Motion in a Circle - PBworks

JJ 2014 H2 PHYSICS (9646) Motion in a Circle

© Physics Dept Page 1 of 13

Motion in a Circle Content 1. Kinematics of uniform circular motion 2. Centripetal acceleration 3. Centripetal force Learning Outcomes Candidates should be able to: (a) express angular displacement in radians. (b) understand and use the concept of angular velocity to solve problems.

(c) recall and use v = rω to solve problems. (d) describe qualitatively motion in a curved path due to a perpendicular force, and

understand the centripetal acceleration in the case of uniform motion in a circle.

(e) recall and use centripetal acceleration a = rω2, a = 2v

r to solve problems.

(f) recall and use centripetal force F = mrω2, F = m2v




Introduction There are many types of motion in the world: Linear motion, projectile motion, circular motion and simple harmonic motion, to name a few, that we would study. Besides travelling in a straight line as studied in Kinematics earlier, we are interested in solving problems that involved motion in a curved path. For example, a car going around a bend, a satellite orbiting Earth, an electron orbiting a nucleus, a roller-coaster ride. In this chapter, we will study the motion of a body moving round a circular path with a uniform (i.e. constant) speed which is described as a uniform circular motion. a) Express angular displacement in radians The diagram shows how angle θ is defined:

rs

arc of radius lengtharc ==θ

where s is the curved distance along the arc of radius r

θ is measured in radians.

Hence one radian (1 rad) is the angle when the arc length is the same as the radius of the arc.

For one complete circle: arc length, s = circumference of circle = 2πr

For one revolution, rr

radiuscircleofncecircumfere

arcofradiuslengtharc πθ 2

=== = 2π rad

To describe the angle turned during motion in a circle, we use the term angular displacement.

If a particle moves from A to B, the change in angular displacement in this case is Δθ = (θ2 - θ1) rad. Angular displacement is measured in radians. Example 1 The laser on a CD player is 5.0 cm from the central hole of the disc. What length of the disc is scanned by the laser when the disc turns through an angle of 0.45 radians? Solution:

rs

arc of radiuslengtharc ==θ Length scanned, s = θ r = (0.45)(5) = 22..33 ccmm

The radian is the angle subtended by an arc length equal to the radius of the circle.

Angular displacement is the angle turned about the centre of the circle.



b) Understand and use the concept of angular velocity to solve problems

If something is spinning, it is not possible to state a unique velocity with which it is travelling, as different parts of it are travelling with different velocities (due to its changing directions). To overcome this problem, the term ‘angular velocity’ is used.

Symbol : ω (small letter ‘omega’) SI Unit of angular velocity : radians per second (rad s-1)

For constant angular velocity, Ttπθω 2 ==

The period T of circular motion is the time taken for one revolution. Since 2π radians is turned in one revolution :

angle turned in one revolutiontime for one revolution =angular velocity

or ωπ2T =

c) Recall and use v = rω to solve problems

Let a particle travel around a circle with uniform angular velocity, ω. Let the length of the arc travelled be s.

Since s = rθ

Linear velocity, v = ( ) ωθθ rdtdr

dtrd

dtds

===

Example 2 The minute hand of a clock is moving at constant angular velocity ω. Which point on the hand, A, B or C is moving the fastest? Solution:

As all points A, B and C lie on the same minute hand, they must all be moving with the same angular velocity ω.

From v = rω, points further from the centre (larger r) will have greater speeds.

Hence, point C is moving the fastest.

Angular velocity ω is the rate of change of angular displacement.

v = rω



Example 3 A train is travelling on a track which is part of a circle of radius 600 m at a constant speed of 50 m s-1. What is its angular velocity? Solution:

v = rω v 50r 600

ω = = = 0.083 rad s-1

Example 4 What is the angular velocity of the Earth about its axis?

d) Describe qualitatively, motion in a curved path due to a perpendicular force,

and understand the centripetal acceleration in the case of uniform motion in a circle

According to Newton’s 1st Law, every object continues in its state of rest or uniform motion in a straight line unless a resultant external force acts on it to change that state.

