Chapter05 Circular motion - Weebly

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5. Circular motion

5.1 Angular displacement and angular

velocity

5.2 Centripetal acceleration

5.3 Centripetal force

By Liew Sau Poh

Objectives

Objectives

a) express angular displacement in radians

b) define angular velocity and period

c) derive and use the formula v = r

d) explain that uniform circular motion

has an acceleration due to the change in

direction of velocity

e) derive and use the formulae for

centripetal acceleration a = v2 / r and a

=r 2

f) explain that uniform circular motion is due

to the action of a resultant force that is

always directed to the centre of the circle

g) use the formulae for centripetal force F =

mv2/r and F = mr 2

h) solve problems involving uniform

horizontal circular motion for a point mass

i) solve problems involving vertical circular

motions for a point mass (knowledge of

tangential acceleration is not required).

Uniform circular motion

Suppose that an

object executes a

v

circular orbit of

radius r with

uniform tangential t = 0

speed v.

5.1 Angular displacement

and angular velocity

s

t=t

(t)

r

v

Uniform circular motion

Uniform circular motion

The instantaneous

position of the

v

object is most

conveniently

specified in terms t = 0

of an angle .

For instance, we

could decide that

v

= 0 corresponds to

the object's

location at t = 0, in t = 0

which case we

would write (t) =

t, where is the

angular velocity of

the object.

s

t=t

(t)

r

v

s

t=t

(t)

r

v

Uniform circular motion

Angular displacement

For a uniformly

rotating object, the

v

angular velocity is

simply the angle

through which the t = 0

object turns in one

second.

Consider the motion

of the object in the

v

time interval

between t = 0

t=0

and t = t.

Here, the object

rotates through an

angle , and traces

out a circular arc of

length s.

s

t=t

(t)

r

v

Angular displacement

Angular displacement

It is fairly obvious

that the arc length s

v

is directly

proportional to the

angle , an angle of t = 0

360 corresponds to

an arc length of 2 r.

Hence, an angle

must correspond to

an arc length of s 2 r

At this stage, it is

convenient to define

v

a new angular unit

known as a radian

t=0

(symbol rad).

s

t=t

(t)

s

t=t

(t)

r

v

s

t=t

(t)

r

r

v

v

360

Angular displacement

Angular displacement

An angle measured

in radians is related

v

to an angle

measured in degrees

via the following t = 0

simple formula:

Thus, 360 2

rad,

v

180

rad,

90 ? rad,

t=0

and 57.296 1 rad.

rad

s

t=t

(t)

r

v

2

360

Angular displacement

When is

measured in

radians,

s

2

r

360

simplifies greatly

to give s = r .

t=t

(t)

r

v

Angular velocity

v

t=0

s

s

t=t

(t)

r

v

Consider the motion of the object in the

short interval between times t and t +

t. In this interval, the object turns

through a small angle

and traces out

a short arc of length s, where s = r .

Angular velocity

Now s/ t (i.e., distance moved per unit

time) is simply the tangential velocity v,

whereas / t (i.e., angle turned

through per unit time) is simply the

angular velocity w. Thus, dividing s =

r by t, we obtain v = rw.

Angular velocity

An object that rotates with uniform

angular velocity w turns through w

radians in 1 second.

Hence, the object turns through 2

radians (i.e., it executes a complete

circle) in T = 2 / w seconds.

Angular velocity

Here, the repetition frequency, f , of the

motion is measured in cycles per

second--otherwise known as hertz

(symbol Hz).

Angular velocity

Note, however, that this formula is only

valid if the angular velocity w is

measured in radians per second.

From now on, in this course, all angular

velocities are measured in radians per

second by default.

Angular velocity

Here, T is the repetition period of the

circular motion. If the object executes a

complete cycle (i.e., turns through 360 )

in T seconds, then the number of cycles

executed per second is f = 1/T = w / 2 .

In other words, w = 2 /T.

Angular velocity

As an example, suppose that an object

executes uniform circular motion,

radius r = 1.2m, at a frequency of f =

50Hz (i.e., the object executes a

complete rotation 50 times a second).

The repetition period of this motion is

simply T = 1/f = 0.02s.

Angular velocity

Furthermore, the angular frequency of

the motion is given by w = 2 f = 314.16

rad/s

Finally, the tangential velocity of the

object is v = r w = 1.2 314.16 = 376.99

m/s.

5.3 Centripetal acceleration

Circular motion

Circular motion

The speed stays

constant, but the

direction changes

The tension

in the string!

Bart swings the tennis ball around his head

in a circle. The ball is accelerating, what

force makes it accelerate?

Centripetal acceleration, aC

v

R

The acceleration in this

case is called

centripetal acceleration

Centripetal acceleration

aC

R

toward the

center

of the circle

v

The acceleration

points toward the

center of the circle

Centripetal acceleration

Centripetal acceleration

An object executing a

circular orbit of radius

Z

Q v

r with uniform

v

tangential speed v

X

possesses a velocity

P

v

r

vector v whose

magnitude is constant,

but whose direction is

continuously

changing.

It follows that the

object must be

accelerating, since

acceleration is the

rate of change of

velocity , and the

velocity is indeed

varying in time.

Y

Z

(vector)

(vector)

Q

v

v

P

v

Y

X

r

(vector)

Centripetal acceleration

Centripetal acceleration

Suppose that the

object moves from

Z

Q v

point P to point Q

v

between times t and t

X

+ t, as shown in the P

v

r

figure above. Suppose,

further, that the object

rotates through

radians in this time

interval.

The vector PX ,

shown in the diagram,

Z

Q v

is identical to the

v

vector QY. Moreover,

X

the angle subtended P

v

r

between vectors PZ

and PX is simply .

Y

Y

Centripetal acceleration

Centripetal acceleration

The vector ZX

represents the

change in vector

velocity, v, between

times t and t + t.

It can be seen that

this vector ZX is

directed towards the

centre of the circle.

From standard

trigonometry, the

length of vector is v

= 2v sin( /2).

Z

Q

v

v

P

v

X

r

Centripetal acceleration

However, for small angles sin

,

provided that is measured in radians.

Hence, v v ,

It follows that a = v/ t = v

/ t=v

, where =

/ t is the angular

velocity of the object, measured in

radians per second.

Y

Z

Q

v

P

v

Y

X

r

Centripetal acceleration

In summary, an object executing a

circular orbit, radius r, with uniform

tangential velocity v, and uniform

angular velocity w = v/r, possesses an

acceleration directed towards the centre

of the circle:- i.e., a centripetal

acceleration:- of magnitude a = vw = v2/r

= rw2.

Centripetal acceleration

centripetal acceleration

5.3 Centripetal force

v

v2

aC =

R

a force is needed to produce this

centripetal acceleration

CENTRIPETAL FORCE

where does this force come from?

Centripetal force

Centripetal force

Suppose that a

weight, of mass m, is

attached to the end

v

of a cable, of length r,

and whirled around m

T

r

such that the weight

executes a horizontal

circle, radius r, with

uniform tangential

velocity v.

As we have just

learned, the weight is

subject to a centripetal

v

acceleration of

m

magnitude v2/r.

T

r

Hence, the weight

experiences a

centripetal force f = m

v2/r.

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