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Rotational M
otion
6-1 Angular Position, Velocity, & Acceleration
Angular Displacement ()
Measured in radians (rad)
this unit has no dimensions
Counterclockwise is positive
6-1 Angular Position, Velocity, & Acceleration
Angular Quantities
oradrev 36021
Angular Velocity ()
Defined in the same way as velocity
Average velocity is displacement divided by time
SI unit
Instantaneous Angular Velocity
6-1 Angular Position, Velocity, & Acceleration
t
1ssrad
tt
0lim
Angular Acceleration ()
Defined like acceleration
Angular acceleration is change in angular velocity divide by time
SI unit
Instantaneous Angular
Acceleration
6-1 Angular Position, Velocity, & Acceleration
t
22
ssrad
tt
0lim
6.2 Rotational Kinematics
All the equations used in kinematics (for conditions of constant acceleration)
Are exactly the same, but with the new quantities
6.2 Rotational Kinematics
)(2
)(
020
2
221
00
021
0
0
xxavv
attvxx
tvvxx
atvv
)(2
)(
020
2
221
00
021
0
0
tt
t
t
6.3 Connections Between Linear & Rotational
For an object moving in a circular path
At any given time, the
object is moving with a
linear velocity tangent
to the arc
The speed would be circumference divide by the time for one revolution
Since would be 2/t
6.3 Connections Between Linear & Rotational
t
rv
2 rvt
Similarly we can determine that
So in a situation where a spinning
object causes linear motion
Circumference of the
tire
Will equal the
displacement
6.3 Connections Between Linear & Rotational
rat
6.4 Rotational K & the Moment of Inertia
In a spinning object, each particle has kinetic energy
Lets assume that we have a rod of uniform mass rotating about its end
6.4 Rotational K & the Moment of Inertia
v
r
Now, just consider a piece of the rod at the very end
To calculate the kinetic energy of the mass
Converting
To rotational
quantities
6.4 Rotational K & the Moment of Inertia
v
r
221 mvK 2
21 )( rmK 2221 )( mrK
So K depends on angular velocity
And it depends on the distribution of mass
This mass distribution is called Moment of Inertia
6.4 Rotational K & the Moment of Inertia
v
r
2221 )( mrK 2
21 IK
Moment of Inertia varies with shape, mass, and axis of rotation (You will need these to solve
problems)
6.4 Rotational K & the Moment of Inertia
6.5 Conservation of Energy
We are adding a new type of Kinetic Energy into our existing Energy Equation
Adding Rotational Kinetic Energy
Where Kr is defined as
6.5 Conservation of Energy
sgLsgL UUKUUK 000
sgLrsgLr UUKKUUKK 0000
221 IKr
So if a solid ball is rolling down a slope
The equation would become
Using the moment of inertia of a solid sphere
And the relationship between v and
6.5 Conservation of Energy
sgLrsgLr UUKKUUKK 0000 Lrg KKU
0
2212
21 mvImgy 2
2122
52
21 )( mvmrmgy 2
2122
51 mvmrmgy 2
2122
51 )( mvmrmgy r
v 2212
51 mvmvmgy 2
107 mvmgy
Example: A yo-yo is released from rest and allowed to drop as the top end of the string is kept stationary. The mass of the yo-yo is 0.056kg its radius is 1.5 cm. Assume it acts as a solid disc rotating around its center. What is the angular speed of the yo-yo after it has dropped 0.50m?
6.5 Conservation of Energy
sgLrsgLr UUKKUUKK 0000
Example: A yo-yo is released from rest and allowed to drop as the top end of the string is kept stationary. The mass of the yo-yo is 0.056kg its radius is 1.5 cm. Assume it acts as a solid disc rotating around its center. What is the angular speed of the yo-yo after it has dropped 0.50m?
What quantities remain in the equation below?
Expanded equation?
Moment of inertia?
