Centripetal and Centrifugal Force - Acceleration

ACCELERATION AND FORCE

Centripetal acceleration occurs with all rotating bodies. Consider a point P rotating

about a center O with constant angular velocity ω (fig. 1).

The radius is the length of the line O-P. The tangential velocity of P is v = ωr. This velocity is constant in magnitude but is continually changing direction.

Let’s remind ourselves of the definition of acceleration.

aC = velocity change / time taken

aC= Δv/t

Velocity is a vector quantity and a change in direction alone is sufficient to produce a change. It follows that a point travelling in a circle is continuously changing its direction, velocity and hence has acceleration

Next let’s remind ourselves of Newton’s second Law of Motion which in its simplest

form states

Force = Mass x acceleration.

It follows that anything with mass travelling in a circle must require a force to produce the acceleration just described.

The force required to make a body travel in a circular path is called CENTRIPETAL FORCE and it always pulls towards the centre of rotation. You can easily demonstrate this for yourself by whirling a small mass around on a piece of string.

Let’s remind ourselves of Newton’s Third Law. Every force has an equal and opposite reaction.

The opposite and equal force of centripetal is the CENTRIFUGAL FORCE

The string is in tension and this means it pulls in both directions. The force pulling the ball towards the middle is the centripetal force and the force pulling on your finger is the centrifugal force. The derivation of the formula for centripetal force and acceleration is done by considering the velocity as a vector.

FIGURE 2

Consider the velocity vector before and after point P has revolved a small angle δθ.

The magnitude of v1 and v2 are equal so let’s denote it simply as v. The direction changes over a small period of time δt by δθ radians. We may deduce the change by using the vector addition rule.

The first vector + the change = Final vector.

The rule is v1 + δv = v2. This is illustrated below.

δv is almost the length of an arc of radius v. If the angle is small, this becomes truer.

The length of an arc is radius x angle so it follows that δv = v×δθ

This change takes place in a corresponding small time δt so the rate of change of velocity is

 = δv / δt = v×δθ/ δt

the acceleration. a_c= dv/ dt = v dθ/ dt = v× ω

since

ω = dθ/ dt = rate of change of angle

Since v = ωr then substitute for v and a_c = ω²×R and this is the centripetal acceleration.

Centripetal acceleration = ω²×r

Since ω = v/r  then substitute for ω and a_c = v²/r

Centripetal acceleration = v²/r

If we examine the vector diagram, we see that as δθ becomes smaller and smaller, so the direction of δv becomes radial and inwards. The acceleration is in the direction of the change in velocity and so centripetal acceleration is radial and inwards. If point P has a mass M, then the force required to accelerate this mass radial inwards is found from Newton's 2nd Law

This  acceleration is named the centripetal acceleration

a_c = v² / r  

    = ω² r 

    = (2 π n_rps)² r   

    = (2 π n_rpm / 60)² r   

    =  (π n_rpm / 30)² r       

where

a_c = centripetal acceleration (m/s², ft/s²) same a_{c =} a

 v = tangential velocity (m/s, ft/s)

r = circular radius (m, ft)

ω = angular velocity (rad/s)

n_rps = revolutions per second (rev/s, 1/s)

n_rpm = revolutions per min (rev/min, 1/min)

If point P has a mass m, then the force required to accelerate this mass radial inwards

is found from Newton's 2nd Law.

Centripetal force = m ω² r

or in terms of velocity v Centripetal force = m v²/r

Centrifugal force is the reaction force and acts radial outwards.

According Newton's second law the centripetal force can be expressed as

F_c = m a_c

    = m v² / r  

    = m ω² r        

    = m (2 π n_s)² r 

    = m (2 π n_rpm / 60)² r 

    =  m (π n_rpm / 30)² r                           

where

F_c = centripetal force (N, lb_f)

m = mass (kg, slugs)

EXAMPLE No.1

Calculate the centripetal acceleration and force acting on an aeroplane of mass

1500 kg turning on a circle 400 m radius at a velocity of 300 m/s.

SOLUTION

Centripetal acceleration = v2/r = 3002/400 = 225 m/s2.

Centripetal force = mass x acceleration = 1500 x 225 = 337.5 kN

 EXAMPLE No.2

A centrifugal clutch is shown in the diagram. The clutch must transmit a torque of

18 Nm at a speed of 142 rev/min. The coefficient of friction ‘μ’ between the drum

and the friction lining is 0.3. The radius to the centre of gravity of each sliding

head is 0.21 m and the inside radius of the drum is 0.25 m. Calculate the required

mass of the sliding heads.

SOLUTION

Torque = 18 Nm, radius = 0.25 m

T = Friction force x radius

Friction force = 18/0.25 = 72 N This is divided between two friction pads so each

must produce 72/2 = 36 N each.

From the law of friction, Friction force = μ x normal force

Normal force = 36/0.3 = 120 N

The normal force acts in a radial direction and must be equal to the centripetal

force.

Centripetal force = mω²r

Equating forces we have mω² r = 120

Speed = 142 rev/min or 142/60 rev/s

Radius to centre of gravity = 0.21 m

Angular velocity ω = 2π x speed = 2π x 142/60 = 14.87 rad/s

mx 14.872x 0.21 =120

m = 120/( 14.872x 0.21) = 2.58 kg for each sliding head.