Physics 1501: Lecture 20, Pg 1 Physics 1501: Lecture 20 Today’s Agenda l Announcements çHW#7: due Oct. 21 l Midterm 1: average ~ 45 % … l Topics çMoments.

Physics 1501: Lecture 20, Pg 1

Physics 1501: Lecture 20Physics 1501: Lecture 20TodayToday’’s Agendas Agenda

AnnouncementsHW#7: due Oct. 21

Midterm 1: average ~ 45 % …

TopicsMoments of InertiaTorque

0 10 20 30 40 50 60 70 80 90 100

Sec. 21-28

0 10 20 30 40 50 60 70 80 90 100

Sec. 1-7


Summary Summary (with comparison to 1-D kinematics)(with comparison to 1-D kinematics)

Angular Linear

And for a point at a distance R from the rotation axis:

x = Rv = Ra = R


Rotation & Kinetic Energy...Rotation & Kinetic Energy...

The kinetic energy of a rotating system looks similar to that of a point particle:

Point ParticlePoint Particle Rotating System Rotating System

v is “linear” velocity

m is the mass.

is angular velocity

I is the moment of inertia

about the rotation axis.


Moment of InertiaMoment of Inertia

Notice that the moment of inertia I depends on the distribution of mass in the system.The further the mass is from the rotation axis, the bigger the moment of inertia.

For a given object, the moment of inertia will depend on where we choose the rotation axis (unlike the center of mass).

We will see that in rotational dynamics, the moment of inertia I appears in the same way that mass m does when we study linear dynamics !

So where


Calculating Moment of Inertia...Calculating Moment of Inertia...

For a discrete collection of point masses we found:

For a continuous solid object we have to add up the mr2 contribution for every infinitesimal mass element dm.

We have to do anintegral to find I : r

dm


Lecture 20: Act 1Lecture 20: Act 1Moment of InertiaMoment of Inertia

Two spheres have the same radius and equal masses. One is made of solid aluminum, and the other is made from a hollow shell of gold. Which one has the biggest moment of inertia about an

axis through its center?

same mass & radius

solid hollow

(a) solid aluminum (b) hollow gold (c) same


Parallel Axis TheoremParallel Axis Theorem Suppose the moment of inertia of a solid object of mass M about an axis through the center of mass is known, = ICM

The moment of inertia about an axis parallel to this axis but a distance R away is given by:

IPARALLEL = ICM + MR2

So if we know ICM , it is easy to calculate the moment of inertia about a parallel axis.


Parallel Axis Theorem: ExampleParallel Axis Theorem: Example

Consider a thin uniform rod of mass M and length D. Figure out the moment of inertia about an axis through the end of the rod.

IPARALLEL = ICM + MD2

L

D=L/2M

xCM

ICMIEND Let’s first find ICM:


Parallel Axis Theorem: ExampleParallel Axis Theorem: Example

L

D=L/2M

xCM

ICMIEND

So

IPARALLEL = ICM + MD2

We know

We get the same result if evaluating:


Direction of Rotation:Direction of Rotation: In general, the rotation variables are vectors (have a 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


Direction of Rotation:Direction of Rotation:The Right Hand RuleThe 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


Rotational Dynamics:Rotational Dynamics:What makes it spin?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 = m a = m r

r F = m r2

r

aa

FF

m

rr^

^

^

^

F

Multiply by r :



r F = mr2use

Define torque: = r F. 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.

r

aa

FF

m

rr^

^

F



So for a collection of many particles arranged in a rigid configuration:

rr1

rr2rr3

rr4

m4

m1

m2

m3

FF4

FF1

FF3

FF2

i I Since the particles are connected rigidly,

they all have the same .



TOT = I

This is the rotational version of FTOT = ma

Torque is the rotational cousin of force:Torque is the rotational cousin of force: The amount of “twist” provided by a force.

Moment of inertiaMoment of inertia I I is the rotational cousin of mass.is the rotational cousin 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.


Lecture 20, Lecture 20, Example Example A rope is wrapped around the circumference of a solid disk

(R=0.2m) of mass M=10kg and an object of mass m=10 kg is attached to the end of the rope 10m above the ground, as shown in the figure.

a) How long will it take for the object to hit the ground ?

b) What will be the velocity of the object when it hits the ground ?

c) What is the tension on the cord ?

