Consider a particle “P” in an object (in XY-plane) moving along a circular path of radius “r” about an axis through “O” , perpendicular to plane of the figure i.e. z-axis. Suppose the particles moves through an angle Dq in time Dt sec.

If DS is its distance for rotating through angle Dq then,

Dq = DS / r

Dividing both sides by Dt, we get Dq / Dt = (DS / r. Dt)

r Dq / Dt = DS/Dt

If time interval Dt is very small , then the angle through which the particle moves is also very

small and therefore the ratio Dq /Dt gives the instantaneous angular speed wins.

i.e.

V = rw

**Tangential velocity**

If a particle “P” is moving in a circle of radius “r”, then its linear velocity at any instant is equal to

tangential velocity which is :

V = rw

**Tangential acceleration**

Suppose an object rotating about a fixed axis changes its angular velocity by Dw in time Dt sec,

then the change in tangential velocity DVt at the end of this interval will be

**DVt = r D w**

Change in velocity in unit time is given by:

**DV / Dt = r. Dw / Dt**

if Dt approaches to zero then DV/Dt will be instantaneous tangential acceleration and Dw/Dt

will be instantaneous angular acceleration “a “.

**at = ra**

## Top comments (0)