Motion, Forces and EnergyLecture 5: Circles and Resistance

mFr

Fr

m

Fr

A particle moving with uniform speed v in a circular path of radius r experiences anacceleration which has magnitude a = v2/r(see Serway Section 4.4 for derivation).

Acceleration is ALWAYS directed towards thecentre of the circle, perpendicular to the velocityvector.

rr mar

mvF

2

When string breaks, ballmoves along tangent.

NB The centripetal force is a familiar force acting in the role of a force that causes a circular motion.

r

Tension as a centripetal force

If the ball has a mass of 0.5kg and the cord is 1.5mlong, we can find the maximum speed of the ballbefore the cord breaks (for a given tension). We assume the ball remains perfectly horizontal duringits motion:.

The TENSION in the cord PROVIDES the CENTRIPETAL FORCE:

mFr

Fr

r

m

Trvso

r

mvT

2

For a tension of say 100 N (breaking strain), theMaximum speed attainable is (100x1.5/0.5)1/2

= 17.3 ms-1.

Conical Pendulum

)1(0cos ymamgT

)2(sin2

r

mvmaT r

tantan2

rgvsorg

v

T

mg

L

A conical pendulum consists of a “bob” revolvingin a horizontal plane.

The bob doesn’t accelerate in the vertical direction:

The centripetal force is provided by the tension component:

(1) / (2) gives

Finally we note that r = L sin so that:

tansingLv

A car moving round a circular curve

r

mvfF sr

2

The centripetal force here is provided bythe frictional force between the car tyresand the road surface:

The maximum possible speed corresponds to themaximum frictional force, fsmax =s n.Since here n = mg, fsmax = s mg

Fr

grv smax

So for s = 0.5 and r = 35 m, vmax = 13.1 ms-1.

Non-uniform circular motion

Sometimes an object may move in a circular path with varying speed. This corresponds to the presence of two forces: a centripetal force and a tangentialforce (the latter of which is responsible for the change in speed of the object asa function of time.

An example of such motion occurs if we whirl a ball tied to the end of a stringaround in a VERTICAL circle. Here, the tangential force arises from gravityacting on the ball.

m

T

R

mgcosmgsin

mg

centre

vtop

vbot

Ttop

Tbot

mg

mg

We must consider bothradial and tangentialforces!

Analysis

cos

cos

sin

sin

2

2

gR

vmT

R

mvmgT

ga

mamg

t

tTangential acceleration

At the bottom of the path where = 0o:

g

R

vmT bottom

bottom

2

g

R

vmT top

top

2

At the bottom of the path where = 180o:

General equation

Therefore, the maximum tensionoccurs at the bottom of the circle,where the cord is most likely to break.

Motion with resistive forces

We’ll take a quick look at fluid resistance (such as air resistance orResistance due to a liquid). Resistive forces can depend on speedIn a complex way, but here we will look at the simplest case:

R = b v ie the resistance is proportional to the speed of the moving object.

mg

R

v

With fluid (air) resistance, the falling object doesnot continue to accelerate, but reaches terminalVelocity (see Terminal Velocity Analysis notes).