-\omega t}.\left ( -\omega \right )+e^{\omega t}\omega \right \}\hat{j}+mg\hat{k}

\sum \vec{F}_{reaction}= \frac{m\omega ^{2}R}{2}\left \{ e^{\omega t}-e^{-\omega t} \right \} \hat{j}+mg\hat{k}.  

The correct option is (b).


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