Rotational Motion Hope you learned polar coordinates
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- Slide 1
- Rotational Motion Hope you learned polar coordinates
- Slide 2
- Rotation The motion of these objects cannot be easily described
by using linear equations of motion. These involve a body that
rotates about an axis that is stationary in some inertial frame of
reference.
- Slide 3
- Purely Rotational Motion No Translational Motion Rotation of a
rigid body about an axis of rotation Rigid body object with
definite shape. It will not deform. The tire is not exactly a rigid
body but work with me Axis of Rotation The line that passes through
a point where all other points move around in circles
- Slide 4
- Purely Rotational Motion
- Slide 5
- Velocity of a point Rotating at 500 rpm 0.5m diameter tire P is
on the rim
- Slide 6
- Velocity of a Point
- Slide 7
- Velocity of a point Rotating at 500 rpm 0.5m diameter tire Q
halfway to the rim
- Slide 8
- Velocity of a Point
- Slide 9
- Angular Quantities Different points on a rotating rigid body
have different linear velocities This makes calculations clunky
Better to use angular velocities
- Slide 10
- Linear Displacement
- Slide 11
- Angular Displacement
- Slide 12
- Angle measured in radians 1 Radian angle subtended by an arc
whose length is equal to the radius
- Slide 13
- Angular Velocity
- Slide 14
- Slide 15
- Instantaneous angular velocity
- Slide 16
- Angular Velocity Convention: Counterclockwise rotation is
positive Clockwise rotation is negative All points in a rotating
object have the same angular velocity !
- Slide 17
- Angular Velocity Define f as the frequency number of complete
revolutions per second Period T time to complete one
revolution
- Slide 18
- Angular Velocity Vector Direction of given by right hand rule
Useful for when the axis of rotation changes
- Slide 19
- Angular Acceleration 500 rpm at time t 0 600 rpm at time t
- Slide 20
- Angular Acceleration Instantaneous acceleration Convention: All
points in a rotating object have the same angular acceleration
!
- Slide 21
- Angular Acceleration Vector Direction of given by right hand
rule Useful for when the axis of rotation changes If and are in the
same direction, objects rotation is speeding up. If in opposite
direction its slowing down
- Slide 22
- Relationship of linear and angular Note: there is still a
radial component of the acceleration
- Slide 23
- Example You are asked to design an airplane propeller to turn
at 2400 rpm. The forward airspeed of the plane is to be 75.0 m/s,
and the speed of the tips of the propeller blades through the air
must not exceed 270 m/s (a) What is the maximum radius the
propeller can have? (b) With this radius, what is the acceleration
of the propeller tip
- Slide 24
- Example v fastest at the tip
- Slide 25
- Example
- Slide 26
- Kinematics of Angular Motion If is constant
- Slide 27
- Kinematics of Angular Motion If is constant
- Slide 28
- Energy in Rotational Motion A rotating rigid body, even with no
translational motion, has kinetic energy Imagine rigid body is made
up of many small particles Where r i is perpendicular distance to
axis of rotation
- Slide 29
- Energy in Rotational Motion Total kinetic energy is the sum of
all the KE of the particles is the same for all particles
- Slide 30
- Moment of Inertia Moment of Inertia is the rotational analogue
of mass It is dependent on the distribution of the mass and the
distance from the axis of rotation
- Slide 31
- Example An engineer is designing a machine part consisting of
three heavy disks linked by lightweight struts. (a) What is the
moment of inertia of this body about an axis through the center of
disk A? (b) What is the moment of inertia about an axis through the
center of disks B and C? (c) If the body rotates about an axis
through A perpendicular to the plane of the diagram with angular
speed =4.0 rad/s, what is its kinetic energy?
- Slide 32
- Example (a) (b)
- Slide 33
- Example (c) Moment of inertia is not an intrinsic property, it
depends on the axis of rotation
- Slide 34
- Moment of Inertia Table
- Slide 35
- Parallel-Axis Theorem Let the moment of inertia about an axis
that passes through the center of mass be Then the moment of
inertia through an axis parallel to the original axis, but
displaced by some distance d is
- Slide 36
- Proof
- Slide 37
- Example A uniform rigid rod has a mass of 3.6kg and length
0.30m. Find the moment of inertia perpendicular through one
end?
