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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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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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Uniform Circular Motion
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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Uniform Circular Motion
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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Uniform Circular Motion
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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F40 BMB, Semester 2, Mr. Tan
Uniform Circular Motion
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.
Uniform Circular Motion
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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
.
Uniform Circular Motion
(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.
Uniform Circular Motion
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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.
Uniform Circular Motion
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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