Physics 111: Lecture 18 Today’s Agenda

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


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


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


Rolling...

If there is no slipping:

v 2v

In the lab reference frame

v

In the CM reference frame

v

Where v = R


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


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


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


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


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


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!


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 :



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



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



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.


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


Torque

= r Fsin

So if = 0o, then = 0

And if = 90o, then = maximum

r

F

rF


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



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.


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


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


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


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


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


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


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)


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


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


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


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


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


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



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


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


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

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Physics 111: Lecture 18 Today’s Agenda