- Slide 38
- Example
- Slide 39
- Brief Review AngularLinear Acceleration a Velocity v
Displacement x Equations (constant acceleration) Kinetic Energy For
a point at distance R away from the axis of rotation x=R v=R
a=R
- Slide 40
- Dynamics of Rotational Motion Torque, Angular Acceleration and
Rotation of Rigid Bodies
- Slide 41
- Rotation of Rigid Bodies Suppose you have a rigid body that is
pinned at point C Force F is applied to point P The object will
begin to rotate about the axis of rotation C ( to the screen)
- Slide 42
- Torque Torque is the quantitative measure of the tendency of
the force to change the rotational motion of a body. Rotational
analog of force. Represented by the symbol Define: Torque is
positive for counter-clockwise rotation and negative for clockwise
rotation Vector form Magnitude Has SI unit N m eq(1)
- Slide 43
- Torque Simple Example To loosen a bolt at point O, a force F is
applied on the wrench at point P. Calculate for the magnitude and
direction of . Two Solutions to = rF tan = F l From right hand rule
direction is outwards to the screen F tan = F sin = Force
perpendicular to r l = r sin = lever arm
- Slide 44
- Torque Practical Application
- Slide 45
- Torque The radial component of the force, F Rad, has no effect
on the rotation of the object. The line of action of the force must
not pass through the axis of rotation (sin 0). Always define torque
with respect to a specific point. [Torque of F relative to point A]
A
- Slide 46
- Torque and Angular Acceleration Applying torque changes the
rotational motion of a body, a change in its acceleration. Assume
the object is made up of small masses For P 1 a pply Newtons Second
law F=ma F tan, 1 =ma=m 1 r 1 1 =F tan, 1 r=m 1 r 1 2 1
- Slide 47
- Torque and Angular Acceleration Torque going through point 1 1
=m 1 r 1 2 Net Torque net = ( m i r i 2 ) net = I where I = ( m i r
i 2 ) Moment of inertia, I, is the rotational analog for mass and
it measures the resistance of an object to a change in angular
acceleration. Equation 2 is rotational analog to Newtons 2 nd law
of motion. Note: valid only for rigid bodies. 1 eq(2)
- Slide 48
- Effect of Internal Torque
- Slide 49
- Not so Simple Example 10-16. A 12.0 kg box resting on the
horizontal, frictionless surface is attached to a 5.00 kg weight by
a thin, light wire that passes over a frictionless pulley. The
pulley has the shape of a uniform solid disk of mass 2.00 kg and a
diameter of 0.500m. After the system is released, find: (a) tension
in the wire on both sides of the pulley (b) the acceleration of the
box (c) the horizontal and vertical components of the force that
the axle exerts on the pulley. 12 kg 5 kg
- Slide 50
- Problem 10-16 F=ma For box 1 T 1 =m 1 a For box 2 m 2 g-T 2 =m
2 a For pulley (for convenience let clockwise rotation be positive)
= I (T 2 -T 1 )R= m 3 R 2 /2 (T 2 -T 1 )=m 3 a/2 Adding all three
equations m 2 g=(m 1 +m 2 +m 3 /2)a (5)(9.8)=(12+5+2/2)a a=2.72 m/s
2 12 kg 5 kg N m1gm1g T1T1 m2gm2g T2T2 T1T1 T2T2 F m3gm3g
- Slide 51
- Problem 10-16 T 1 =m 1 a=(12)(2.72) =32.6N T 2 =m 2
(g-a)=(5)(9.8- 2.72)=35.4N Force on pulley F x =-T 1 =32.6N F y
=-(T 2 +m 3 g) =35.4+(2)(9.8) =55.0N 12 kg 5 kg m1gm1g T1T1 m2gm2g
T2T2 T1T1 T2T2 F m3gm3g
- Slide 52
- Rigid Body Rotation about a Moving Axis We now combine
rotational and translational motion Motion of a rigid body can
always be divided into translation of the center of mass and
rotation about the center of mass
- Slide 53
- Rolling without Slipping Linear velocity at the rim has equal
magnitude but opposite direction to the velocity of the center of
mass. Where R is the radius of the car
- Slide 54
- Rolling without Slipping Velocity at point of contact is zero!
We use static friction
- Slide 55
- Example A primitive yoyo is made by wrapping a string around
cylinder with mass 1kg and radius 10cm. Holding one end of the
string stationary, the cylinder is released from rest. The string
unwinds without slipping. Find the speed of the center of mass of
the cylinder after it has dropped a distance 0.25m.
