Dynamics of Uniform Circular Motion

An object moving on a circular path of radius r at a constant speed v

As motion is not on a straight line, the direction of the velocity vector is not constant

The motion is circular

v

v

vr

Compare to: 1D – straight line2D – parabola

Velocity vector is always tangent to the circle

Velocity direction constantly changing, but magnitude remains constant

Vectors r and v are always perpendicular

Since the velocity direction always changes, this means that the velocity is not constant (though speed is constant), therefore the object is accelerating

v

ra

The acceleration ar points radially inward. Like velocity its direction changes, therefore the acceleration is not constant (though its magnitude is) Vectors ar and v are also perpendicular

The speed does not change, since ar acceleration has no component along the velocity direction

Why is the acceleration direction radially inward?

2v

ra

1v

1v

2v

ra

Since

12

12 vv

tta

This radial acceleration is called the centripetal acceleration

This acceleration implies a ``force’’

racp

2v

r

mmaF cpcp

2v The ``centripetal force’’

(is not a force)

The centripetal force is the net force required to keep an object moving on a circular path

Consider a motorized model airplane on a wire which flies in a horizontal circle, if we neglect gravity, there are only two forces, the force provided by the airplane motor which tends to cause the plane to travel in a straight line and the tension force in the wire, which forces the plane to travel in a circle – the tension is the ``centripetal force’’

motorF

T

r

mTmaF rr

2v

Consider forces in radial direction (positive to center)

Time to complete a full orbit

Tension)(not Periodv

2

speed

ncecircumferencecircumfere2

r

T

rD

The Period T is the time (in seconds) for the object to make one complete orbit or cycle

Find some useful relations for v and ar in terms of T

cpcp aT

r

rT

r

ra

T

r

2

222 412v

2v

Example

A car travels around a curve which has a radius of 316 m. The curve is flat, not banked, and the coefficient of static friction between the tires and the road is 0.780. At what speed can the car travel around the curve without skidding?

r

v

mg

fs

FN

mg

FN

fs

y

r

mgF

mgF

maF

N

N

yy

0

r

mf

maF

s

rr

2v

Now, the car will not skid as long as Fcp is less than the maximum static frictional force

hrmi

sm

sm

sm

2max

2

110m 1609

mi 1

hr 1

s 36001.49

1.49m) 316)(80.9)(780.0(v

vv

2

gr

mgr

mFf

r

m

s

sNss

ExampleTo reduce skidding, use a banked curve. Consider same conditions as previous example, but for a curve banked at the angle

mg

FN

rFN

mg

r

fs

y

Choose this coordinate system since ar

is radial

fs

0sincos

mgfF

maF

sN

yy

Since acceleration is radial only

Since we want to know at what velocity the car will skid, this corresponds to the centripetal force being equal to the maximum static frictional force

Nsss Fff max Substitute into previous equation

sincos)sin(cos

sincos

sNsN

NsN

mgFmgF

mgFF

r

mF

r

mfF

maF

sN

sN

rr

2

2

v)cos(sin

vcossin

Substitute for FN and solve for v

)sin(cos

)cos(sinv

v)cos(sin

sincos

2

s

s

ss

rg

r

mmg

Adopt r = 316 m and = 31°, and s=0.780 from earlier

Compare to example 6-9 where s=0

hrmi

sm 200 89.7v

hrmi

sm 5.961.43tan

)(cos

)(sinv

rgrg