Circular motion-centripetal acceleration

[allok]

hi

This latex code is giving me some problems. I write one thing, it displays something completely different

In circular motion velocity only changes direction but not size

change of velocity - \Delta v
Change of angle - \Delta T
Velocity - V
Centripetal acceleration - a

delta(V) = V * delta(T)

When delta(T) approaches its limit (goes to zero), change of velocity has same direction as acceleration vector?

We compute the magnitude of velocity change with :

Delta(v) = v * Delta(T)

I see this being true when change of angle approaches its limit ( goes to zero ), since then length of circular arc ( with radius begin velocity vector ) equals \Delta v. But that is not true if delta(T) is not approaching limit. So how can we use formula

delta(v) = V * delta(T)

in cases were delta(T) is not approaching zero, since I assume length of circle arc is quite different than delta(T) if delta(T) doesn't go to zero?

cheers

 

[mezarashi]

I think your original assumption may be incorrect. If you want to relate the linear velocity of the object on the circle circumference with the change in angle, it would be:

s = r\theta
\frac{ds}{dt} = r\frac{d\theta}{dt}
v = r\frac{d\theta}{dt}

where s is the circle arclength, r is the circle radius, and theta is the angle. The \frac{d\theta}{dt} would be equal to your \Delta T.

 

[allok]

I think your original assumption may be incorrect. If you want to relate the linear velocity of the object on the circle circumference with the change in angle, it would be:

s = r\theta
\frac{ds}{dt} = r\frac{d\theta}{dt}
v = r\frac{d\theta}{dt}

where s is the circle arclength, r is the circle radius, and theta is the angle. The \frac{d\theta}{dt} would be equal to your \Delta T.

I don't get it. First of all, v = r\frac{d\theta}{dt} is not same formula as delta(v)=v*delta(theta)

I still don't understand why delta(v) = v * delta(theta) would give us correct result when delta(theta) is anything but d\theta ?