• Uniform Circular Motion

When an object follows a circular path at a constant speed, the motion of the object is called `color{red}(ul"uniform circular motion.")`

Suppose an object is moving with uniform speed v in a circle of radius `R` as shown in Fig. Since the velocity of the object is changing continuously in direction, the object undergoes acceleration.

Let `vecr` and `vec(r′)` be the position vectors and `vecv` and `vec(v′)` the velocities of the object when it is at point `P` and `P′` as shown in `color{blue}{Fig. (a).}`

By definition, velocity at a point is along the tangent at that point in the direction of motion. The velocity vectors `vecv` and `vec(v′)` are as shown in `color{blue}{Fig. (a1).}`

`Δvecv` is obtained in Fig. (a2) using the triangle law of vector addition.

• `vecv` `bot` `vecr` and so is `vec(v′)` `bot` `vec(r′)`. Therefore, `Δvecv` `bot` `Δvecr`. Since average acceleration is along `Δvecv` the average acceleration `bara` `bot` `Deltavecr`.

In Fig. (c), `Δt->0` and the average acceleration becomes the instantaneous acceleration. It is directed towards the centre.

• Thus, we find that the acceleration of an object in uniform circular motion is always directed towards the centre of the circle.

The magnitude of `veca` is, by definition, given by

`|veca|=lim_(Deltat->0) |Deltavecv|/(Deltat)`

Let the angle between position vectors `vecr` and `vec(r')` be `Delta theta`.

From the triangles CPP' and GHI,

`|Deltavecv|/v = |Deltavecr|/R`

`=> |Deltavecv|=v |Deltavecr|/R`

Therefore,

`|veca|= lim_(Deltat->0) |Deltavecv|/(Deltat) = lim_(Deltat->0) (v|Deltavecr|)/(RDeltat)=v/R lim_(Deltat->0) |Deltavecr|/(Deltat)`

If Δt is small, Δθ will also be small and then arc PP′ can be approximately taken to be `|Δvecr|` :

`|Δvecr| =~vDeltat`

`|Deltavecr|/(Deltat)=~v`

Or, `lim_(Deltat->0) |Deltavecr|/(Deltat)=v`

Therefore, the centripetal acceleration `a_c` is :

`a_c=(v/R)v=v^2/R`

The acceleration is always directed towards the centre. This is why this acceleration is called centripetal acceleration.

• Since v and R are constant, the magnitude of the centripetal acceleration is also constant. However, the direction changes — pointing always towards the centre. Therefore, a centripetal acceleration is not a constant vector.

The time rate of change of angular displacement :

`omega=(Delta theta)/(Deltat)`

Now, if the distance travelled by the object during the time Δt is Δs, i.e. PP′ is Δs, then :

`v=(Deltas)/(Deltat)`

but `Δs = R Δθ`. Therefore :

`v=R (Delta theta)/(Deltat) = R omega`

`v=R omega`

We can express centripetal acceleration `a_c` in terms of angular speed :

`a_c=v^2/R=(omega^2R^2)/R=omega^2R`

`a_c=omega^2R`

`text(Time Period)`

The time taken by an object to make one revolution is known as its time period T.

`T=(2piR)/v`

`text(Frequency)`

The number of revolution made in one second is called its frequency.

`nu=1/T`

In terms of frequency `nu`, we have

`omega=2pinu`

`v=2piRnu`

`a_c=4pi^2nu^2R`

Suppose an object is moving with uniform speed v in a circle of radius `R` as shown in Fig. Since the velocity of the object is changing continuously in direction, the object undergoes acceleration.

Let `vecr` and `vec(r′)` be the position vectors and `vecv` and `vec(v′)` the velocities of the object when it is at point `P` and `P′` as shown in `color{blue}{Fig. (a).}`

By definition, velocity at a point is along the tangent at that point in the direction of motion. The velocity vectors `vecv` and `vec(v′)` are as shown in `color{blue}{Fig. (a1).}`

`Δvecv` is obtained in Fig. (a2) using the triangle law of vector addition.

• `vecv` `bot` `vecr` and so is `vec(v′)` `bot` `vec(r′)`. Therefore, `Δvecv` `bot` `Δvecr`. Since average acceleration is along `Δvecv` the average acceleration `bara` `bot` `Deltavecr`.

In Fig. (c), `Δt->0` and the average acceleration becomes the instantaneous acceleration. It is directed towards the centre.

• Thus, we find that the acceleration of an object in uniform circular motion is always directed towards the centre of the circle.

The magnitude of `veca` is, by definition, given by

`|veca|=lim_(Deltat->0) |Deltavecv|/(Deltat)`

Let the angle between position vectors `vecr` and `vec(r')` be `Delta theta`.

From the triangles CPP' and GHI,

`|Deltavecv|/v = |Deltavecr|/R`

`=> |Deltavecv|=v |Deltavecr|/R`

Therefore,

`|veca|= lim_(Deltat->0) |Deltavecv|/(Deltat) = lim_(Deltat->0) (v|Deltavecr|)/(RDeltat)=v/R lim_(Deltat->0) |Deltavecr|/(Deltat)`

If Δt is small, Δθ will also be small and then arc PP′ can be approximately taken to be `|Δvecr|` :

`|Δvecr| =~vDeltat`

`|Deltavecr|/(Deltat)=~v`

Or, `lim_(Deltat->0) |Deltavecr|/(Deltat)=v`

Therefore, the centripetal acceleration `a_c` is :

`a_c=(v/R)v=v^2/R`

The acceleration is always directed towards the centre. This is why this acceleration is called centripetal acceleration.

• Since v and R are constant, the magnitude of the centripetal acceleration is also constant. However, the direction changes — pointing always towards the centre. Therefore, a centripetal acceleration is not a constant vector.

The time rate of change of angular displacement :

`omega=(Delta theta)/(Deltat)`

Now, if the distance travelled by the object during the time Δt is Δs, i.e. PP′ is Δs, then :

`v=(Deltas)/(Deltat)`

but `Δs = R Δθ`. Therefore :

`v=R (Delta theta)/(Deltat) = R omega`

`v=R omega`

We can express centripetal acceleration `a_c` in terms of angular speed :

`a_c=v^2/R=(omega^2R^2)/R=omega^2R`

`a_c=omega^2R`

`text(Time Period)`

The time taken by an object to make one revolution is known as its time period T.

`T=(2piR)/v`

`text(Frequency)`

The number of revolution made in one second is called its frequency.

`nu=1/T`

In terms of frequency `nu`, we have

`omega=2pinu`

`v=2piRnu`

`a_c=4pi^2nu^2R`

Q 3085691567

An insect trapped in a circular groove of radius `12 cm` moves along the groove steadily and completes 7 revolutions in 100 s. (a) What is the angular speed, and the linear speed of the motion? (b) Is the acceleration vector a constant vector ? What is its magnitude ?

This is an example of uniform circular motion. Here `R = 12 cm`. The angular speed` ω` is given by

`ω = (2π)/T = 2π × 7/100 = 0.44 rad//s`

The linear speed v is :

`v =ω R = 0.44 s^(-1) × 12 cm = 5.3 cm s^(-1)`

The direction of velocity v is along the tangent to the circle at every point. The acceleration is directed towards the centre of the circle. Since this direction changes continuously, acceleration here is not a constant vector. However, the magnitude of acceleration is constant:

`a = ω^2 R = (0.44 s^(–1))^2 (12 cm)`

`= 2.3 cm s^(-2)`