In the previous section we completed a discussion on projectile motion. In this section, we will see Uniform circular motion.
If we want to say that an 'object is in uniform circular motion', the following two conditions should be satisfied:
(i) The object should be moving in a circular path
♦ The center of that circle should not change
♦ The radius of that circle should not change
(ii) The speed of travel along that path should be constant
• We know that, when a particle travels along a circular path, the direction of velocity at any point is tangential to the circle at that point
• So we can say: when a particle travels along a circular path, it's velocity changes continuously
■ If there is change in velocity, there must be an acceleration.
• We want to find the magnitude and direction of this acceleration
We will write the steps:
1. Consider fig.4.36(a) below
An object is travelling along the circumference of a circle shown in green colour.
2. The origin O of the coordinate axes is made to coincide with the center C of the circle
• This enables us to draw the position vectors easily
3. Let at any instant when the stop watch shows 't' s, the position of the object be P
• Let after an interval of 'Δt' s, the position of the object be P'
Then we can write the following 2 points:
(i) Position vector of the object at P is $\mathbf\small{\overrightarrow{OP}}$
♦ It is denoted as $\mathbf\small{\vec{r}}$
(ii) Position vector of the object at P' is $\mathbf\small{\overrightarrow{OP'}}$
♦ It is denoted as $\mathbf\small{\vec{r'}}$
4. When we know the initial and final postion vectors, we can find the displacement vector
• We have seen the method for doing it in a previous section.
• Using that method, the displacement vector for our present case is $\mathbf\small{\overrightarrow{PP'}}$
♦ It is denoted as $\mathbf\small{\overrightarrow{\Delta r}}$
5. So we obtained the displacement vector. We now consider velocities:
• Let the velocity at P be $\mathbf\small{\vec{v}}$
• Let the velocity at P' be $\mathbf\small{\vec{v'}}$
• They are marked in the fig.a
We see the following 3 points:
(i) $\mathbf\small{\vec{v}}$ and $\mathbf\small{\vec{v'}}$ have the same length
♦ This is because, though they have different directions, magnitudes are the same
(ii) $\mathbf\small{\vec{v}}$ is tangential to the circle at P
♦ So obviously, $\mathbf\small{\vec{v}}$ is perpendicular to $\mathbf\small{\vec{r}}$
(iii) $\mathbf\small{\vec{v'}}$ is tangential to the circle at P'
♦ So obviously, $\mathbf\small{\vec{v'}}$ is perpendicular to $\mathbf\small{\vec{r'}}$
6. Next we consider 'change in velocity':
■ Change in velocity is (final velocity - initial velocity)
• We have seen how to find it in the case of velocity vectors. We saw it in a previous section.
Application of that method is shown in fig.b
(i) We drag $\mathbf\small{\vec{v}}$ and $\mathbf\small{\vec{v'}}$ to a convenient place. This is shown in fig.b
(iii) We arrange them in such a way that their tails coincide
(iv) Finally we draw the required $\mathbf\small{\vec{\Delta v}}$ from the head of $\mathbf\small{\vec{v}}$ to head of $\mathbf\small{\vec{v'}}$. This vector is shown in cyan color
7. We obtained $\mathbf\small{\vec{\Delta v}}$ by dragging the concerned vectors away to a convenient place.
• But where does this $\mathbf\small{\vec{\Delta v}}$ actually act?
♦ It cannot act at P. Because at P, $\mathbf\small{\vec{v}}$ is acting
♦ It cannot act at P'. Because at P', $\mathbf\small{\vec{v'}}$ is acting
• Since the magnitudes of $\mathbf\small{\vec{v}}$ and $\mathbf\small{\vec{v'}}$ are equal, their difference $\mathbf\small{\vec{\Delta v}}$ will be acting at a point midway between P and P'
• So we drag $\mathbf\small{\vec{\Delta v}}$ and place it's tail at the midpoint of PP'
■ On doing so, we find that, $\mathbf\small{\vec{\Delta v}}$ is acting towards the center of the circle. It is shown in fig.c
This is a very useful result. We can write:
■ In uniform circular motion, the 'change in velocity vector' acts towards the center of the circular path
8. Next we consider acceleration
• In the above step, we have calculated the change in velocity $\mathbf\small{\vec{\Delta v}}$.