Solve (remember v=r)
Cancel and insert values
6.5 Conservation of Energy
sgLrsgLr UUKKUUKK 0000 Lrg KKU
0
2212
21 mvImgy 2
2122
21
21 )( mvmrmgy 2
2122
41 mvmrmgy 2
2122
41 )( rmmrmgy 22
2122
41 mrmrmgy 22
43 mrmgy 2243 rgy 2243 )015(.)5)(.8.9( s
rad17
6.6 Torque
Rotational Dynamics – the causes of rotational motion
How do we make an object spin
Apply a force away from the pivot
6.6 Torque
The ability to spin increases with force and the distance from the pivot point
If the force is parallel to the distance, no rotation occurs
6.6 Torque
It is the perpendicular component of force that causes rotation
This quantity is called Torque ()
And is measure in Nm
6.6 Torque
sinFr
As with angular quantities
Counterclockwise torque is positive
Clockwise torques are defined as negative
6.6 Torque
r
F
r
F
6.7 Torque and Angular Acceleration
If an unbalanced torque is applied to an object
Consider a point where the force is applied
The acceleration of that point is
6.7 Torque and Angular Acceleration
m
Fa
Since
We can write this equation as
Combining we get
Now, math tricks, multiply by r/r
The of the fraction is torque, and the bottom is inertia
Usually written as
6.7 Torque and Angular Acceleration
m
Fa
ra
mr
F
rm
F
r
a
2mr
Fr
mr
F
r
r
I
I
Applying Newton’s second law to rotational motion we get
6.7 Torque and Angular Acceleration
I
6.8 Static Equilibrium
Static Equilibrium
How do we calculate the forces needed to support a bridge?
The bridge is at static (not moving) equilibrium
6.8 Static Equilibrium
Using a diagram of the bridge
Two conditions for equilibrium
1. The sum of the forces equals zero
6.8 Static Equilibrium
0F
P1 P2
W
021 WPP
2.Sum of the torques equals zero
For torque we need a pivot point – the actual point does not matter because the object is not rotating
Use a pivot point that eliminates a variable
6.8 Static Equilibrium
0
The equation becomes
6.8 Static Equilibrium
0FP1 P2
W
0sinsin 222 wwWrrP
Try the problem with some variable
The bridge is 100 m long with two pylons that are 10 m from each end. The mass of the bridge is 10,000 kg. What is the force on each pylon?
6.8 Static Equilibrium
P1 P2
W
The bridge is 100 m long with two pylons that are 10 m from each end. The mass of the bridge is 10,000 kg. What is the force on each pylon?
First Condition of Equilibrium
6.8 Static Equilibrium
P1 P2
W
0)8.9)(10000(21 PP NPP 9800021
The bridge is 100 m long with two pylons that are 10 m from each end. The mass of the bridge is 10,000 kg. What is the force on each pylon?
Second Condition of Equilibrium
6.8 Static Equilibrium
P1 P2
W
0sinsin22 wwWrrP 022 wWrrP 0)40)(98000()80(2 P NP 490002
The bridge is 100 m long with two pylons that are 10 m from each end. The mass of the bridge is 10,000 kg. What is the force on each pylon?
Combine equations
6.8 Static Equilibrium
P1 P2
W
NPP 9800021 NP 98000490001 NP 490001
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is 0.20. How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder)
6.8 Static Equilibrium
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is 0.20. How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder)
Diagram?
6.8 Static Equilibrium
Wall
ladder
Dude
=60o
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is 0.20. How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder)
Force Diagram?
6.8 Static Equilibrium
Wall
ladder
Dude
=60o
N1
f
WL
WD
N2
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is 0.20. How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder)
Force Equations?
6.8 Static Equilibrium
Wall
ladder
=60o
N1
f
WL
WD
N2
DLy WWNF 1
2NfFx
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is 0.20. How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder)
Pivot Point for Torque?
6.8 Static Equilibrium
Wall
ladder
=60o
N1
f
WL
WD
N2
DLy WWNF 1
2NfFx
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is 0.20. How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder)
Distances?
6.8 Static Equilibrium
Wall
ladder
=60o
N1
f
WL
WD
N2
DLy WWNF 1
2NfFx
20m 10m
r
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is 0.20. How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder)
Angles?
6.8 Static Equilibrium
Wall
ladder
=60o
N1
f
WL
WD
N2
DLy WWNF 1
2NfFx
20m 10m
r
=60o
=30o=30o
=30o
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is 0.20. How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder)
Torque Equation?
