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Basic Engineering

Mechanics B

Semester 2 (2015/2016)

F40 BMB

Circular Motion I

Lesson 1

(Reference1: Chapter 5, page 119-122)


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1.1 Uniform Circular Motion

Understand and apply circular motion tovarious problems.

Objectives:

Outline


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F40 BMB, Semester 2, Mr. Tan

The acceleration is always directed radially inward. Therefore, it iscalled acentripetal(meaning center seeking)acceleration.

A particle that travels around a circle or a circular arcat constant (uniform) speed v, is in uniform circularmotion.

Uniform Circular Motion

The magnitude of this accelerationais,

rva

2

= (centripetal acceleration)

whereris the radius of the circle andvis the speed of the particle.

10.1


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Tis called theperiod of revolution, or simply theperiod, of the motion.It is the time for a particle to go around a closed path exactly once.

In addition, the particle travels the circumference of the circle(a distance of 2r) in time,

v

r

T

2=

(period).

Another important quantity that measure the number of revolution persecondisangular speed(angular frequency),.

In addition, ,

where f, frequency, is the number of complete revolutions (rev) persecond, that is, 1 rev/s = 1 Hz.

Tf 1=

The unit for angular speed is radian per second (rad/s),T

2=


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Figures show an object is in uniform circular motion when at timetoit was at pointO, and at a later timetat pointP.

Proof of Centripetal Acceleration

A careful observation & investigation at thechange in the velocity vector shows that theacceleration is towards thecenter of the circle.


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So for the two triangles, we have:

r

tv

v

v

=

Thus, the centripetal acceleration is given by:

In the limit thattis very small, the arc length OPis approximately a straight line whose length

distance is the distancevttraveled by the object.

rv

tvac

2

=

=

The direction is toward the center of the circle.

Proof of Centripetal Acceleration


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The proof starts from the velocity vector.

Another method to proof Centripetal Acceleration

Finally, through time derivative ofv, obtained:

jvivjvivv yx )cos()sin( +=+=r

jr

vi

r

va sincos

22

+

=

r

where, we have:

r

y

=sin r

x

=cosand


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Example 1

Solutions:

The wheel of a vehicle has a radius of 28 cm and is being rotated at 830revolutions per minute (rpm) on a tire-balancing machine. Determine the

speed, in m/s, at which the outer edge of the wheel is moving. What isthe centripetal acceleration of the point?

Ans: 24.34 m/s;

2116 m/s2.



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Example 2

Solutions:

A grinding wheel has a diameter of 0.36 m. A point on the edge of thegrinding wheel rotates at 2500 rpm.

Ans: 0.024 s; 1.23 104

m/s2

.


(b) What is the magnitude of the points acceleration in m/s2?

(a) What is the period of the grinding wheels rotation?


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Example 3

Solutions:

The bobsleigh track at the 1994 Winter Olympics inLillehammer, Norway, contained turns with radii of

33.0 m and 24.0 m. Find the centripetal accelerationat each turn for a speed of 34.0 m/s, a speed that wasachieved in the two-man event. Express the answeras multiples ofg= 9.80 m/s2.

Ans: 35.0 m/s2

; 48.2 m/s2

; 3.57g; 4.92g.



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Example 4

Solutions:

(a) How fast are we moving now since we are riding the Earth as itorbits the Sun?

(a)

(b) What is our centripetal acceleration?

(b)

Ans: 2.99 104 m/s.

Ans: 5.96 103 m/s2.


Given the distance of the Earth to the Sun is 150 millions km and it takesa year (365 days) to complete one orbit.

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