• We know that, if we divide this $\mathbf\small{\vec{\Delta v}}$ by the time duration Δt, we will get average acceleration $\mathbf\small{\bar{\vec{a}}}$
• That is., $\mathbf\small{\bar{\vec{a}}=\frac{\vec{\Delta v}}{\Delta t}}$
9. We are dividing the vector by a scalar. In such a division, the direction of the vector will not change
• So the direction of average acceleration $\mathbf\small{\bar{\vec{a}}}$ is same as the direction of $\mathbf\small{\vec{\Delta v}}$. We can write:
■ In uniform circular motion, the 'average acceleration' acts towards the center of the circular path
10. So we get the direction of average acceleration. Next we want it's magnitude
• For that, first we want the angle between $\mathbf\small{\vec{v}}$ and $\mathbf\small{\vec{v'}}$
We can find it using the following steps:
(i) The angle between $\mathbf\small{\vec{r}}$ and $\mathbf\small{\vec{r'}}$ is Δθ.
(ii) The angle between $\mathbf\small{\vec{r}}$ and $\mathbf\small{\vec{v}}$ will be always 90°.
(iii) The angle between $\mathbf\small{\vec{r'}}$ and $\mathbf\small{\vec{v'}}$ will be always 90°.
(iv) So the angle between $\mathbf\small{\vec{v}}$ and $\mathbf\small{\vec{v'}}$ will be Δθ. This is shown in fig.b
[A detailed proof can be seen here]
11. Now, triangle OPP' in fig.a is an isosceles triangle because OP = OP' = radius of the circle
♦ So base angles at P and P' are equal
♦ We get: ∠P = ∠P' = [0.5 × (180-Δθ)]
• Similarly, triangle IGH in fig.b is an isosceles triangle because GH = GI = $\mathbf\small{|\vec{v}|}$ = $\mathbf\small{|\vec{v'}|}$
♦ So base angles at I and H are equal
♦ We get: ∠I = ∠H = [0.5 × (180-Δθ)]
12. So all the angles in the two triangles OPP' and IGH are equal.
■ Thus they are similar triangles. Some notes on similar triangles can be seen here.
• We can take ratios:
$\mathbf\small{\frac{side\: opposite\: \Delta \theta \: in\: OPP' }{side\: opposite\: \Delta \theta \: in\: IGH}}$ = $\mathbf\small{\frac{side\: opposite\: [0.5 \times(180-\Delta \theta)] \: in\: OPP' }{side\: opposite\: [0.5 \times(180-\Delta \theta)] \: in\: IGH}}$
• Thus we get: $\mathbf\small{\frac{PP'}{IH}=\frac{OP}{GI}}$
$\mathbf\small{\Rightarrow \frac{|\vec{\Delta r}|}{|\vec{\Delta v}|}=\frac{|\vec{r}|}{|\vec{v}|}}$
$\mathbf\small{\Rightarrow |\vec{\Delta v}|=\frac{|\vec{v}|\times |\vec{\Delta r}|}{|\vec{r}|}=\frac{|\vec{v}|\times |\vec{\Delta r}|}{R}}$
• Where R is the radius of the circle
13. Thus we get the 'magnitude of change in velocity'
■ If we divide this magnitude by the Δt during which the change occurs, we will get the magnitude of the average acceleration
• So we can write: $\mathbf\small{|\bar{\vec{a}}|=\frac{|\vec{\Delta v}|}{\Delta t}}$
Substituting for $\mathbf\small{|\vec{\Delta v}|}$ from (12), we get Eq.4.26: $\mathbf\small{|\bar{\vec{a}}|=\frac{|\vec{v}|\times |\vec{\Delta r}|}{R\times \Delta t}}$
14. Thus we obtained the magnitude of average acceleration. Our next aim is to find $\mathbf\small{|\vec{a}|}$, the magnitude of instantaneous acceleration
• For that, we make 'Δt' closer and closer to zero.