- Slide 56
- Example at all times
- Slide 57
- Example A uniform rod of length 1.0m with weight 2.0 kg is free
to rotate around a frictionless pin going through one end. (a)What
is its angular speed when it reaches its lowest position? (b) What
is the tangential speed of the center of mass and the tangential
speed of the lowest point when the rod is in the vertical
position?
- Slide 58
- Example
- Slide 59
- Conceptual Problem Several objects roll without slipping down a
ramp of height H, all starting from rest at the same moment. You
have a hoop, a basketball, a solid marble, a solid cylinder and a
greased box that slides without friction. In what order will they
reach the bottom of the ramp?
- Slide 60
- Conceptual Problem
- Slide 61
- Where c is a coefficient dependent on the objects shape
- Slide 62
- Conceptual Problem Larger c means lower v. Box c=0 Solid Sphere
c=2/5 Solid Cylinder c=1/2 Spherical Shell c=2/3 Thin Cylinder
c=1
- Slide 63
- Work and Power Similar to linear, a force applied on a rigid
object will cause it to rotate a certain distance . If torque
applied is constant
- Slide 64
- Work and Power Work-energy theorem for rotation Power for
rotation
- Slide 65
- Example An electric motor exerts a constant torque of 10 N.m on
a grindstone. The moment of inertia of the grindstone about its
shaft is around 2.0 kg.m 2. If the system starts from rest, find
the work done by the motor in 8 seconds and the kinetic energy at 8
s. What was the average power delivered by the motor?
- Slide 66
- Example
- Slide 67
- Angular Momentum Rotational analog to linear momentum Magnitude
Where l is the perpendicular distance
- Slide 68
- Angular Momentum For a slice of an extended body
- Slide 69
- Angular Momentum We then compute the angular momentum for other
slices, computation gets complicated for irregularly shaped
objects. If the body rotates about an axis of symmetry If the axis
of rotation is not a line of symmetry, is not parallel to the axis
of rotation
- Slide 70
- Example Estimate the angular momentum of a 6.0 kg bowling ball
with a radius of 12cm spinning at 10 rev/s?
- Slide 71
- Example
- Slide 72
- Conservation of Angular Momentum When the net external torque
of a system is zero, the total angular momentum of the system is
conserved (constant).
- Slide 73
- Example A man on a turntable holds two 5.0 kg weights at arms
length away while he is spinning at one revolution per second. What
is his angular velocity when he tucks the weights nearer to his
body? The dumbbells are 1.0m away from the axis of rotation
initially and 0.20m at the end.
- Slide 74
- Example
- Slide 75
- Example A horizontal circular platform rotates about a vertical
frictionless axle. The platform has a mass M=100.0 kg and a radius
of R=2.0m. A student of m=60.0 kg walks from the rim towards the
center. If the angular speed when the student is at the rim is 2.0
rad/s, what is the angular speed when the student is at
r=0.50m
- Slide 76
- Example
- Slide 77
- Earths Rotation
- Slide 78
- Leap Seconds The Earths Rotation is Slowing down (0.5 billion
years ago a day was 22 hours long) Tidal drag one of main
reasons
- Slide 79
- Leap Seconds Glacial Rebound helps reduce the slowing But it
wont be there forever Source: http://what-if.xkcd.com/26/
- Slide 80
- Problem Giancoli 8-20 A small rubber wheel is used to drive a
large pottery wheel, and they are mounted so that their circular
edges touch. The small wheel has a radius of 2.0cm and accelerates
at a rate of 7.2rad/s 2 and is in contact with the pottery wheel
(radius 25.0cm) without slipping. Calculate (a) the angular
acceleration of the pottery wheel, and (b) the time it takes the
pottery wheel to reach its required speed of 65rpm.
- Slide 81
- Young and Freedman 9.59 A thin uniform rod of mass M and length
L is bent at its center so that the two segments are perpendicular
to each other. Find its moment of inertia about an axis
perpendicular to its plane and passing through (a) the point where
the two segments meet and (b) the mid point of the line connecting
the two ends.
- Slide 82
- Serway 10-31 Find the net torque on the wheel about the axle O
if a= 10.0 cm and b=25.0 cm
- Slide 83
- Giancoli 8-65 An asteroid of mass 1.0x10 5 kg travelling at a
speed of 30 km/s relative to the Earth, hits the Earth at the
equator tangentially, and in the direction of Earths rotation. Use
angular momentum to estimate the percent change in the angular
speed of the Earth as a result of the collision. M E =5.97x10 24 kg
R E =6.38x10 6 m