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`
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`