• So P' will become closer and closer to P
A sample is shown in fig.4.37(a) below:
• In fig.a, P' is closer to P than in fig.4.36(a) that we saw at the beginning of this section
• All the steps are repeated:
♦ $\mathbf\small{\vec{\Delta v}}$ is obtained in fig.b
♦ It is shown in cyan color
♦ In fig.c, the tail of the cyan vector is placed midway between P and P'
• We find that, here also, the cyan vector is directed towards the center of the circle
15. What will happen if we make Δt smaller and smaller
Ans: The interval Δt will become so small that it can no longer be called an 'interval'
• Instead, we will have to call it an 'instant'
• Then, the acceleration calculated in such a small Δt is not average acceleration $\mathbf\small{\bar{\vec{a}}}$
■ It is the instantaneous acceleration $\mathbf\small{\vec{a}}$
• So we can write: $\mathbf\small{|\vec{a}|= \lim \limits_{\Delta t \to 0} \frac{|\vec{\Delta v}|}{\Delta t}}$
• Substituting for $\mathbf\small{|\vec{\Delta v}|}$ from (12), we get:
$\mathbf\small{|\vec{a}|= \lim \limits_{\Delta t \to 0} \frac{|\vec{v}|\times|\vec{\Delta r}|}{R \times \Delta t}}$
16. This limit can be worked out as follows:
(i) Consider the travel from P to P' in fig.4.37(a).
• When Δt tends to zero, P' is very close to P
• Then Δθ is very small
(ii) The object is travelling with a uniform speed of $\mathbf\small{|\vec{v}|}$ along the circular path
• So it will cover a circular distance of [$\mathbf\small{|\vec{v}|}$ × Δt] in Δt seconds
• Obviously, this distance will be equal to arc length PP'
• So we can write: arc length PP' = [$\mathbf\small{|\vec{v}|}$ × Δt]
(iii) When Δt tends to zero, that is., when Δt is very small, Δθ is also very small and so, arc length PP' will be nearly equal to chord length PP'
• But chord length PP' = $\mathbf\small{\vec{\Delta r}}$
(iv) Using the results in (ii) and (iii), we can write:
chord length PP' = $\mathbf\small{\vec{\Delta r}}$ = [$\mathbf\small{|\vec{v}|}$ × Δt]
(v) So at the limiting state shown in (15), we can use [$\mathbf\small{|\vec{v}|}$ × Δt] instead of $\mathbf\small{\vec{\Delta r}}$
• So the result in (15) becomes: $\mathbf\small{|\vec{a}|=\frac{|\vec{v}|\times [|\vec{v}|\times \Delta t]}{R \times \Delta t}}$
■ Thus we get Eq.4.27: $\mathbf\small{|\vec{a}|=\frac{|\vec{v}|^2}{R}}$
17. In the limiting state, $\mathbf\small{\vec{r}}$ nearly coincides with $\mathbf\small{\vec{r'}}$. This is shown in fig.4.37(c)
• $\mathbf\small{\vec{r}}$ is perpendicular to $\mathbf\small{\vec{v}}$ because, $\mathbf\small{\vec{v}}$ is tangential to the circle at P
• The instantaneous acceleration at P is directed towards the center from P
19. This acceleration is called centripetal acceleration.
• The word 'centripetal' comes from the Greek term which means 'center seeking'.
• This term is very suitable because, as we saw earlier, this acceleration is always directed towards the center.
• It may be noted that the instantaneous acceleration $\mathbf\small{\vec{a}}$ is not a constant vector because, it's direction changes continuously.
