Physics UNIFORM CIRCULAR MOTION
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### Topics Covered

• Uniform Circular Motion

### 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

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 ?

Solution:

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)