# Did not understand this regarding angular velocity

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

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.

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).

• I have edited the question . Please see if it is understandable – 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.