If we want to say that an 'object is in uniform circular motion', the following two conditions should be satisfied:
(i) The object should be moving in a circular path
♦ The center of that circle should not change
♦ The radius of that circle should not change
(ii) The speed of travel along that path should be constant
• We know that, when a particle travels along a circular path, the direction of velocity at any point is tangential to the circle at that point
• So we can say: when a particle travels along a circular path, it's velocity changes continuously
■ If there is change in velocity, there must be an acceleration.
• We want to find the magnitude and direction of this acceleration
We will write the steps:
1. Consider fig.4.36(a) below
 |
| Fig.4.36 |
2. The origin O of the coordinate axes is made to coincide with the center C of the circle
• This enables us to draw the position vectors easily
3. Let at any instant when the stop watch shows 't' s, the position of the object be P
• Let after an interval of 'Δt' s, the position of the object be P'
Then we can write the following 2 points:
(i) Position vector of the object at P is $\mathbf\small{\overrightarrow{OP}}$
♦ It is denoted as $\mathbf\small{\vec{r}}$
(ii) Position vector of the object at P' is $\mathbf\small{\overrightarrow{OP'}}$
♦ It is denoted as $\mathbf\small{\vec{r'}}$
4. When we know the initial and final postion vectors, we can find the displacement vector
• We have seen the method for doing it in a previous section.
• Using that method, the displacement vector for our present case is $\mathbf\small{\overrightarrow{PP'}}$
♦ It is denoted as $\mathbf\small{\overrightarrow{\Delta r}}$
5. So we obtained the displacement vector. We now consider velocities:
• Let the velocity at P be $\mathbf\small{\vec{v}}$
• Let the velocity at P' be $\mathbf\small{\vec{v'}}$
• They are marked in the fig.a
We see the following 3 points:
(i) $\mathbf\small{\vec{v}}$ and $\mathbf\small{\vec{v'}}$ have the same length
♦ This is because, though they have different directions, magnitudes are the same
(ii) $\mathbf\small{\vec{v}}$ is tangential to the circle at P
♦ So obviously, $\mathbf\small{\vec{v}}$ is perpendicular to $\mathbf\small{\vec{r}}$
(iii) $\mathbf\small{\vec{v'}}$ is tangential to the circle at P'
♦ So obviously, $\mathbf\small{\vec{v'}}$ is perpendicular to $\mathbf\small{\vec{r'}}$
6. Next we consider 'change in velocity':
■ Change in velocity is (final velocity - initial velocity)
• We have seen how to find it in the case of velocity vectors. We saw it in a previous section.
Application of that method is shown in fig.b
(i) We drag $\mathbf\small{\vec{v}}$ and $\mathbf\small{\vec{v'}}$ to a convenient place. This is shown in fig.b
(iii) We arrange them in such a way that their tails coincide
(iv) Finally we draw the required $\mathbf\small{\vec{\Delta v}}$ from the head of $\mathbf\small{\vec{v}}$ to head of $\mathbf\small{\vec{v'}}$. This vector is shown in cyan color
7. We obtained $\mathbf\small{\vec{\Delta v}}$ by dragging the concerned vectors away to a convenient place.
• But where does this $\mathbf\small{\vec{\Delta v}}$ actually act?
♦ It cannot act at P. Because at P, $\mathbf\small{\vec{v}}$ is acting
♦ It cannot act at P'. Because at P', $\mathbf\small{\vec{v'}}$ is acting
• Since the magnitudes of $\mathbf\small{\vec{v}}$ and $\mathbf\small{\vec{v'}}$ are equal, their difference $\mathbf\small{\vec{\Delta v}}$ will be acting at a point midway between P and P'
• So we drag $\mathbf\small{\vec{\Delta v}}$ and place it's tail at the midpoint of PP'
■ On doing so, we find that, $\mathbf\small{\vec{\Delta v}}$ is acting towards the center of the circle. It is shown in fig.c
This is a very useful result. We can write:
■ In uniform circular motion, the 'change in velocity vector' acts towards the center of the circular path
8. Next we consider acceleration
• In the above step, we have calculated the change in velocity $\mathbf\small{\vec{\Delta v}}$.