Uniform circular motion is the motion of an object moving in circular path at constant speed.

If an object is travelling with constant angular velocity in a circle, the direction of

motion is continually changing (i.e. not travelling in a straight line). This implies that :

• the velocity is changing but the speed is constant.

• there must be a resultant force acting on the object, but directed towards the

centre of the circle.

Solution: Angular velocity of the Earth’s rotation about its axis

ωπ2T =

hr 242

T2 ππω == = 0.26 rad hr-1

However, we typically state ω in rad s-1.

Hence s 3600 24

2T

2×

==ππω = 7.3 x 10-5 rad s-1



Analysis of Uniform Circular Motion

Consider an object moving in uniform circular motion at constant angular velocity, ω, from A to B. If no resultant force acts on the object, by Newton’s 1st law, the object will travel from A to C at constant velocity, v. However, since the object travels along the curved path from A to B, this implies that a resultant force F must have acted on the object to change its velocity. Instead of arriving at C, the object arrives at B. This means that the object must have been “pulled” towards the centre of the circle "by this resultant force" F.

Points to note:

1. Since the object is moving at constant speed, this means that there is no component of acceleration in the direction of motion, implying there is no component of resultant force acting along the tangential path of the object.

[Conversely, if the speed is changing (not constant), there will be a component of resultant force acting tangentially to the circle. This component causes a change in speed.]

2. However, since the velocity keeps changing (i.e. only the direction of motion is changing), this must imply that there must be a resultant force acting on the object. The only direction that this resultant force can act is towards the centre of the circle. This resultant force is called centripetal force.

3. Since the object experiences a resultant force towards the centre of the circle, by Newton’s 2nd law, there must be an acceleration towards the centre of the circle. This acceleration is called centripetal acceleration, which is a vector with units m s-2.

In uniform circular motion, the speed of the object does not change, but the direction is changing (i.e. velocity is changing). Hence, the centripetal acceleration does not change the magnitude of the velocity, but only changes its direction.

4. The centripetal force is NOT an additional force acting on the object. It is the resultant of all the forces acting on the object and acts towards the centre of the circle. It can be a combination of forces like tension, friction, gravitational force, electric force or nuclear force. It is called centripetal force because of what it is doing - moving object in a circular path. It does not appear in any free body diagrams as a fundamental force, rather it is a resultant force.

5. The centripetal force is always perpendicular to the direction of motion. Hence, this force does not do any work on the object.

Centripetal force is the resultant force acting on an object in uniform circular motion and is directed towards the centre of the circle.

A

B v

v

O

C

ω

F

F



Example 5 [N93/I/6] Which of the following statements is correct for a particle moving in a horizontal circle with constant angular velocity? A the linear momentum is constant but the kinetic energy varies B the kinetic energy is constant but the linear momentum varies C both kinetic energy and linear momentum are constant D neither the linear momentum nor the kinetic energy is constant E the speed and the linear velocity are both constant. Solution:

Ans Remarks about statement

A

Incorrect. Linear momentum is a vector and since the direction is changing, the linear momentum varies (i.e. it is NOT constant). Kinetic energy is a scalar. As the object is moving at constant angular velocity, the kinetic energy is constant.

B

Correct statement.

C

Incorrect. KE is constant, but linear momentum (vector) is always changing due to the changing direction of velocity.

D Incorrect. KE is constant.

E

Incorrect. Speed is constant, but velocity (vector) is changing.

e) Recall and use centripetal acceleration a = rω2, a = 2v


Centripetal acceleration is given by [the proof of this equation is not necessary].

Centripetal acceleration, a = 2

2

ωrrv

=

1. The direction of the centripetal acceleration at any instant is in a straight line towards the centre of the circle (circular path).

2. For circular motions at constant speed, the smaller the curvature (radius r) & larger the speed v, the larger the centripetal acceleration, a.

As there is no displacement in the direction of the resultant force (resultant force is always perpendicular to the velocity), there is no work done on the object moving in uniform circular motion. Hence there is no change in the kinetic energy of the object.



