ROTATIONAL MOTION Y. Edi Gunanto. Gerak Rotasi Dalam Fisika dibedakan 2 jenis gerak benda : 1....

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Transcript of ROTATIONAL MOTION Y. Edi Gunanto. Gerak Rotasi Dalam Fisika dibedakan 2 jenis gerak benda : 1....
ROTATIONAL MOTIONROTATIONAL MOTION
Y. Edi GunantoY. Edi Gunanto
Gerak Rotasi Dalam Fisika dibedakan 2 jenis gerak benda :
1. Gerak Translasi :
the motion of an object through space. (change of location)
2. Gerak Rotasi :
the spinning of an object around an axis. (change of orientation)
Objects may have purely translational motion, purely rotational motion, or both.
A rigid body (benda tegar) is composed of particles in fixed positions.
In a rigid body undergoing purely rotational motion, all points of the body move in circles centered around a line called the axis of rotation.
Rotational Motion
Key to learning rotational motion:
THE CONCEPTS FROM TRANSLATIONAL MECHANICS (E.G. VELOCITY, ACCELERATION, FORCE
ENERGY, & MOMENTUM) HAVE ANALOGOUS ROTATIONAL
CONCEPTS.
Rotational Kinematics
Angular displacement (Angular displacement ()) is the rotational is the rotational analog of displacement (∆x).analog of displacement (∆x).
Angular displacements are measured in radians.
= Arc length/radius
= r/r = 1 radian
r
r
= 1 rad= 1 radFor one complete For one complete revolution:revolution: = 2= 2r/r = 2r/r = 2 radians radiansSo, 2So, 2 rad = 360° rad = 360°
l = R
Rotational Kinematics
Answers
180
/2
57.3
20
Find the following:
rad = ___ degrees
90° = ___ rad
1 rad = ___ degrees
10 rev = ___ rad
Rotational KinematicsRotational Kinematics
Average angular velocity (Average angular velocity (avgavg) ) is the is the rotational analog of velocity (v).rotational analog of velocity (v).
Velocity = Displacement / TimeVelocity = Displacement / Time
v = ∆x / tv = ∆x / t
Ang. velocity = Ang. displacement / TimeAng. velocity = Ang. displacement / Time
avgavg = = / t / t
Rotational KinematicsRotational Kinematics
What is the angular velocity of the points What is the angular velocity of the points on a record rotating at 45 rpm?on a record rotating at 45 rpm?
= 45 rev = 45 rev xx (2 (2 rad / 1 rev) = 90 rad / 1 rev) = 90 rad rad
= = / t / t
= 90= 90 rad / 60 s rad / 60 s
= 3= 3/2 rad/s/2 rad/s
Rotational KinematicsRotational Kinematics
Angular acceleration (Angular acceleration ()) is the rotational is the rotational analog of acceleration (a).analog of acceleration (a).
Acceleration = Change in velocity / TimeAcceleration = Change in velocity / Time
a = ∆v / ta = ∆v / t
Ang. accel. = Change in Ang.veloc. / Ang. accel. = Change in Ang.veloc. / TimeTime
= = / t / t
Rotational Kinematics
A record begins spinning from rest and reaches a rotation of 45 rpm in three
seconds. What is the angular acceleration of the points on the record?
= / t
= (3/2 rad/s  0) / 3 s
= /2 rad/s2
Rotational Kinematics
Rotational Motion = o + t
= (1/2)(o + )t
avg = (1/2)(o + )
= ot + (1/2)t2
2 = o2 + 2
Linear Motion
v = vo + at
∆x = (1/2)(vo + v)t
vavg = (1/2)(vo + v)
∆x = vot + (1/2)at2
v2 = vo2 + 2a∆x
Problem 1
Given: ω = 27.5 rad/s at t = 0
α = 10 rad/s2
Line PQ lies on the disc along the x axis
Find:
a) ω at t = 0.30 s?
b) Angle PQ makes with x – axis at this time?
EXAMPLE EXAMPLE 11
Part a Solution:
Use: ωz = ωiz + αzt for t = 0.3 s
ωz = (27.5 rad/s) + (10 rad/s2)(0.3s)
= 24.5 rad/s
In addition to stated givens we know
Θi= 0° at t = 0s
Given: ω = 27.5 rad/s at t = 0
α = 10 rad/s2
Line PG lies on the disc
along the x axis
Part b
Part b Solution:
Use: Θ = Θi + ωizt + ½ αzt2 for t = 0.3 s
Θ = 0 + (27.5 rad/s)(0.3s) + ½ (10 rad/s2)(0.3s)2 = 7.8 rad
# revolutions?: 7.8 rad (1rev/2π rad) = 1.24 rev
The disc has gone through one revolution plus 0.24 into the next. So the angle at that time with the x –axis is:
0.24 (360°/rev) = 87°
In addition to stated givens we know
Θi= 0° at t = 0s
Rolling Motion (without slipping)
Rolling without slipping depends on static friction between the wheel, ball, etc. and the ground.
The friction is static because the rolling object’s point of contact with the ground is at rest at each moment.
Kinetic friction comes in if the object skids, that is, slides.
VERY IMPORTANT CONCEPT!
Rolling Without Slipping Involves Both Rotation and Translation.
Static friction: wheel does not slide
Reference frame attached to ground
Center of mass (CM)
Reference frame attached to wheel
Reference frame attached to the ground
Reference frame attached to the wheel
Torque
Torque (t ) deals with the dynamics, or causes of rotational motion.
Dynamics in linear motion dealt with force. In rotational motion TORQUE is
analogous to FORCE in linear motion. That is, force through a distance
F
Definition of torque (magnitude)*
RF = RFsin
* Torque is a vector.
