Question

In circular motion, assuming `bar v = bar w xx bar r` , obtain an expression for the resultant acceleration of a particle in terms of tangential and radial component.

Answer 1

Acceleration of a particle,

`a=lim_(deltat->0) ((delta v)/(delta t)) ... delta t ->0; delta t ne 0`

`therefore a=(dv)/(dt)`

`But, v = r omega`

`therefore a=d/dt (r omega)`

`=r (d omega)/dt + omega (dr)/(dt)`

`because "r is constant"`

`(dr)/(dt) = 0`

`therefore a=r (d omega)/(dt)`

`because (d omega)/(dt) = alpha`

`alpha = r alpha`

Given that:

`bar "v"= bar omega xx bar "r"`

Differentiating w.r.t. time

`(d bar v)/(dt)=d/dt (bar omega xx bar "r")`

`(d barv)/(dt) = (d baromega)/(dt) xx barr + baromega xx (d barr)/(dt)`

`(d barv)/(dt) = baralpha xx barr + baromega xx barv`

`therefore bara = bara_T + bara_r`

Where,

a = Linear acceleration

a_T = Tangential component of linear acceleration

a_r = Radial component of linear acceleration