5. Circular motion By Liew Sau Poh

Content 5.1 Angular displacement and angular

velocity 5.2 Centripetal acceleration 5.3 Centripetal force

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

Objectives 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).

5.1 Angular displacement and angular velocity

Uniform circular motion Suppose that an object executes a circular orbit of radius r with uniform tangential speed v.

(t) t = 0

v

v

r

s

t = t

Uniform circular motion The instantaneous position of the object is most conveniently specified in terms of an angle .

(t) t = 0

v

v

r

s

t = t

Uniform circular motion For instance, we could decide that = 0 corresponds to the object's location at t = 0, in which case we would write (t) =

t, where is the angular velocity of the object.

(t) t = 0

v

v

r

s

t = t

Uniform circular motion For a uniformly rotating object, the angular velocity is simply the angle through which the object turns in one second.

(t) t = 0

v

v

r

s

t = t

Angular displacement Consider the motion of the object in the time interval between t = 0 and t = t. Here, the object rotates through an angle , and traces out a circular arc of length s.

(t) t = 0

v

v

r

s

t = t

Angular displacement It is fairly obvious that the arc length s is directly proportional to the angle , an angle of 360 corresponds to an arc length of 2 r. Hence, an angle must correspond to an arc length of rs

3602

(t) t = 0

v

v

r

s

t = t

Angular displacement At this stage, it is convenient to define a new angular unit known as a radian (symbol rad). (t) t = 0

v

v

r

s

t = t

Angular displacement An angle measured in radians is related to an angle measured in degrees via the following simple formula:

3602rad

(t) t = 0

v

v

r

s

t = t

Angular displacement Thus, 360 2 rad, 180 rad, 90 ½ rad, and 57.296 1 rad.

(t) t = 0

v

v

r

s

t = t

Angular displacement When is measured in radians, simplifies greatly to give s = r .

rs3602

(t) t = 0

v

v

r

s

t = t

Angular velocity 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 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 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, 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 Here, the repetition frequency, f , of the motion is measured in cycles per second--otherwise known as hertz (symbol Hz).

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

Bart swings the tennis ball around his head in a circle. The ball is accelerating, what force makes it accelerate?

The tension in the string!

Circular motion

R v The speed stays constant, but the direction changes

The acceleration in this case is called

centripetal acceleration

Centripetal acceleration, aC

R

v

aC

The acceleration points toward the center of the circle

Centripetal acceleration

toward the center

of the circle

Centripetal acceleration An object executing a circular orbit of radius r with uniform tangential speed v possesses a velocity vector v whose magnitude is constant, but whose direction is continuously changing.

Z

X P

Y Q v

v

v r

Centripetal acceleration It follows that the object must be accelerating, since acceleration (vector) is the rate of change of velocity (vector), and the velocity (vector) is indeed varying in time.

Z

X P

Y Q v

v

v r

Centripetal acceleration Suppose that the object moves from point P to point Q between times t and t + t, as shown in the figure above. Suppose, further, that the object rotates through radians in this time interval.

Z

X P

Y Q v

v

v r

Centripetal acceleration The vector PX , shown in the diagram, is identical to the vector QY. Moreover, the angle subtended between vectors PZ and PX is simply .

Z

X P

Y Q v

v

v r

Centripetal acceleration The vector ZX represents the change in vector velocity, v, between times t and t + t.

Z

X P

Y Q v

v

v r

Centripetal acceleration 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

X P

Y Q v

v

v 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.

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.

5.3 Centripetal force

Centripetal acceleration centripetal acceleration

a force is needed to produce this centripetal acceleration CENTRIPETAL FORCE where does this force come from?

2

Cva =R

Centripetal force Suppose that a weight, of mass m, is attached to the end of a cable, of length r, and whirled around such that the weight executes a horizontal circle, radius r, with uniform tangential velocity v.

m v

T r

Centripetal force As we have just learned, the weight is subject to a centripetal acceleration of magnitude v2/r. Hence, the weight experiences a centripetal force f = m v2/r.

m v

T r

Centripetal force What provides this force? Well, in the present example, the force is provided by the tension T in the cable. Hence, T = mv2/r .

m v

T r

Centripetal force Suppose that the cable is such that it snaps whenever the tension in it exceeds a certain critical value Tmax.

m v

T r

Centripetal force It follows that there is a maximum velocity with which the weight can be whirled around: namely,

m v

T r

mrTv max

max

Centripetal force If v exceeds vmax then the cable will break. As soon as the cable snaps, the weight will cease to be subject to a centripetal force.

m v

T r

Centripetal force So it will fly off; with velocity vmax along the straight-line which is tangential to the circular orbit it was previously executing.

m v

T r

Question 1 A 2.0 kg mass swinging at the end of a 0.50 m string is traveling 3.0 m/s. What is the a. centripetal acceleration of the mass? b. centripetal force on the mass?