R x F = RFsin
These two methods of calculation are completely equivalent.
Example 2
Torque on a compound wheel.
Two thin cylindrical wheels, of radii R1 = 30 cm and R2 = 50 cm, are attached to each other. Calculate the net torque on the twowheel system due to the two forces shown, each of magnitude 50 N.
F2
Rotational Dynamics; Torque and Rotational Dynamics; Torque and Rotational InertiaRotational Inertia
Linear Acceleration: Linear Acceleration: directlydirectly proportional to proportional to forceforce.. inversely inversely proportional to proportional to inertiainertia..
Angular Acceleration:Angular Acceleration: directlydirectly proportional to proportional to torquetorque.. inverselyinversely proportional to the proportional to the moment of moment of
inertiainertia..
Torque and Force
= I
F = ma
Moment of inertia
Example 3
Two Weights on a Bar: Different Axis, Different I.
Two small “weights” of mass 5.0 kg and 7.0 kg are mounted 4.0 m apart on a light rod (whose mass can be ignored). Calculate the moment of inertia of the system (a) when rotated about an axis half way between the weights, and (b) when the system rotates about an axis 0.50 m to the left of the 5.0 kg mass.
Example 4
A heavy pulley.
A 15.0N force (represented by FT) is applied to a cord wrapped around a pulley of mass M = 4.0 kg and radius Ro = 33.0 cm. The pulley is observed to accelerate uniformly from rest to reach an angular speed of 30.0 rad/s. If there is a frictional torque (at the axle), t = 1.10 mN, determine the moment of inertia of the pulley. The pulley is assumed to rotate about its center.
Example 5
Rotating rod.
A uniform rod of mass M and length l can pivot freely (i.e., we ignore friction) about a hinge or pin attached to the case of a large machine. The rod is held horizontally and then released. At the moment of release determine (a) the angular acceleration of the rod (b) the linear acceleration of the tip of the rod. Assume the force of gravity acts at the CM of the rod.
Determining Moments of Inertia
The are two methods of determining moments of inertia: By experiment (See Example 4) By calculus
Using Calculus
For many bodies, or systems of particles, the moment of inertia can be calculated directly, as in Example 3.
Many bodies can be considered as a continuous distribution of mass. In this case we can write:
I = R2 dm
where dm represents the mass of an infinitesimal particle of the body and R is the perpendicular distance of the particle from the axis of rotation.
EXAMPLE 6
Show that the moment of inertia of a uniform rod length L :
a. throught the centre of mass
b. throught the end of rod
The ParallelAxis and The ParallelAxis and PerpendicularAxis TheoremsPerpendicularAxis Theorems
ParallelAxis Theorem:ParallelAxis Theorem: if the moment of if the moment of inertia inertia II is the moment of inertia of a body of is the moment of inertia of a body of total mass total mass MM about any axis, and about any axis, and IICMCM is the is the
moment of inertia about an axis passing moment of inertia about an axis passing through the center of mass and parallel to the through the center of mass and parallel to the first axis but a distance first axis but a distance hh away, then away, then
I = ICM + Mh2
Angular Momentum and Its Angular Momentum and Its ConservationConservation
Linear momentum: Linear momentum: p = p = mmvv
Angular momentum: Angular momentum: L = L = IIww
The SI units for The SI units for LL are kg.m are kg.m22/s./s.
Angular and Linear Angular and Linear MomentumMomentum
= I = dL dt
F = ma =dp dt
Conservation of Angular Conservation of Angular MomentumMomentum
Law of Conservation of Angular Momentum
The total angular momentum of a rotating body remains constant if the net external torque acting on it is zero.
When there is zero net torque acting on a body, and the body is rotating about a fixed axis or about an axis through its CM such that the direction doesn’t change, we can write
Iw = Iowo = constant
When the skater pulls her arms in, she is decreasing her moment of inertia.
Example 7
Clutch design. You are designing a clutch assembly for a piece of machinery. The clutch assembly consists of two cylindrical plates, of mass MA = 6.0 kg and MB = 9.0 kg, with equal radii Ro = 0.60 m. They are initially separated. Plate MA is accelerated from rest to an angular velocity w1 = 7.2 rad/s in a time Δt = 2.0 s. Calculate (a) the angular momentum of MA, and (b) the torque required to have accelerated MA from rest to w1.
Clutch design (cont.).
Next, plate MB, initially at rest but free to rotate without friction, is allowed to fall vertically so it is in firm contact with plate MA. Before contact, MA was rotating at constant w1, with no friction or other torque exerted on it. Upon contact, the two plates both rotate at a constant angular velocity w2, which is considerably less than w1. (c) Why does this happen, and what is w2?
Rotational Kinetic Rotational Kinetic EnergyEnergy
Linear Kinetic Energy: Linear Kinetic Energy: KK = ½ = ½ mvmv22
Rotational Kinetic Energy: Rotational Kinetic Energy: K K = ½ = ½ IwIw22
WorkWork
Linear work: W = ½ mv22 – ½ mv1
2 = ΔK
Rotational Work: Wrot = ½ Iw22 – ½ Iw1
2 = ΔKrot
Example 8
Rotating rod.
A rod of mass M is pivoted on a frictionless hinge at one end. The rod is held at rest horizontally and then released. Determine the angular velocity of the rod when it reaches the vertical position, and the speed of the rod’s tip at this moment.
Rotational Plus Translational Motion
The EquationsThe Equations
Kinetic Energy = KCM + Krot
K = ½ MvCM + ½ ICMw22
Example 9
Sphere rolling down an incline.
What will be the speed of a solid sphere of mass M and radius Ro when it reaches the bottom of an incline if it starts from rest at a vertical height H and rolls without slipping? Compare to the result of sliding down the incline.
TERIMA KASIH