Example 6 The maximum speed of the blades on rotary lawn-mowers is restricted to avoid danger from flying stones. If the rate of rotation of the blades in one particular model is 3500 revolutions per minute and the blade has a radius of 0.23 m, find

a) the angular velocity of the blade. b) the linear velocity of the tip of the blade. c) the centripetal acceleration of the tip of the blade.

Solution: a) Blade moves 3500 revolutions per minute.

Hence, time for ONE revolution =

f) Recall and use centripetal force F = mrω2, F = mr

v 2 to solve problems.

From Newton’s 2nd Law, if the mass remains constant, then resultant force, F = ma.

In circular motion, centripetal acceleration, a = 2

2

rr

v ω=

Hence, resultant force towards the centre of circle (centripetal force) is given by:

Centripetal force, F = 22

rmr

mv ω=

Procedure to solve circular motion questions:

Step 1: Identify the object and the centre of the circular motion Step 2: Draw the free body diagram of the object in circular motion Step 3: Resolve all forces and find the resultant force pointing towards the

centre of the circular motion Step 4: equate this resultant force provides the centripetal force.

b) v = rω

c) a = 2v

r or a 2rω=



Example 7 For each object, draw all the relevant forces, and write down the expression for the centripetal force (free body diagrams).

a) of mass m tied to a string, describing a circular path in a vertical plane o

Resultant force to the centre,

F = T - mg

∴ T – mg = r

mv 2

Note: There are two types of circular motion possible in this vertical circle - uniform speed and non-uniform speed. The speeds of objects in a vertical plane are generally not constant (and difficult to maintain at constant speed), hence is not uniform circular motion. This implies the centripetal acceleration cannot be the same magnitude throughout the motion.

b) A stunt driver driving his car inside a vertical circular track.

v

v

r

r

Object at bottom of circle

Object at top of circle

Object at side of circle

mg

T v

v v

Object at bottom of circle

Object at top of circle

Object at side of circle

r

r

v

v



c) A conical pendulum (mass m) describing a circle in a

horizontal plane.

Notice that the horizontal component of the tension will now provide the centripetal force while the vertical component balances the weight of the pendulum. Their vector sum is not zero since there must be a resultant force as the centripetal force.

Solution: Forces towards the centre of circle:

rmvT

2

=θsin

Example 8 The diagram represents a cyclist making a left turn on a rough surface at constant speed, v, as viewed from behind. The total mass of the bicycle and rider is m and their combined centre of gravity is at G. If R is the resultant force of the normal reaction and the frictional force, which vector diagram represents the directions of the forces acting on the bicycle and its rider?

Example 9 A satellite of mass 800 kg is orbiting the Moon in a circular path of radius 1760 km. The pull of the moon on the satellite is 1300 N. Find the speed of the satellite and its period of rotation. Solution: In this example, the only force acting on the satellite is the force by the moon on the satellite, which is already directed towards the centre of the circle (i.e. moon).

rmvF

2

== 1300 ( )800

10176013001300 3×==

mrv = 1690 m s-1

Period of rotation, ⎟⎠⎞

⎜⎝⎛

×

=⎟⎠⎞

⎜⎝⎛

==

31017601690

222 ππωπ

rv

T = 6540 s

Forces in the vertical:

mgT =θcos



Banking of roads The part of the road that is curved is usually tilted at an angle (i.e. banked). Banking is beneficial to the safety of a car moving over a curved road. This is because on a banked curve, the normal force exerted by the road on the car contributes to the centripetal force required for the car to turn. If the tilt angle is just right, the normal force will be able to provide the entire centripetal force. This will allow the car to negotiate the curve even if there is no friction between its tyres and the road. Example 10 If a roadway is banked at the proper angle, a car can round a corner without any assistance from friction between the tyres and the road. Find the appropriate banking angle for a 900 kg car travelling at 20.5 m s-1 in a turn of radius 85.0 m Solution: Resolving forces vertically: Resolving forces horizontally:

tan θ = Example 11 (Airplane taking a bank) An aircraft is travelling at a constant speed of 180 m s-1 in a horizontal circle of radius 20 km.

