Section II 7 Motion in a Circle

7/28/2019 Section II 7 Motion in a Circle

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Motion in a circle

7. Motion in a Circle Content

7.1 Kinematics of uniform circular motion

7.2 Centripetal acceleration

7.3 Centripetal force

Learning Outcomes

(a) express angular displacement in radians.

(b) understand and use the concept of angular velocity to solve

problems.

(c) recall and use v = r 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 = r2 , a = v2/r.

(f) recall and use centripetal force F = m r2 , F = mv2/r.


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Linear Quantities

We began our study of linear motion by describing where something

was relative to an origin.


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Rotational Quantities

In a similar way, for rotational motion, we can describe how far

something has rotated by telling how far it has rotated relative to an

origin or a reference line.

One degree by tradition equal to the angle of a circle divided by 360

One radian (rad) is defined as the angle subtended at the centre of a

circle by an arc equal to the radius i.e = length of arc(s)/radius of

circle(r) , s = r


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Angular Velocity

When we studied linear motion, we immediately asked how fast an

object is moving.

Likewise, we can now ask how fast something is rotating.

We call this the angular velocity and use the Greek letter omega to

indicate it

Angular speed , is defined as the angle swept out by the radius per

second.

Angular velocity is the angular speed in a given direction

unit: rad/s


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Angular Acceleration

In linear motion, how fast linear velocity is changing is the linear

acceleration, a

Likewise, how fast angular velocity is changing is called the angular

acceleration and we use the Greek letter alpha to indicate it

unit: rad/s


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Rotational Kinematics

We have now defined angle, angular velocity, and angular acceleration

exactly in the same way that we defined their linear analogues,

displacement, velocity, and acceleration:

Unit: rad/s

Unit: rad/s

Unit: m/s

Unit: m/s


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Angular Kinematic Equation

When we studied linear motion, we derived the "big three Kinematics

Equations":

Linear Kinematic

Equation

Angular Kinematic

Equation


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Linear vs Angular quantities

v = s/t = r/t = r since s = r, and dividing both sides by t

a = v/t = r/t = r

a = v/r = r, (we shall derive this a little later)


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Example

The drum of a spin dryer has a radius of 20 cm and rotates at 600revolutions per minute (rpm).

a) show that the angular speed of the drum is about 63 rad/s

b) calculate for a point on the edge of the drumi) its linear speed

ii) its acceleration towards the centre of the drum

Solution

a) 600 rpm is 10 revolutions per second.

The time for 1 rev is 0.10 s.Each rev is 2 rad, so the angular speed = /t = 2/0.10 = 63 rad/s

b) i) v = r =0.20 x 63 = 12.6 m/s

ii) a = v2/r = (12.6)2/0.20 = 800 m/s


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Torque

A rotational force, also known as a torque, depends upon the force and

where that force is applied

torque = moment arm x force.

Moment arm is defined as the perpendicular distance from the axis of

rotation to the line of action of the force

=(rsin)F = rFsin

SI unit for torque is m.N


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Moment Arm


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?

What can have a steady speed but a changing velocity?

What accelerates towards something, yet never gets closer to it?

Does velocity need to get bigger or smaller in order to change?

If the speed is constant, does it always mean that there is noacceleration?


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Circular Motion

An object moves in a straight line if the net force on it acts in the

direction of motion, or the net force is zero.

If a net force acts at an angle to the direction of motion, then the object

will move in a curved path.

If a net force always acts at an angle perpendicular to the velocity, then

the object will move in a circular motion.


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Uniform Circular Motion

Uniform circular motion is the motion of an object moving in a perfect

circle with a constant or uniform speed.

V


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Uniform circular motion is one of many forms of rotational motion which

moves at a constant speed.

An object moving in uniform circular motion would cover the same linear

distance in each second of time.

When moving in a circle, an object traverses a distance around the perimeter

of the circle.


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If your car were to move in a circle with a constant speed of 5 m/s, then

the car would travel 5 meters along the perimeter of the circle in each

second of time.

R - Radius

T - Period


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Period & Frequency


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Speed vs Velocity

Objects moving with uniform circular motion will have a constant speed.

But does this mean that they will have a constant velocity?

Speed and velocity refer to two distinctly different quantities. Speed is a

scalar quantity and velocity is a vector quantity.

Velocity, being a vector, has both a magnitude and a direction.


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Direction of Velocity

The magnitude of the velocity vector is the

instantaneous speed of the object.

The direction of the velocity vector at any

instant is in the direction of a tangent line

drawn to the circle at the object's location


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Direction of Velocity

Tangent - VA is 90 with OA Tangent - VB is 90 with OB


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Circular Motion Acceleration

An object moving in uniform circular motion is moving in a circle with

uniform or constant speed.

The magnitude of the velocity remains constant but the direction of the

velocity continuously changes as the object moves in a circle.

A change in direction of velocity constitutes an acceleration just as the

change of magnitude of velocity does.


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Circular Motion Acceleration

For this reason, it can be safely concluded that an object moving in a circle

at constant speed is indeed accelerating.

It is accelerating because the direction of the velocity vector is changing.