• We know that, if we divide this $\mathbf\small{\vec{\Delta v}}$ by the time duration Δt, we will get average acceleration $\mathbf\small{\bar{\vec{a}}}$
• That is., $\mathbf\small{\bar{\vec{a}}=\frac{\vec{\Delta v}}{\Delta t}}$
9. We are dividing the vector by a scalar. In such a division, the direction of the vector will not change
• So the direction of average acceleration $\mathbf\small{\bar{\vec{a}}}$ is same as the direction of $\mathbf\small{\vec{\Delta v}}$. We can write:
■ In uniform circular motion, the 'average acceleration' acts towards the center of the circular path
10. So we get the direction of average acceleration. Next we want it's magnitude
• For that, first we want the angle between $\mathbf\small{\vec{v}}$ and $\mathbf\small{\vec{v'}}$
We can find it using the following steps:
(i) The angle between $\mathbf\small{\vec{r}}$ and $\mathbf\small{\vec{r'}}$ is Δθ.
(ii) The angle between $\mathbf\small{\vec{r}}$ and $\mathbf\small{\vec{v}}$ will be always 90°.
(iii) The angle between $\mathbf\small{\vec{r'}}$ and $\mathbf\small{\vec{v'}}$ will be always 90°.
(iv) So the angle between $\mathbf\small{\vec{v}}$ and $\mathbf\small{\vec{v'}}$ will be Δθ. This is shown in fig.b
[A detailed proof can be seen here]
11. Now, triangle OPP' in fig.a is an isosceles triangle because OP = OP' = radius of the circle
♦ So base angles at P and P' are equal
♦ We get: ∠P = ∠P' = [0.5 × (180-Δθ)]
• Similarly, triangle IGH in fig.b is an isosceles triangle because GH = GI = $\mathbf\small{|\vec{v}|}$ = $\mathbf\small{|\vec{v'}|}$
♦ So base angles at I and H are equal
♦ We get: ∠I = ∠H = [0.5 × (180-Δθ)]
12. So all the angles in the two triangles OPP' and IGH are equal.
■ Thus they are similar triangles. Some notes on similar triangles can be seen here.
• We can take ratios:
$\mathbf\small{\frac{side\: opposite\: \Delta \theta \: in\: OPP' }{side\: opposite\: \Delta \theta \: in\: IGH}}$ = $\mathbf\small{\frac{side\: opposite\: [0.5 \times(180-\Delta \theta)] \: in\: OPP' }{side\: opposite\: [0.5 \times(180-\Delta \theta)] \: in\: IGH}}$
• Thus we get: $\mathbf\small{\frac{PP'}{IH}=\frac{OP}{GI}}$
$\mathbf\small{\Rightarrow \frac{|\vec{\Delta r}|}{|\vec{\Delta v}|}=\frac{|\vec{r}|}{|\vec{v}|}}$
$\mathbf\small{\Rightarrow |\vec{\Delta v}|=\frac{|\vec{v}|\times |\vec{\Delta r}|}{|\vec{r}|}=\frac{|\vec{v}|\times |\vec{\Delta r}|}{R}}$
• Where R is the radius of the circle
13. Thus we get the 'magnitude of change in velocity'
■ If we divide this magnitude by the Δt during which the change occurs, we will get the magnitude of the average acceleration
• So we can write: $\mathbf\small{|\bar{\vec{a}}|=\frac{|\vec{\Delta v}|}{\Delta t}}$
Substituting for $\mathbf\small{|\vec{\Delta v}|}$ from (12), we get Eq.4.26: $\mathbf\small{|\bar{\vec{a}}|=\frac{|\vec{v}|\times |\vec{\Delta r}|}{R\times \Delta t}}$
14. Thus we obtained the magnitude of average acceleration. Our next aim is to find $\mathbf\small{|\vec{a}|}$, the magnitude of instantaneous acceleration
• For that, we make 'Δt' closer and closer to zero.
