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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

Тyma Gaidash
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simplename
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    If by "deceleration" you mean retardation, then it means the rate of decrease of speed, so is a scalar.

    For circular motion with radius $r$ and linear velocity $v,,$ $v$ is the transverse velocity, so the speed is $r\left|\frac{\mathrm d\theta}{\mathrm dt}\right|.$ If the angular velocity is nonnegative, then the speed is $r\frac{\mathrm d\theta}{\mathrm dt}$ and the retardation is $-r\frac{\mathrm d^2\theta}{\mathrm dt^2}.$

     – ryang Jun 05 '23 at 06:29
  • is retardation the same as angular acceleration then, is it also a scalar? and i can calculate it as α = Δω/Δt – simplename Jun 05 '23 at 15:56

1 Answers1

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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.

Narasimham
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