Physics 151: Lecture 21, Pg 1 Physics 151: Lecture 21 Today’s Agenda l Topics çMoments of InertiaCh. 10.5 çTorqueCh. 10.6, 10.7

Physics 151: Lecture 21, Pg 1

Physics 151: Lecture 21Physics 151: Lecture 21Today’s AgendaToday’s Agenda

TopicsMoments of Inertia Ch. 10.5Torque Ch. 10.6,

10.7


Lecture 22, Lecture 22, ACT 1ACT 1Rotational DefinitionsRotational Definitions

Your goofy friend likes to talk in physics speak. She sees a disk spinning and says “ooh, look! There’s a wheel with a negative and with antiparallel and !!”

Which of the following is a true statement ?

(a)(a) The wheel is spinning counter-clockwise and slowing down.

(b) (b) The wheel is spinning counter-clockwise and speeding up.

(c)(c) The wheel is spinning clockwise and slowing down.

(d) The wheel is spinning clockwise and speeding up


Example:Example:

A wheel rotates about a fixed axis with a constant angular acceleration of 4.0 rad/s2. The diameter of the wheel is 40 cm. What is the linear speed of a point on the rim of this wheel at an instant when that point has a total linear acceleration with a magnitude of 1.2 m/s2?

a. 39 cm/sb. 42 cm/sc. 45 cm/sd. 35 cm/se. 53 cm/s

See text: 10.1


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 !

K 12

2I I m ri ii

2

See text: 10.4

So where


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.

See text: 10.5


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

ICM ML1

122

We know

IEND ML ML

ML

1

12 2

1

32

22So

which agrees with the result from the board.

ICMIEND

See text: 10.5


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

See text: 10.1


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 = ma = mr

r

aa

FF

m

rr^

^

^

See text: 10.6 and 10.7

^

F

Multiply by r :

rF = mr2

video



rF = mr2use

Define torque: = rF.

is the tangential force Ftimes the lever arm r.

I=

I=

2mr=I

r

aa

FF

m

rr^

^

F


Torque has a direction:+ z if it tries to make the system

spin CCW. - z if it tries to make the system

spin CW.



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

rr1

rr2rr3

rr4

m4

m1

m2

m3

FF4

FF1

FF3

FF2

r F m ri ii

i ii

i, 2

i I

ii I

Since the particles are connected rigidly,they all have the same .

I=tot




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 II 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.



TorqueTorque

See Figure 10.13

= rF

= r Fsin r sin F

Recall the definition of torque:

rr

r

FF

F

Fr

r = “distance of closest approach”

See text: 10.6, 10.7, 11.2

Fr


Example:Example:

You throw a Frisbee of mass m and radius r so that it is spinning about a horizontal axis perpendicular to the plane of the Frisbee. Ignoring air resistance, the torque exerted about its center of mass by gravity is :

a. 0.b. mgr.c. 2mgr.d. a function of the angular velocity.e. small at first, then increasing as the

Frisbee loses the torque given it by your hand.

See text: 10.1


TorqueTorque

= r Fsin

So if = 0o, then = 0

And if = 90o, then = maximum

rr

FF

rr

FF

See Figure 10.13



Lecture 21, Lecture 21, Act 1Act 1TorqueTorque

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


Lecture 21, Lecture 21, ACT 2ACT 2 A uniform rod of mass M = 1.2kg and length L = 0.80 m,

lying on a frictionless horizontal plane, is free to pivot about a vertical axis through one end, as shown. If a force (F = 5.0 N, = 40°) acts as shown, what is the resulting angular acceleration about the pivot point ?

a. 16 rad/s2

b. 12 rad/s2

c. 14 rad/s2

d. 10 rad/s2

e. 33 rad/s2


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

See text: 11.2

DIRECTION: =rr X FF

MAGNITUDE: =rr FF sin sin


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

See text: 11.2


How Much WORK is Done ?How Much WORK is Done ?

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 = FRdcos()

= FRdcos(90-) = FRdsin()

= FRsin() 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

See text: 10.8


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:

K Wf i net 12

2 2I

See text: 10.8


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 though 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?

See example 10.15

F

RM

See text: 10.8

= 792.5 rad/s


Lecture 21, Lecture 21, ACT 2ACT 2 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


Example 2Example 2 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.

M

m

h =10 m

T

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

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

a) What is the tension on the cord ?

1.7 s

11m/s

32 N


Example: Rotating RoadExample: Rotating Road

A uniform rod of length L=0.5m and mass m=1 kg is free to rotate on a frictionless pin passing through one end as in the Figure. The rod is released from rest in the horizontal position. What is

a) angular speed when it reaches the lowest point ?

b) initial angular acceleration ?

c) initial linear acceleration of its free end ?

See example 10.14

See text: 10.8

L

m = 7.67 rad/sa)

= 30 rad/s2b)

c) a = 15 m/s2


Lecture 22, Lecture 22, ACT 2ACT 2 A campus bird spots a member of an opposing football

team in an amusement park. The football player is on a ride where he goes around at angular velocity at distance R from the center. The bird flies in a horizontal circle above him. Will a dropping the bird releases while flying directly above the person’s head hit him?

a. Yes, because it falls straight down.b. Yes, because it maintains the acceleration of the bird

as it falls.c. No, because it falls straight down and will land behind

the person.d. Yes, because it mainatins the angular velocity of the

bird as it falls.e. No, because it maintains the tangential velocity the

bird had at the instant it started falling.


Example:Example:

A mass m = 4.0 kg is connected, as shown, by a light cord to a mass M = 6.0 kg, which slides on a smooth horizontal surface. The pulley rotates about a frictionless axle and has a radius R = 0.12 m and a moment of inertia I = 0.090 kg m2. The cord does not slip on the pulley. What is the magnitude of the acceleration of m?

a. 2.4 m/s2

b. 2.8 m/s2

c. 3.2 m/s2

d. 4.2 m/s2

e. 1.7 m/s2

See text: 10.1


Recap of today’s lectureRecap of today’s lecture

Chapter 10, Calculating moments of inertiaTourqueRight Hand Rule

Documents

Physics 151: Lecture 21, Pg 1 Physics 151: Lecture 21 Today’s Agenda l Topics çMoments of InertiaCh. 10.5 çTorqueCh. 10.6, 10.7