It says $V\sin \alpha$ (component of velocity perpendicular to OP) is the cause of angular displacement. How is it?

if only $V\cos \alpha$ existed, we need not turn over head to always to look at a particle. What does this mean?

Also, Can we say that $V$ is the linear speed? Because that is the one tangential in direction.

How is $PQ = OP \cdot \Delta \theta$ ?

enter image description here


2 Answers 2


Regarding the second part of your question: How is $PQ = OP* \Delta \theta$ ?

The author is probably assuming that the angle $\Delta \theta$ is very small. In that case as you see in the following image (where the angle is denoted as $\theta$, the $\sin\theta$ is approximately equal to the arc.

enter image description here

The arc length is equal to $R\cdot\theta $, where R is the radius (which is equal to one for this cycle. So you end up having (in the general case) that $R \sin\theta \approx R\theta\Rightarrow \sin\theta \approx \theta$ (where $\theta$ is in radian.

Now, in your case, like you noticed, $PQ$ should be given by $$PQ= OP\sin\Delta \theta$$

However for an infinitesimally small $\Delta \theta$ you can use the small angle approximation $ \sin\Delta\theta \approx \Delta\theta$ and the above results in:

$$PQ= OP\cdot \Delta \theta$$

Unfortunately regarding the first part, I could not understand exactly what you were asking. I would be glad to update the question if it becomes clearer what is point 1 and point 2, and what is it that you don't understand.

Regarding the first part there is still ambiguity. I will try to reply to what I understand :

It says $V\sin\alpha$ ( component of velocity perpendicular to OP) is the cause of angular displacement.How is it ?

if only $V\cos\alpha$ existed we need not turn over head to always to look at a particle. What does this mean ?

It's easier to reply about $V\cos\alpha$. If you see in your image, $V\cos\alpha$ is parallel to $OP$. The observer is standing on O. If only $V\cos\alpha$ existed then the observer would only see the object to travel either away or towards the observer (point O).

Regarding the other part $V\sin\alpha$, $V$ can be decomposed in many way. One way is as $V\sin\alpha$ and $V\cos\alpha$. $V\sin\alpha$ is perpendicular to $V\cos\alpha$. So $V\sin\alpha$ would make the observer at O rotate his head in order to follow the object at P.

Also , Can we say that V is the linear speed ? Because that is the one tangential in direction.in

I am not sure what you ask with "V is the linear speed?". Velocity is a vector and always point to one direction. It is always tangential to the trajectory of an object when its moving (this is a consequence of velocity being the rate of change of displacement).

  • $\begingroup$ I have edited the question . Please see if it is understandable $\endgroup$
    – S.M.T
    Commented Jan 29, 2021 at 14:56

P is a particle subjected to the linear velocity vector V.

this vector has two orthagonal components, $V_{sin} \alpha \ and \ V_{cos}\alpha , $

$ V_{cos} \alpha, $ will move the particle up radially but has no effect on its angular movement because it is orthogonal to the angular rotation.

$ V_{sin }, $ on the other hand, can and is the only component that causes angular rotation of particle P because it is perpendicular to OP, Your figure does not show this clearly and that may be the cause of your confusion.

I guess your text means if there was only $ V_{cos}\alpha$ it meant your particle will only move our and up like a rocket and you'd need to look back to see the particle.

And the answer to your other part is yes, V is the linear speed.

And $PQ=OP*\Delta \theta $, not $ \ PQ= OP* V_{sin} \alpha$ because the linear speed is not necessarily always slanted at an angle of $ \alpha $, it can change and cause a change in rotation speed. So how much turn we have after sometime depends on the curve of the slant with the tangent to rotation.


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