This acceleration is known as Centripetal Acceleration or Radial

Acceleration

What is the magnitude and direction?

Given, definition of a = v/t


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Direction of

Centripetal Acceleration


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Magnitude of

Centripetal Acceleration


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Centripetal Force

An object moving in a circle experiences an acceleration ie a = v2/r = 2r

According to Newton's second law of motion, an object which experiences

an acceleration must also be experiencing a net force. Hence F = mv2/r

F is directly proportional to the mass and velocity squared and

inversely proportional to the radius

The direction of the net force is in the same direction as the acceleration

which is directed towards the center of the circle.

If the string holding a ball in circular motion breaks, the force is removed

and the ball does exactly what Newton's 1st law predicts

The ball is not flung outwards

The force needed to keep an object in a circular path depends on the mass,speed and radius


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Centripetal Force

This net force is known as Centripetal Force

Centripetal force is not a new type a force.

The word centripetal means center-seeking whereby the force points

toward the center of the circle.

The centripetal force may be due to gravity, the tension in a string etc.


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Examples of Circular Motion

Centripetal force in the following scenarios.

Car Making a Turn A bucket spun in a circle Moon Orbiting the Earth


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Vertical Circular Motion


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Vertical Circular Motion


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Example

An aircraft in a display team makes a turn in a horizontal circle of

radius 500 m. It is travelling at a speed of 100 m/s.

Calculate a) the cetripetal acceleration of the aircraft

b) the centripetal force acting on the pilot if the mass of thepilot is 80 kg.

Solution

a) a = v2/r = 20 m/s

b) F = ma = 1600 N


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Example 1

A 900 kg car moving at 10 m/s takes a turn around a circle with a radius

of 25.0 m. Determine the acceleration and the net force acting upon the

car.

(Answer: a = 4 m/s, Fnet = 3600 N)


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Example 2

A 95 kg half-back makes a turn on the football field. The half-back

sweeps out a path which is a portion of a circle with a radius of 12

meters. The half-back makes a quarter of a turn around the circle in 2.1

seconds. Determine the speed, acceleration and net force acting upon

the half-back.

(Answer: v = 8.97 m/s, a = 6.71 m/s, Fnet = 637 N)


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Example 3

A Lincoln Continental and a Yugo are making a turn. The Lincoln is

four times more massive than the Yugo. If they make the turn at the

same speed, then how do the centripetal forces acting upon the two cars

compare? Explain.


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Example 4

Determine the centripetal force acting upon a 40 kg child who makes 10

revolutions around the Cliffhanger in 29.3 seconds. The radius of the

barrel is 2.90 meters.

(Answer: Fnet = 533 N)


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Banked Curved


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Banked Curve

Without friction, the roadway still exerts a normal force n perpendicular to

its surface.

And the downward force of the weight w = mg is present.

Those two forces add as vectors to provide a resultant or net force Fnet

which points towards the center of the circle; this is the centripetal force.


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Banked Curve

This gives the angle necessary for a banked curve that will allow a car to travel

in a curve of radius r with constant speed v and require no friction force.

All of the force required for the circular motion is provided by the horizontal

component of the normal force

A banked curve is designed for one specific speed.

If the banked curve is icy so there is no frictional force at all then travelling at

higher than design speed means the car will slide out, up, and over the edge

Travelling at lower than design speed means the car will slide in, down, and off

the bank.


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Banked Curve


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Derivation

Referring to the last figure, because there is no vertical motion of the

vehicle, the vertical component of the normal force must be equal and

opposite to the weight of the vehicle

i.e n cos

= mg But the force providing the centripetal acceleration is given by

Fc = mac = mv2/r

All this force is now provided by the horizontal component of the

normal force

n sin = mv2/r

but tan = n sin /n cos = mv2/rmg = v2/g

hence tan = v2/g


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Example

A bend of radius 300 m is to be constructed on a freeway where the

maximum speed is 100 kph. Determine the banking angle necessary

such that for a car travelling at 90 kph there is no reliance on friction to

provide the force necessary for the centripetal acceleration.

Solution

v = 90 kph = 25 ms-1

tan = v2/rg = 0.2129

= 12.0


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Exercise

The Daytona International Speedway at Daytona Beach Florida, USA

has bends of radius 305 m which are banked at an angle of 31.

1) At what speed should a NASCAR racing car travel around these

bends so that there is no reliance on frictional force to provide theforce necessary to cause the centripetal acceleration? Give your answer

in kph

2) Racing cars often travel around these bends in the vicinity of 250

kph (70 m/s). If the mass of such a car is 800 kg, what is the frictional

force between the tyres and the track if it is travelling around the bend

at 70 m/s?


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Solution

1) tan = v2/g, therefore v = 153 kph

2) force required to provide the centripetal force is

Fc = mac = mv2/r = 12,852 N

The vertical component of the normal force is balanced out bythe weight of the car since, n cos = mg

The horizontal component of the normal force is

n sin

= mg/cos

x sin

= mg tan

=4711 N

Therefore the frictional force FFrrequired is

FFr= 12852 4711 = 8140 N

Documents

Section II 7 Motion in a Circle