An object is moving in a circular path of radius $r$. It has a linear (tangential) velocity $v$.
The linear (tangential) deceleration of the object is $d$. In other words, d is the rate of decrease of $v$.
How can I calculate its stopping distance i.e. how far along this circular path does it go before coming to a stop?
For linear motion ($r=0$), this would be $v^2 = 2as$, where $s$ is the stopping distance.
For circular motion, how do we calculate this? Does the deceleration have to be resolved into a radial and tangential component?
thank you
HINTS:
It goes the exactly same way as in straight line motion when there is no radial velocity component.
$$ \frac{ds}{dt}= v= \omega\cdot r$$
Differentiate
$$\frac{d^2s}{dt^2}= \frac{dv}{dt}= \alpha\cdot r=d $$
Integrate
$$v= d\cdot t + v_0$$
Integrate again
$$ s=s_0+ v_0t+ \frac{d\cdot t^2}{2}$$
Note, $d<0$ for stopping time in retardation. A quadratic equation is to be solved to find arc length traversed.