• So P' will become closer and closer to P
A sample is shown in fig.4.37(a) below:
 |
| Fig.4.37 |
• All the steps are repeated:
♦ $\mathbf\small{\vec{\Delta v}}$ is obtained in fig.b
♦ It is shown in cyan color
♦ In fig.c, the tail of the cyan vector is placed midway between P and P'
• We find that, here also, the cyan vector is directed towards the center of the circle
15. What will happen if we make Δt smaller and smaller
Ans: The interval Δt will become so small that it can no longer be called an 'interval'
• Instead, we will have to call it an 'instant'
• Then, the acceleration calculated in such a small Δt is not average acceleration $\mathbf\small{\bar{\vec{a}}}$
■ It is the instantaneous acceleration $\mathbf\small{\vec{a}}$
• So we can write: $\mathbf\small{|\vec{a}|= \lim \limits_{\Delta t \to 0} \frac{|\vec{\Delta v}|}{\Delta t}}$
• Substituting for $\mathbf\small{|\vec{\Delta v}|}$ from (12), we get:
$\mathbf\small{|\vec{a}|= \lim \limits_{\Delta t \to 0} \frac{|\vec{v}|\times|\vec{\Delta r}|}{R \times \Delta t}}$
16. This limit can be worked out as follows:
(i) Consider the travel from P to P' in fig.4.37(a).
• When Δt tends to zero, P' is very close to P
• Then Δθ is very small
(ii) The object is travelling with a uniform speed of $\mathbf\small{|\vec{v}|}$ along the circular path
• So it will cover a circular distance of [$\mathbf\small{|\vec{v}|}$ × Δt] in Δt seconds
• Obviously, this distance will be equal to arc length PP'
• So we can write: arc length PP' = [$\mathbf\small{|\vec{v}|}$ × Δt]
(iii) When Δt tends to zero, that is., when Δt is very small, Δθ is also very small and so, arc length PP' will be nearly equal to chord length PP'
• But chord length PP' = $\mathbf\small{\vec{\Delta r}}$
(iv) Using the results in (ii) and (iii), we can write:
chord length PP' = $\mathbf\small{\vec{\Delta r}}$ = [$\mathbf\small{|\vec{v}|}$ × Δt]
(v) So at the limiting state shown in (15), we can use [$\mathbf\small{|\vec{v}|}$ × Δt] instead of $\mathbf\small{\vec{\Delta r}}$
• So the result in (15) becomes: $\mathbf\small{|\vec{a}|=\frac{|\vec{v}|\times [|\vec{v}|\times \Delta t]}{R \times \Delta t}}$
■ Thus we get Eq.4.27: $\mathbf\small{|\vec{a}|=\frac{|\vec{v}|^2}{R}}$
17. In the limiting state, $\mathbf\small{\vec{r}}$ nearly coincides with $\mathbf\small{\vec{r'}}$. This is shown in fig.4.37(c)
• $\mathbf\small{\vec{r}}$ is perpendicular to $\mathbf\small{\vec{v}}$ because, $\mathbf\small{\vec{v}}$ is tangential to the circle at P
• The instantaneous acceleration at P is directed towards the center from P
19. This acceleration is called centripetal acceleration.
• The word 'centripetal' comes from the Greek term which means 'center seeking'.
• This term is very suitable because, as we saw earlier, this acceleration is always directed towards the center.
• It may be noted that the instantaneous acceleration $\mathbf\small{\vec{a}}$ is not a constant vector because, it's direction changes continuously.
In the next section, we will see another expression for centripetal acceleration.
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