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LGB 10203 / ENGINEERING SCIENCE CHAPTER 3 – ROTATIONAL AND CIRCULAR MOTION
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CHAPTER 3 – ROTATIONAL AND CIRCULAR MOTION
A. Uniform Circular Motion
A body is in uniform circular motion if it moves in a circle with constant speed. The magnitude of the velocity remains constant in this case, but the direction of the velocity continuously changes as the object moves around the circle.
The direction of the velocity is tangential to the circle, that is, perpendicular to the radius.
The net force acting upon such an object is directed towards the center of the circle.
Centripetal Acceleration: - Although the speed of a particle in uniform circular motion is
constant, it has acceleration because the direction of the velocity is constantly changing. Since acceleration is the rate of change of velocity, the particles accelerate.
- Centripetal acceleration, a = v ω Or a = r ω²
Or a = 𝒗𝟐
𝒓
Centripetal Force: - When a body is in circular motion, it accelerates towards the
centre of the circle. - According to Newton’s Second Law, F = ma, the body is able to
move in a circle due to the action of a resultant force in the direction of the acceleration, which is towards the centre of the circle.
- Centripetal force, F = ma F = m v ω F = m r ω²
F = 𝒎𝒗𝟐
𝒓
- No work is done by the centripetal force because it is perpendicular to the direction of the displacement.
B. Angular Displacement, Velocity and Acceleration
Angular displacement, θ - When the particle moves from A to B, the radius rotate through a
small angle θ. The angle θ is known as the angular
displacement of the particle in the time interval t. - An object’s angular displacement Δθ is the difference in its final
and initial angles.
- θ = θf – θi ; S.I. unit: radians (rad).
LGB 10203 / ENGINEERING SCIENCE CHAPTER 3 – ROTATIONAL AND CIRCULAR MOTION
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Angular Velocity, ω - The angular velocity ω of the particle is defined as the rate of change of angular
displacement. - Angular velocity is a vector quantity. - The average angular speed (ωav) of a rotating rigid body during the time interval Δt is
defined as the angular displacement Δθ divided by Δt.
- ; S.I.unit: radian per second (rad/s).
C. Angular Motion equations and Tangential Quantities The equations for uniform linear acceleration have rotational analogue which are:
ttt if
if
av
LGB 10203 / ENGINEERING SCIENCE CHAPTER 3 – ROTATIONAL AND CIRCULAR MOTION
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Example 3.1 The initial angular velocity of a body which rotates with uniform angular acceleration is 11 rads-1. After 2.0 s, its angular velocity is 19 rads-1. Calculate
a) The angular acceleration b) The angular displacement after 2.0 s
Solutions;
a) Using 𝝎𝟏 = 𝝎𝟎 + α 𝒕
α = 𝝎𝟏−𝝎𝟎
𝒕 = α =
𝟏𝟗 −𝟏𝟏
𝟐 rad 𝒔−𝟐
= 4.0 rad s-2 b) Using θ = 𝝎𝟎𝒕 +
𝟏
𝟐 α 𝒕𝟐
θ = (11)(2)+ 𝟏
𝟐 (4) (𝟐)𝟐
= 30 rad
D. Rotational Work and Kinetic Energy The kinetic energy of a rotating object is analogous to linear kinetic energy and can be
expressed in terms of the moment of inertia and angular velocity. The total kinetic energy of an extended object can be expressed as the sum of the
translational kinetic energy of the center of mass and the rotational kinetic energy about the center of mass. For a given fixed axis of rotation, the rotational kinetic energy can be expressed in the form
The expressions for rotational and linear kinetic energy can be developed in a parallel manner from the work-energy principle.
Consider the following parallel between a constant torque exerted on a flywheel with moment of inertia I and a constant force exerted on a mass m, both starting from rest.
LGB 10203 / ENGINEERING SCIENCE CHAPTER 3 – ROTATIONAL AND CIRCULAR MOTION
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The speed of the other particles in the rigid body are 𝒗 = 𝒓𝝎
Rotational kinetic energy of the rigid body = sum of the kinetic energy of n particles
= 1
2𝑚1𝑣1
2 + 1
2𝑚2𝑣2
2 + 1
2𝑚3𝑣3
2 + 1
2𝑚4𝑣4
2 + … … . +1
2𝑚𝑛𝑣𝑛
2
= 1
2𝑚1𝑟1
2𝜔2 + 1
2𝑚2𝑟2
2𝜔2 + 1
2𝑚3𝑟3
2𝜔2 + 1
2𝑚4𝑟4
2𝜔2 + … … . +1
2𝑚𝑛𝑟𝑛
2𝜔2
= 1
2(𝑚1𝑟1
2 + 𝑚2𝑟22 + 𝑚3𝑟3
2 + 𝑚4𝑟42 + … … . +𝑚𝑛𝑟𝑛
2)𝜔2
= 𝟏
𝟐𝑰𝝎𝟐 (I =mr2 ; v=rω)
When a cylinder of mass m rolls on a horizontal surface, it has both translational kinetic
energy and rotational kinetic energy.
