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


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 $\endgroup$ – Srijan M.T Jan 29 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.