Physics 111: Lecture 18, Pg 1
Physics 111: Lecture 18
Today’s Agenda
More about rolling
Direction and the right hand rule
Rotational dynamics and torque
Work and energy with example
Physics 111: Lecture 18, Pg 2
Rotational v.s. Linear Kinematics
Angular Linear
constant=
t0 +=
200 t
21
t ++=
ttanconsa
atvv 0 +=
200 at
21
tvxx ++=
And for a point at a distance R from the rotation axis:
x = Rv = Ra = R
Physics 111: Lecture 18, Pg 3
Rolling Motion
Cylinders of different I rolling down an inclined plane:
h
v = 0 = 0 K = 0
RK = - U = Mgh
22 Mv21
I21
K +=
v = R
M
Roll objectsdown ramp
Physics 111: Lecture 18, Pg 4
Rolling...
If there is no slipping:
v 2v
In the lab reference frame
v
In the CM reference frame
v
Where v = R
Physics 111: Lecture 18, Pg 5
Rolling...
22 Mv21
I21
K +=
( ) 2222 Mv121
Mv21
MR21
K +=+=
Use v = R and I = cMR2 .
So: ( ) MghMv121 2 =+ 1
1gh2v
+=
The rolling speed is always lower than in the case of simple slidingsince the kinetic energy is shared between CM motion and rotation.
We will study rolling more in the next lecture!
hoop: c = 1
disk: c = 1/2
sphere: c = 2/5
etc...c c
c c
Physics 111: Lecture 18, Pg 6
Direction of Rotation: In general, the rotation variables are vectors (have direction) If the plane of rotation is in the x-y plane, then the convention
is
CCW rotation is in the + z direction
CW rotation is in the - z direction
x
y
z
x
y
z
Physics 111: Lecture 18, Pg 7
Direction of Rotation:The Right Hand Rule
To figure out in which direction the rotation vector points, curl the fingers of your right hand the same way the object turns, and your thumb will point in the direction of the rotation vector!
We normally pick the z-axis to be the rotation axis as shown.= z
= z
= z
For simplicity we omit the subscripts unless explicitly needed.
x
y
z
x
y
z
Physics 111: Lecture 18, Pg 8
Example:
A flywheel spins with an initial angular velocity 0 = 500 rad/s. At t = 0 it starts to slow down at a rate of 0.5 rad/s2. How long does it take to stop?
Realize that = - 0.5 rad/s2.
0 t Use to find when = 0 :
t
0
min././ 716s1000srad50srad500t 2 So in this case
Physics 111: Lecture 18, Pg 9
Lecture 18, Act 1Rotations
A ball rolls across the floor, and then starts up a ramp as shown below. In what direction does the angular acceleration vector point when the ball is on the ramp?
(a) down the ramp (b) into the page(c) out of the page
Physics 111: Lecture 18, Pg 10
Lecture 18, Act 1Solution
When the ball is on the ramp, the linear acceleration a is always down the ramp (gravity).
a
The angular acceleration is therefore counter-clockwise.
Using your right hand rule, is out of the page!
Physics 111: Lecture 18, Pg 11
Rotational Dynamics:What makes it spin?
Suppose a force acts on a mass constrained to move in a circle. Consider its acceleration in the direction at some instant:a = r
Now use Newton’s 2nd Law in the direction:F = ma = mr
rF = mr2
r
a
F
m
r
^
F
Multiply by r :
Physics 111: Lecture 18, Pg 12
Rotational Dynamics:What makes it spin?
rF = mr2use
Define torque: = rF. is the tangential force F
times the lever arm r.
Torque has a direction:+ z if it tries to make the system
spin CCW.- z if it tries to make the system
spin CW.
I=
I=
2mr=I
r
a
F
m
r
F
Physics 111: Lecture 18, Pg 13
Rotational Dynamics:What makes it spin?
So for a collection of many particles arranged in a rigid configuration:
r1
r2r3
r4
m4
m1
m2
m3
F4
F1
F3
F2
ii
2ii
iii rmFr
,
i I
ii I
Since the particles are connected rigidly,they all have the same .
INET
Physics 111: Lecture 18, Pg 14
Rotational Dynamics:What makes it spin?
NET = I
This is the rotational analogue of FNET = ma
Torque is the rotational analogue of force: The amount of “twist” provided by a force.
Moment of inertia I is the rotational analogue of mass. If I is big, more torque is required to achieve a given
angular acceleration. Torque has units of kg m2/s2 = (kg m/s2) m = Nm.
Physics 111: Lecture 18, Pg 15
Torque
= r F sin = r sin F
= rpF
Equivalent definitions!
rp = “distance of closest approach”
= rF
Recall the definition of torque:
r
rp
F
F
Fr
Physics 111: Lecture 18, Pg 16
Torque
= r Fsin
So if = 0o, then = 0
And if = 90o, then = maximum
r
F
rF
Physics 111: Lecture 18, Pg 17
Lecture 18, Act 2Torque
In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same.
(a) case 1 (b) case 2
(c) same L
L
F F
axis
case 1 case 2
Physics 111: Lecture 18, Pg 18
Lecture 18, Act 2Solution
Torque = F x (distance of closest approach)
LF F
case 1 case 2
L
Torque is the same!
The applied force is the same.The distance of closest approach is the same.