M

m

h =10 m

T


Lecture 20, Lecture 20, Example Example SolutionSolution

M

m

h =10 m

T

v2 - vo2 = 2 a h

v = (2 a h) 1/2

v = a / tt = v / a

mg - T = a m

a = R I = =T RT = I / R = (1/2) MR2 a / R2

T = Ma/2

mg - Ma/2 = a ma = mg / (m+M/2) = 2g/3

a = 2g/3 = 2 x 9.8 m/s2 / 3 = 6.5 m/s2

v = (2 a h)1/2 =(2 x 6.5m/s2 10m)1/2 =11m/st = v / a = 11 m/s / 6.5 m/s2 = 1.7 sT = Ma/2 = 10kg 6.5m/s2 / 2 = 32 N


TorqueTorque

= r F

= r Fsin

= r sin F

Recall the definition of torque:

rr

r

FF

F

Fr

r = “distance of closest approach”


TorqueTorque

= r Fsin

So if = 0o, then = 0

And if = 90o, then = maximum

rr

FF

rr

FF


Lecture 20, Lecture 20, Act 2Act 2TorqueTorque

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)(a) case 1

(b)(b) case 2

(c)(c) same

L

L

F F

axis

case 1 case 2


Torque and the Torque and the Right Hand Rule: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 r

FF

x

y

z


Review: The Cross ProductReview: 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:

AA X BB = CC

The length of CC is given by:

C = ABsin

The direction of CC is perpendicular to the plane defined by AA and BB, and inthe direction defined by the right-handrule.

AA

BB

CC


The Cross ProductThe Cross Product

The cross product of unit vectors:

ii x ii == 0 0 ii x j j == k k ii x kk = = -j -j

jj x i i == - -k k jj x jj = = 0 0 jj x kk = = i i

kk x i i == j j kk x jj = = -i -i kk x kk = = 00

A X B = (AX i i + + AY jj + Azkk) ) X (BX i i + + BY jj + Bzkk) )

= (AX BX i i x x i i ++ AX BY i i xx jj ++ AX BZ i i xx kk))

+ (AY BX jj x x i i ++ AY BY jj xx jj + + AY BZ jj xx kk))

+ (AZ BX kk x x i i ++ AZ BY kk xx j j ++ AZ BZ kk xx kk))

ii

jj

kk


The Cross ProductThe Cross Product

Cartesian components of the cross product:

C C = AA X BB

CX = AY BZ - BY AZ

CY = AZ BX - BZ AX

CZ = AX BY - BX AY

AA

BB

CC

Note: B X A = - A X B


Torque & the Cross Product:Torque & the Cross Product:

rr

FF

x

y

z

So we can define torque as:

= r r x FF

= r F sin

X = y FZ - z FY

Y = z FX - x FZ

Z = x FY - y FX


WorkWork

Consider the work done by a force FF acting on an object constrained to move around a fixed axis. For an infinitesimal angular displacement d:

dW = FF.drdr = F R dcos()

= F R dcos(90-) = F R dsin()

= F R sin() ddW = d

We can integrate this to find: W = Analogue of W = F •r W will be negative if and have opposite sign !

R

FF

dr=Rddaxis


Work & Kinetic Energy: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:


Example: Disk & StringExample: Disk & String

A massless string is wrapped 10 times around a disk of mass M=40 g and radius R=10cm. The disk is constrained to rotate without friction about a fixed axis through its center. The string is pulled with a force F=10N 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

RM


Disk & String...Disk & String...

WNET = W = 62.8 J = K

Recall thatIfor a disk about

its central axis is given by:

So

= 792.5 rad/s

RM

W = = F x r . = (10 N)(0.10 m)(10*2) = 62.8 J


Lecture 20, Lecture 20, ACT 3ACT 3Work & EnergyWork & 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 are made of identical material (i.e. their density = M/V is the same). Both disks rotate freely around axes though their centers, and start at rest.Which disk has the biggest angular velocity after the pull ?

(a)(a) disk 1

(b)(b) disk 2

(c)(c) same FF

1 2

Physics 1501: Lecture 20, Pg 1 Physics 1501: Lecture 20 Today’s Agenda l Announcements çHW#7: due Oct. 21 l Midterm 1: average ~ 45 % … l Topics çMoments.

Documents

rotation axis

axis parallel

moment of inertia notice

biggest moment of inertia

direction of rotation

r slide

parallel axis theorem

plane of rotation