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This question is an exact duplicate of:

A particle is moving on a circular path with constant speed. Which of the following is true?

(a)it posses radial acceleration.

(b)it posses radial velocity.

(c)it posses tangential acceleration.

(d)it does not posses tangential velocity.

The answer is option (a).

Now,my question is how the other 3 factors in option b,c,and d are becoming zero?why the body will have only radial acceleration?why only option (a) is true?

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marked as duplicate by Kyle Kanos, Daniel Griscom, ACuriousMind, ja72, user36790 Jan 5 '16 at 15:01

This question was marked as an exact duplicate of an existing question.

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    $\begingroup$ Possible duplicate by OP: physics.stackexchange.com/q/227572/2451 $\endgroup$ – Qmechanic Jan 5 '16 at 7:58
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    $\begingroup$ Hi user302630. Welcome to Phys.SE. Please don't repost a closed question in a new entry. Instead, you are supposed to edit the original question within the original entry. $\endgroup$ – Qmechanic Jan 5 '16 at 8:02
  • $\begingroup$ @Qmechanic I am really sorry for that. But will you please help me with my question. I am unable to fully understand it from the answer i received $\endgroup$ – user302630 Jan 5 '16 at 11:11
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Calculations and formulas are not necessary here. Simple logical considerations will do:

(a)it posses radial acceleration.

Yes. This is the component that makes it change direction all the time to cause an orbit and not just a travel along a straight line. If there was no radial component, the velocity would never change sideways.

(b)it posses radial velocity.

No, because in a circular orbit, the distance to the centre must always stay the same. Any radial velocity (velocity towards the centre) would change this distance.

(c)it posses tangential acceleration.

No, because the speed is constant. If there was any tangential acceleration (acceleration in the same direction as the motion / velocity) the velocity in this direction would also change.

(d)it does not posses tangential velocity.

It does indeed have tangential velocity - otherwise it would not move around the centre but could only move straight towards the centre or stay still.

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Remember the speed is constant, in the natural coordinate system

$$\vec{v}=\frac{d\vec{r}}{dt} = \frac{d\vec{r}}{ds}\frac{d\vec{s}}{dt}=v\vec{\tau}$$

where $\vec{\tau}$ is the tangential direction,

so you see no matter whether v changes or not, that is, the spead varies, there is no radial velocity.

$$\vec{a}=\dot{\vec{v}}=\dot{v}\vec{\tau}+v\dot{\vec{\tau}} $$

and $\dot{\vec{\tau}}=\dot{\theta}\vec{n}$, where $\vec{n}$ is the radial direction.

So we can have, $$\vec{a}=\dot{v}\vec{\tau}+\frac{v^2}{r}{\vec{n}} $$

since $\dot{v}=0$, you can get your answer to this question

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  • $\begingroup$ I am still not clear why only option (a) is given correct in my book $\endgroup$ – user302630 Jan 5 '16 at 6:50
  • $\begingroup$ @user302630 the velocity vector can be split into radial and tangential components, the tangential component as you have stated in your question is a constant (tangential $v$= constant) so tangential $a=0$, radial acceleration is $v^2/r$ $\endgroup$ – Oswald Jan 5 '16 at 6:56
  • $\begingroup$ @TheGhostOfPerdition how the tangential component is becoming zero? $\endgroup$ – user302630 Jan 5 '16 at 7:01
  • $\begingroup$ @TheGhostOfPerdition and what about the radial velocity?why is that zero? I am very much confused $\endgroup$ – user302630 Jan 5 '16 at 7:02
  • $\begingroup$ remember that $a=dv/dt$, if $v=$constant $a=0$. The direction of change of the particle's position is tangential to the radius, hence radial velocity is zero. $\endgroup$ – Oswald Jan 5 '16 at 8:05
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The position vector of the particle is $\vec{r}=r\hat{r}$

The unit vector along the radius vector $\hat{r}=\{\cos \theta, \sin \theta\} \tag{1}$

the unit vector along $\theta$ is $\hat{\theta}=\{-\sin\theta, \cos \theta\} \tag{2}$

$$\vec{v}=\frac{d\vec{r}}{dt}=\frac{d(\vec{r}\hat{r})}{dt}=\vec{r}\frac{d\hat r}{dt}+\hat r\frac{dr}{dt} \tag{3}$$

Now since the particle is moving in uniform circular motion i.e radius $r$ is fixed $\frac{dr}{dt}=0$, so eq(3) becomes

$$\frac{d\vec{r}}{dt}=\vec{r}\frac{d\hat r}{dt}$$

from eq(1) $$\frac{d\hat r}{dt}=\{\sin \theta, \cos \theta \}\dot{\theta}=\hat {\theta}\dot \theta $$

So, $$\vec{v}=\vec{r}\dot \theta \hat {\theta}$$

So, you can see that $\vec{v}$ only has a $\hat{\theta}$ component, and no radial component. This should answer your (b) and (d)

For the acceleration $$\vec{a}=\frac{d\vec{v}}{dt}$$

Follow the same procedure above, you will find the answers for (a) and (c)

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