Total kinetic energy = translational kinetic energy + rotational kinetic energy
= 1
2𝑚𝑣2 +
1
2𝐼𝜔2 (I =
1
2 mr2 , ω =
v
r)
= 1
2𝑚𝑣2 +
1
2(
1
2 mr2 ) (
𝑣
𝑟)2
=𝟑
𝟒 𝒎𝒗𝟐
Example 3.2 A solid cylinder initially rolls on a horizontal table climbs up an n inclined plane. If the horizontal speed is 5ms-1, the angle of inclination is 370 and ignoring friction, determines the maximum distance made by the cylinder up the inclined plane.
ω
v = rω
r
370
h s
v
ω
LGB 10203 / ENGINEERING SCIENCE CHAPTER 3 – ROTATIONAL AND CIRCULAR MOTION
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Solutions: Apply conservation of mechanical energy to initial and final positions. Total mechanical energy = KE (linear) + KE (rotation) + PE
Initial total energy, 𝟏
𝟐𝒎𝒗𝟐 +
𝟏
𝟐𝑰𝝎𝟐 + 𝟎
Final total energy, 0 + 0 + mgh Using conservation of energy,
0 + 0 + mgh = 𝟏
𝟐𝒎𝒗𝟐 +
𝟏
𝟐𝑰𝝎𝟐 + 𝟎
(𝑰 =𝟏
𝟐 𝒎𝒓𝟐 , 𝝎 =
𝒗
𝒓)
mgh = 𝟏
𝟐𝒎𝒗𝟐 +
𝟏
𝟐(
𝟏
𝟐 𝒎𝒓𝟐 ) (
𝒗
𝒓)𝟐
h = 𝟑𝒗𝟐
𝟒𝒈 =
𝟑 𝒙 𝟓𝟐
𝟒 𝒙 𝟗.𝟖𝟏
= 1.91 m
Distance along the plane, s = 𝒉
𝒔𝒊𝒏 𝜽 =
𝟏.𝟗𝟏
𝟎.𝟔
= 3.19 m
E. Rotational Inertia In linear motion, the inertia of a body is its resistance to change from its state of rest or
motion. The quantity in rotational motion that is analogous to inertia in linear motion is moment of
inertia. The moment of inertia of a rigid body is its resistance to change from its state of rest or rotational motion.
The moment of inertia of a rigid body about an axis of rotation is defined as the sum of the products of the mass and the square of the distance from the axis of rotation of particles that make up the rigid body.
Moment of Inertia, I = (𝑚1𝑟12 + 𝑚2𝑟2
2 + 𝑚3𝑟32 + 𝑚4𝑟4
2 + … … . +𝑚𝑛𝑟𝑛2)
= ∑ 𝒎𝒊𝒓𝒊𝟐𝒏
𝒊=𝟏
Factors that affect the moment of inertia of a rigid body are: - Mass of the body - Distribution of the mass or shape of the body - The position of the axis of rotation
LGB 10203 / ENGINEERING SCIENCE CHAPTER 3 – ROTATIONAL AND CIRCULAR MOTION
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F. Newton’s Law of Gravity Newton’s proposed his law of universal gravitation, which we can state as follows:
“Every particle in the universe attracts every other particle with a force that is proportional to the product of their masses and inversely proportional to the
square of the distance between them. This force acts along the line joining the two particles.”
The magnitude of the gravitational force, F can be written as
Where m1 and m2 are the masses of the two particles, r is the distance between them, and G is a universal constant which must be measured experimentally and has the same numerical value for all objects.
The accepted value today is G = 6.67 x 10-11 N m2 / kg2
Example 3.3 Determine the net force on the Moon (mM = 7.35 x 1022kg) due to the gravitational attraction of both the Earth (mE = 5.98 x 1024kg) and the Sun (mS = 1.99 x 1030 kg). Assuming they are at right angles to each other as in figure below. (Given; rME = 3.84 x 105 km, rMS = 1.50 x 108 km)
LGB 10203 / ENGINEERING SCIENCE CHAPTER 3 – ROTATIONAL AND CIRCULAR MOTION
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G. Weight and Gravitational Force
Weight is the gravitational force acting on a body mass. Transforming Newton's Second Law related to the weight as a force due to gravity can be expressed as
W = m g
where
w = weight
m = mass (kg)
g = acceleration of gravity (m/s2)
LGB 10203 / ENGINEERING SCIENCE CHAPTER 3 – ROTATIONAL AND CIRCULAR MOTION
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The handling of mass and weight depends on the systems of units that is used. The most common systems of units are the - International System – SI - British Gravitational System – BG - English Engineering System - EE
H. Kepler’s Law
More than a half century before Newton proposed his three laws of motion and his law of
universal gravitation, the German astronomer Johannes Kepler had worked out a detailed
description of the motion of the planets about the Sun and now we refer to as Kepler’s laws of
planetary motion.
Kepler’s Law of planetary motion:
1. Kepler’s First Law: the path of each planet about the Sun is an ellipse with the Sun at one
focus.
2. Kepler’s Second Law: Each planet moves so that an imaginary line drawn from the Sun to
the planet sweeps out equal areas in equal periods of time.
3. Kepler’s Third Law: The ratio of the squares of the periods T of any two planets revolving
about the Sun is equal to the ration of the cubes of their mean distances s from the Sun.
LGB 10203 / ENGINEERING SCIENCE CHAPTER 3 – ROTATIONAL AND CIRCULAR MOTION
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