6.8 Static Equilibrium
Wall
ladder
N1
f
WL
WD
N2
DLy WWNF 1
2NfFx
20m 10m
r
=60o
=30o=30o
=30o
111 sinsinsinsin rNrWrWfr DDDLLLff
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is 0.20. How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder)
Values?
6.8 Static Equilibrium
Wall
ladder
N1
f
WL
WD
N2
DLy WWNF 1
2NfFx
20m 10m
r
=60o
=30o=30o
=30o
111 sinsinsinsin rNrWrWfr DDDLLLff )30sin()20()30sin()8.9)(75()30sin()10)(8.9)(40()60sin()20( 1Nrf D 0105.36719603.17 1 Nrf D
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is 0.20. How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder)
Value for N1?
6.8 Static Equilibrium
Wall
ladder
N1
f
WL
WD
N2
DLy WWNF 1
2NfFx
20m 10m
r
=60o
=30o=30o
=30o
111 sinsinsinsin rNrWrWfr DDDLLLff )30sin()20()30sin()8.9)(75()30sin()10)(8.9)(40()60sin()20( 1Nrf D 0105.36719603.17 1 Nrf D
7353920 1 N 11271 N
0)1127(105.36719603.17 Drf 0112705.36719603.17 Drf
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is 0.20. How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder)
Value for f?
6.8 Static Equilibrium
Wall
ladder
N1
f
WL
WD
N2
2NfFx
20m 10m
r
=60o
=30o=30o
=30o
7353920 1 N 11271 N
0112705.36719603.17 Drf
1Nf )1127)(2(.f 4.225f
0112705.3671960)4.225(3.17 Dr 0112705.36719603899 DrmrD 7.14
6.9 Center of Mass and Balance
An object is balanced, if the net torque is zero
If the rod above is uniform, then when the pivot is in the middle, half the mass is on one side and half on the other
This balance point is called the center of mass
6.9 Center of Mass and Balance
Wall
mrmr 21
21
6.10 Conservation of Angular Momentum
When an object spins, it has angular momentum
Defined as
The conservation of angular momentum is treated just like the conservation of linear momentum
6.10 Conservation of Angular Momentum
v
IL
Conservation of L
So the equation for conservation becomes
6.10 Conservation of Angular Momentum
LL 0 BBAABBAA IIII 0000
Example: A small mass, m, attached to the end of a string revolves in a circle on a frictionless tabletop. The other end of the string passes through a hole in the table. Initially, the mass revolves with a speed v0=2.4m/s in a circle of radius r0=0.80m. The string is then pulled slowly through the hole so that the radius is reduced to r=0.48m. What is the speed, v, of the mass now?
6.10 Conservation of Angular Momentum
Example: A small mass, m, attached to the end of a string revolves in a circle on a frictionless tabletop. The other end of the string passes through a hole in the table. Initially, the mass revolves with a speed v0=2.4m/s in a circle of radius r0=0.80m. The string is then pulled slowly through the hole so that the radius is reduced to r=0.48m. What is the speed, v, of the mass now?
Using conservation of momentum (only one object)
Mass on the end of a string so
Masses cancel and we can convert to v
Enter numbers and solve6.10 Conservation of Angular Momentum
LL 0 II 00 20
20 mrmr 2
020 rr r
vrv rr 22
0 0
0 rvvr 00 v)48(.)4.2)(8(. smv 4
Example: A record player with a mass of 1.5 kg is spinning at 33.3 rev/min. The radius of the turntable is 15cm. A bug with a mass of 0.25 kg lands 10cm from the center of the record. What is the new velocity of the record player?
6.10 Conservation of Angular Momentum
Example: A record player with a mass of 1.5 kg is spinning at 33.3 rev/min. The radius of the turntable is 15cm. A bug with a mass of 0.25 kg lands 10cm from the center of the record. What is the new velocity of the record player?
The record player is a solid disk, the bug is a point mass
Convert rev/min to rad/s
Substitute and solve
6.10 Conservation of Angular Momentum
LL 0 BBRRRR III 00 BBBRRRRRR rmrmrm 22212
21
0
srad
srevradrev 49.3))((3.33 60
min12min
BR 2)10)(.25(.
2)15)(.5.1(2
1)49.3(2)15)(.5.1(2
1 BR 0025.0169.0589. v 0025.0169.0589. 0194.0589. srad04.3 min0.2904.3 rev
srad