(a) Find the centripetal acceleration of the aircraft. (b) Draw the free body diagram of the aircraft. (c) Calculate the angle of the bank, θ, of the aircraft.

Solution:

(a) Centripetal acceleration, a = rv 2

= 3

2

20x10180

= 1.6 m s-2

O θ



(b)

(c) Since the plane is in equilibrium in the vertical direction,

Resultant force (vertically), Fy = may = 0 L cos θ = mg -------------------------------(1)

Since plane is performing circular motion, there must be a resultant force towards the centre of the circle (i.e. centripetal force),

Centripetal force, r

mvF2

=

r

mvθL2

=sin ----------------------------(2)

Equations (2) / (1): grvθ

2

=tan

( )( )⎟⎟⎠

⎞⎜⎜⎝

⎛

×=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

3

211

10209.81180tantan

grvθ

2

= 9.4°

Example 12 (The rotating spaceman) A spaceman in training is rotated in a seat at the end of a horizontal rotating arm of length 5 m. If he can withstand accelerations up to 9g, what is the maximum number of revolutions per second permissible? Calculate the period of the rotating arm in maintaining this no. of revolutions. Solution:

Using a = rω2 9g = 5ω2 5

9g=ω = 4.2 rad s-1

Maximum number of revolutions in 1 second = ππ

ω22

2.4= = 0.67

Period of rotation, T = 4.222 π

ωπ= = 1.5 s

O θ

mg

L



Things I have learnt today / Things I found interesting / Questions I still have Reference Textbooks: 1. Comprehensive Physics for ‘A’ level (3rd edition), K.F. Chan, Charles Chew & S.H. Chan.

Federal Study Aids. 2. Physics (2nd edition), Robert Hutchings. Bath Advanced Science. 3. Calculations for A-level Physics (3rd edition), T.L. Lowe & J.F. Rounce. Stanley Thornes

(Publishers) Ltd. 4. Physics (2nd edition), James S. Walker Extra Reading

What is the maximum G force a human can survive? That depends on the duration thereof. If subjected to constant acceleration, it would be doubtful if someone could withstand 9g for more than a few minutes. After a few minutes at that rate, one faints from the lack of blood to the brain, and presumably after that, some damage will occur, like heart failure, brain hemorrhage and so on. But for very short durations, very high accelerations can be supported, although some damage can result. Colonel John Paul Stapp of the US Air Force did several experiments, strapping himself to a rocket sled, and determined that 32g was an acceleration someone could walk away from, which then became the acceleration used in the design of fighter jet seat. Stapp subjected himself to 45g and above, but since he did that so often, ended up with damage to his eyes. The British Formula One racer David Purley crashed in 1977, his car going from 173 km/h (107 mph) to 0 in only 66 cm (two feet) (which means he hit a wall and the car structure compressed to decelerate him). He broke many bones, but survived. This deceleration of 178g is believed to be the highest ever survived by a human being. [Excerpt from http://answers.yahoo.com/question/index?qid=20070101085015AAok4JJ]



Summary

1) A body travelling at constant speed v in a circular path has acceleration towards the

centre of its path. This acceleration is called a centripetal acceleration.

2) The centripetal force is the resultant force acting towards the centre of the circular path. It is perpendicular to the direction of motion (tangential to the circular path), and hence, no work (as defined) is done by the centripetal force.

3) Procedure to solve circular motion questions: Step 1: Identify the object and the centre of the circular motion

Step 2: Draw the free body diagram of the object in circular motion Step 3: Resolve all forces and find the resultant force pointing towards the

centre of the circular motion Step 4: This resultant force provides the centripetal force.

Examples of Applications: 1. Washing Machine

In washing machine, the drums provide the necessary centripetal force (reaction of the drum acting on the clothes) to keep the clothes in circular motion. In the drums there are holes but they cannot provide the centripetal force. When the drum is rotated with high speed, water particles acquire linear speed and the speed is high so that they move in a spiral path. The radius of the circular path is increasing corresponding to the speed. Thus, it is the lack of necessary centripetal force that causes the water particle to move away from the centre through the holes along a tangential path and not along the radius.

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Motion in a Circle - PBworks