Physics 111: Lecture 18, Pg 19
Torque and the Right Hand Rule:
The right hand rule can tell you the direction of torque:Point your hand along the direction from the axis to the
point where the force is applied.Curl your fingers in the direction of the force.Your thumb will point in the direction
of the torque.
r
F
x
y
z
Physics 111: Lecture 18, Pg 20
The Cross Product
We can describe the vectorial nature of torque in a compact form by introducing the “cross product”.The cross product of two vectors is a third vector:
A X B = C
The length of C is given by: C = AB sin
The direction of C is perpendicular to the plane defined by A and B, and inthe direction defined by the right handrule.
A
B
C
Physics 111: Lecture 18, Pg 21
The Cross Product
Cartesian components of the cross product:
C = A X B
CX = AY BZ - BY AZ
CY = AZ BX - BZ AX
CZ = AX BY - BX AY
A
B
C
Note: B X A = - A X B
Physics 111: Lecture 18, Pg 22
Torque & the Cross Product:
r
F
x
y
z
So we can define torque as:
= r X F = rF sin
X = rY FZ - FY rZ = y FZ - FY z
Y = rZ FX - FZ rX = z FX - FZ x
Z = rX FY - FX rY = x FY - FX y
Physics 111: Lecture 18, Pg 23
Comment on = I
When we write = I we are really talking about the z component of a more general vector equation. (Recall that we normally choose the z-axis to be the the rotation axis.)
z = Izz
We usually omit the z subscript for simplicity.
z
z
z
Iz
Physics 111: Lecture 18, Pg 24
Example
To loosen a stuck nut, a (stupid) man pulls at an angle of 45o on the end of a 50 cm wrench with a force of 200 N. What is the magnitude of the torque on the nut?If the nut suddenly turns freely, what is the angular
acceleration of the wrench? (The wrenchhas a mass of 3 kg, and its shape is that of a thin rod).
L = 0.5 m
F = 200 N
45o
Physics 111: Lecture 18, Pg 25
Example
L = 0.5m
F = 200 N45o
Torque = LFsin = (0.5 m)(200 N)(sin 45) = 70.7 Nm
Wrench w/ bolts
If the nut turns freely, = I We know and we want , so we need to figure out I.
222 kgm250m50kg331ML
31 ..I
= 283 rad/s2
So = / I = (70.7 Nm) / (0.25 kgm2)
Physics 111: Lecture 18, Pg 26
Work
Consider the work done by a force F acting on an object constrained to move around a fixed axis. For an infinitesimal angular displacement d:
dW = F.dr = FR dcos()
= FR dcos(90-) = FR dsin()
= FR sin() d dW = d
We can integrate this to find: W = Analogue of W = F •r W will be negative if and have opposite signs!
R
F
dr = R ddaxis
Physics 111: Lecture 18, Pg 27
Work & Kinetic Energy:
Recall the Work/Kinetic Energy Theorem: K = WNET
This is true in general, and hence applies to rotational motion as well as linear motion.
So for an object that rotates about a fixed axis:
NET2i
2f W
21K I
Physics 111: Lecture 18, Pg 28
Example: Disk & String
A massless string is wrapped 10 times around a disk of mass M = 40 g and radius R = 10 cm. The disk is constrained to rotate without friction about a fixed axis though its center. The string is pulled with a force F = 10 N until it has unwound. (Assume the string does not slip, and that the disk is initially not spinning).
How fast is the disk spinning after the string has unwound?
F
R M
Physics 111: Lecture 18, Pg 29
Disk & String...
The work done is W = The torque is = RF (since = 90o)The angular displacement is
2 rad/rev x 10 rev.
F
R M
So W = (.1 m)(10 N)(20rad) = 62.8 J
Physics 111: Lecture 18, Pg 30
Disk & String...
WNET = W = 62.8 J = K 12
2I
Recall thatIfor a disk aboutits central axis is given by:
I 12
2MR
K MR W
12
12
2 2So
4 4 62 804 12 2
WMR
Jkg
.. .
= 792.5 rad/s
R M
Flywheel, pulley, & mass
Physics 111: Lecture 18, Pg 31
Lecture 18, Act 3Work & Energy
Strings are wrapped around the circumference of two solid disks and pulled with identical forces for the same distance. Disk 1 has a bigger radius, but both have the same moment of inertia. Both disks rotate freely around axes though their centers, and start at rest.Which disk has the biggest angular velocity after the pull ?
(a) disk 1 (b) disk 2
(c) same FF
1 2
Physics 111: Lecture 18, Pg 32
Lecture 18, Act 3Solution
FF
1 2
d
The work done on both disks is the same!W = Fd
The change in kinetic energy of each will therefore also be the same since W = K.
But we know K 12
2I
So since I1 = I2
1 = 2
Physics 111: Lecture 18, Pg 33
Spinning Disk Demo:
We can test this with our big flywheel.
22 mv21
21KW I
I
m
negligiblein this case
2WI
In this case, I = 1 kg - m2
W = mgh = (2 kg)(9.81 m/s2)(1 m) = 19.6 J
= 6.26 rad/s ~ 1 rev/s
Physics 111: Lecture 18, Pg 34
Recap of today’s lecture
More about rolling (Text: 9-6)
Direction and the right hand rule (Text: 10-2)
Rotational dynamics and torque (Text: 9-2, 9-4)
Work and energy with example (Text: 9-5)
Look at textbook problems Chapter 9: # 21, 23, 25, 49, 91, 119