physics 1501: lecture 21 today ’ s agenda

Physics 1501: Lecture 21, Pg 1

Physics 1501: Lecture 21Physics 1501: Lecture 21TodayToday’’s Agendas Agenda

AnnouncementsHW#8: due Oct. 28

Honors’ studentssee me after class

Midterm 1: average ~ 45 % … Topics

TorqueRotational energyRolling motion

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.


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


Rotational Dynamics:Rotational Dynamics:What makes it spin?What makes it spin?

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.


Comment onComment on == II When we write = I we are really talking about

the z component of a more general vector equation. (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


ExampleExample

To loosen a stuck nut, a man pulls at an angle of 45o on the end of a 50cm 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=0.5m

F=200N

45o


Example: solutionExample: solution

L=0.5m

F=200N

45o

Torque = Lfsin = (0.5m)(200N)(sin(45)) = 70.7 Nm

If the nut turns freely (no more friction), = I We know and we want , so we need to figure out I.

= 283 rad/s2

So = / I = (70.7Nm) / (.25kgm2)


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


Rotational WorkRotational Work

Work done by a constant torque in turning an object by an angle isConsider a wheel of radius R with a

rope wrapped around itYou pull the rope with a constant

tension TW = F d cos

Here d and F are parallelW = F d

Total displacement is d = s =R soW = F R

But FR = since F and R are perpendicularW =

F

s

F

R


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 21, Lecture 21, ACT 1ACT 1Work & 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


Rolling MotionRolling Motion

Now consider a cylinder rolling at a constant speed.

VCM CM

The cylinder is rotating about CM and its CM is moving at constantspeed (vCM). Thus its total kinetic energy is given by :



Consider again a cylinder rolling at a constant speed.

VCM

P

Q

CM

At any instant the cylinder is rotating about point P. Its kinetic energy is given by its rotational energy about that point.

KTOT = 1/2 IP 2



We can find IP using the parallel axis theorem

V = VCM

P

Q

CM

IP = ICM + MR2

KTOT = 1/2 (ICM + MR2 ) 2

KTOT = 1/2 ICM 2 + 1/2 M (R22 ) = 1/2 ICM 2 + 1/2 M VCM2 !

V = 2VCM

V = 0

V = r


Connection with CM motionConnection with CM motion For a system of particles, the kinetic energy is :

= KREL = KCM

For a solid object rotating about it’s center of mass, we now see that the first term becomes:

Substituting

but


Connection with CM motion...Connection with CM motion... So for a solid object which rotates about its center of mass and whose CM is moving:

VCM


Rolling MotionRolling Motion Cylinders of different I rolling down an inclined plane:

h

v = 0

= 0

K = 0

R

K = - U = Mgh

v = R

M


Rolling...Rolling...

If there is no slipping (due to friction):

v 2v

In the lab reference frame

v

In the CM reference frame

v

Where v = R


Rolling...Rolling...

Use v= R and I = cMR2 .

So:

The rolling speed is always lower than in the case of simple

sliding since the kinetic energy is shared between CM motion

and rotation.

hoop: c=1

disk: c=1/2

sphere: c=2/5

etc...cc

cc


Lecture 21, Lecture 21, ACT 2ACT 2Rolling MotionRolling Motion

A race !!

Two cylinders are rolled down a ramp. They have the same radius but different masses, M1 > M2. Which wins the race to the bottom ?

A) Cylinder 1

B) Cylinder 2

C) It will be a tie

M1

h

M?

M2

Active Figure


Example :Example : Rolling MotionRolling Motion A cylinder is about to roll down an inclined plane. What is its

speed at the bottom of the plane ?

M

h

Mv ?

Ball has radius R

M M

M

M

M


Example :Example : Rolling MotionRolling Motion Use conservation of energy.

Ei = Ui + 0 = Mgh

Ef = 0 + Kf = 1/2 Mv2 + 1/2 I 2

= 1/2 Mv2 + 1/2 (1/2MR2)(v/R)2

Mgh = 1/2 Mv2 + 1/4 Mv2

v2 = 4/3 g h

v = ( 4/3 g h ) 1/2


Consider a roller coaster.

We can get the ball to go around the circle without leaving the loop.

Note:Radius of loop = RRadius of ball = r


How high do we have to start the ball ?Use conservation of energy.Also, we must remember that the minimum

speed at the top is vtop = (gR)1/2

E1 = mgh + 0 + 0E2 = mg2R + 1/2 mv2 + 1/2 I2

= 2mgR + 1/2 m(gR) + 1/2 (2/5 mr2)(v/r)2

= 2mgR + 1/2 mgR + (2/10)m (gR) = 2.7 mgR

1

2


How high do we have to start the ball ?

E1 = mgh + 0 + 0E2 = 2.7 mgRmgh = 2.7 mgRh = 2.7 Rh = 1.35 D

(The rolling motion added an extra 2/10 R to the height: without it, h = 2.5 R)

1

2

physics 1501: lecture 21 today ’ s agenda

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