Answer 1 a) ac = v2/r

ac = (3.0 m/s)2/(0.5 m) ac = 18 m/s2

b) Fc = mac = 36 J

Question 2 A student swings a ball in a circle of radius 70 cm in the vertical plane. The angular velocity of the ball is 10 rad s 1. a) What is the velocity of the ball? b) How long does the ball take to

complete one revolution?

Answer 2 a) v = r = (10)(0.70) = 7.0 m s-1

b)

22vrT

s63.0102

7)70.0(2T

Question 2 (continue) A student swings a ball in a circle of radius 70 cm in the vertical plane. The angular velocity of the ball is 10 rad s 1. The student releases the ball when it is at A, which is 130 cm above the ground, and the ball travels vertically upwards. Calculate a) the maximum height, above the

ground, the ball will reach; b) the time taken for the ball to hit the

ground after its release from A.

a) v2 = u2 + 2as

0 = (7)2 + 2(-9.8) s => s = 2.50 m => max. height = 2.5 + 1.30 = 3.8 m

b) Overall: A => max. height => ground: s = ut + ½ at2 -1.30 = 7t ½ (9.8)t2 time, t = 1.59 s

Ball on a string The tension in the string provides the necessary centripetal force to keep the ball going in a circle.

Path of ball if the string breaks

Top view

Example What is the tension in a string used to twirl a 0.3 kg ball at a speed of 2 ms-1 in a horizontal circle of 1 meter radius? Answer: Force = mass x acceleration [ m ac ] Acceleration, ac = v2 / R = (2)2/ 1 = 4 ms-2

Force, F = mac = 0.3 4 = 1.2 N

Example If the string is not strong enough to handle this tension (1.2 N) it will break and the ball goes off in a straight line.

Vertical Circular Motion If a pail is whirled in a vertical circle, the speed of the pail varies along its circular motion mg

v

T r

O

Vertical Circular Motion The cetripetal acceleration, a = v2/r towards O. Using F = ma, T mg cos = mv2/r Tension, T = mv2/r + mg cos

mg

v

T r

O

Vertical Circular Motion T = mv2/r + mg cos At lowest point, = 0

T = mv2/r + mg At highest point, =

T = mv2/r mg When = /2 or 3 /2

T = mv2/r; (cos = 0)

mg

v

T r

O

Vertical Circular Motion T = mv2/r + mg cos At highest point, = T = mv2/r mg The pail well not drop if mv2/r mg > 0 mv2/r > mg v > (gr)1/2

mg

v

T r

O

Conical Pendulum An object is moving in a horizontal circle (forming a conical pendulum) For horizontal circular motion, a = v2/r (towards centre) mg

T

r

l

Conical Pendulum For horizontal circular motion, a = v2/r (towards centre)

F = ma = mv2/r For horizontal component, T sin = mv2/r = mr 2 = m(l sin ) 2 T = ml 2

mg

T

r

l Ty

Tx

Ty = T cos , Tx = T sin

Conical Pendulum For vertical component, ml 2 cos = mg cos = g / l 2

= cos-1(g/l 2)

mg

T

r

l Ty

Tx

Ty = T cos , Tx = T sin

Negotiating a flat (level) turn The centripetal force is provided by the friction force between the road and tires. this force is reduced if the road is wet or icy

Banked Turns

31 degree bank

Velodrome

Banked turns Since the road is banked (not horizontal) the force of the road on the box is not vertical. Part of the force on the box from the road points toward the center of the circle. This provides the centripetal force. No friction is necessary to keep the box in the circle.

R

N

FCENT

The red object will make the turn only if there is enough friction on it 0r it goes straight the apparent outward force is called the centrifugal force it is NOT A REAL force! an object will not move in a circle until something makes it!

object on the dashboard

straight line object naturally

follows

Centrifugal force: Definition Force represents the effects of inertia that arise in connection with rotation and which are experienced as an outward force away from the center of rotation

Silly Silo (Rotor)

weight

Friction between Bart and wall

wall pushing in on Bart

The inward wall force keeps Bart in the circle. Friction keeps him from falling down.

Summary: Circular motion = / t

v = r Angular

Displacement & Angular Velocity

a = v2/r a = v a = r 2

Centripetal Acceleration

F = mv2/r = mv = mr 2 Motion in horizontal circle Motion in vertical circle

Centripetal Force