rotational motion and equilibrium angular quantities of rotational motion rotational kinematics...

ROTATIONAL MOTIONAND EQUILIBRIUM

Angular Quantities of Rotational Motion

Rotational Kinematics

Torque

Center of Gravity

Moment of Inertia

Rotational Kinetic Energy

Angular Momentum

Rotational Equilibrium

ANGULAR SPEED AND ANGULAR ACCELERATION:

= angular speed rad/s deg/s rev/s

= angular acceleration rad/s2 deg/s2 rev/s2

rotatingdrum

axis

t

t

ANGULAR CONVERSIONS:

1 rev = 2 rad = 360 deg

rad = 180 deg

Convert 246o to radians.

Convert 16.4 rev to degrees.

246 24618

1.370

o oo

rad

36016.4 16.4 5904

oorev rev

rev

Convert 246o to radians.

Convert 16.4 rev to degrees.

212

2 2 2

o o

o

o o

t

t

ROTATIONAL KINEMATICS:

Uniform angular acceleration:

= instantaneous angular position

= instantaneous angular speed

EXAMPLE:

Through what angle does a wheel rotate if it begins from restand accelerates at 3 o/s2 for 15 seconds? Also calculate its finalangular speed.

2

2

212

2deg12

deg

337deg

45de

0 0 3 15

0 3

g/

15

o o

s

o

s

t t

s

s

t

s

ANSWER:

Through what angle does a wheel rotate if it begins from restand accelerates at 3 o/s2 for 15 seconds? Also calculate its finalangular speed.

ROTATIONAL AND LINEAR QUANTITIES:

axis

r

s r

v r

a r

Angular quantities must be expressed interms of radians.

EXAMPLE: A point on the rim 5cmfrom the axis of rotation moves through anangle of 15 degrees. How far has it traveled?

ANSWER: A point on the rim 5cm fromthe axis of rotation moves through an angleof 15 degrees. How far has it traveled?

0.05 15180

0.13oo

s m m

EXAMPLE:A wheel of radius r is rolling along the ground without slipping so thatits center moves with a constant speed v? What are the speeds of thepoints at its top and at its bottom?

Since it is not slipping the speed at the center must be v = 2r/T.

The rotational speed must be = 2/T.

The speed of a point on the rim, relative to the center, must be r = 2r/T. This is equal to v.

At its bottom the rim is moving backwards relative to the center so thatits velocity relative to the ground if v -v = 0.

At its top the rim is moving forward relative to the center to that itsvelocity is v + v = 2v. 2v

v

TORQUE:

Torque, , is produced when a force acts on a body along a line that isdisplaced from a pivot point or axis of rotation.

LINE OF ACTION: The line of action of a force is the line alongwhich the force acts.

MOMENT ARM: The moment arm of a force is the perpendiculardistance from the pivot or axis of rotation to the line of action of theforce. Bar

line of action

Forcemoment arm

dF

l

pivot

sin

sin

Fd

d l

Fl

The torque produced by this force is trying torotate the bar in a counter-clockwise direction.

HOW TO CALCULATE A TORQUE:

1. In a sketch, locate the pivot point or axis of rotation.

2. Locate the force producing the torque and draw its line ofaction.

3. Draw a line segment from the pivot to the line of actionforming a 90o angle with the line of action. This is the momentarm.

4. Using the right-triangle formed by the object, force andmoment arm, calculate the length of the moment arm.

5. The magnitude of the torque is the product of the force andthe moment arm.

6. Determine the direction of rotation the torque is trying toinduce as being either clockwise or counter-clockwise.

EXAMPLE:

Calculate the torque produced bythe force acting on the rod as shown.

pivot

F = 16N

57o

1.25m

1 - Draw Line of Action

EXAMPLE:

Calculate the torque produced bythe force acting on the rod as shown.

pivot

F = 16N

57o

1.25m

line of action

1 - Draw Line of Action2 - Draw Moment Arm

EXAMPLE:

Calculate the torque produced bythe force acting on the rod as shown.

pivot

F = 16N

57o

1.25m

dline of action

1 - Draw Line of Action2 - Draw Moment Arm3 - Calculate Moment Arm

1.25 sin 1.0557o md m

EXAMPLE:

Calculate the torque produced bythe force acting on the rod as shown.

pivot

F = 16N

57o

1.25m

dline of action

1 - Draw Line of Action2 - Draw Moment Arm3 - Calculate Moment Arm4 - Calculate Torque

1.25 sin 1.0557o md m

EXAMPLE:

Calculate the torque produced bythe force acting on the rod as shown.

pivot

F = 16N

57o

1.25m

dline of action

= F·d = 16N·1.05m = 16.8Nm

1 - Draw Line of Action2 - Draw Moment Arm3 - Calculate Moment Arm4 - Calculate Torque5 - See Rotation

1.25 sin 1.0557o md m

EXAMPLE:

Calculate the torque produced bythe force acting on the rod as shown.

Rotation will be counter-clockwise

pivot

F = 16N

57o

1.25m

dline of action

= F·d = 16N·1.05m = 16.8Nm

CENTER OF GRAVITY:

The center of gravity of an object is the point, (xcg , ycg), from whichthe weight acts.

i i i i i icg cg

i i i i

x m g y m gx y

m g m g

For most ordinary objects all gi are equal. Thus we can write:

i i i icg cg

i i

x m y mx y

m m

When g is considered constant cg is usually referred to as center of mass.

EXAMPLE:

A system of three masses, m1, m2 andm3 are arranged as shown. Calculatethe coordinates of the center of massof this system.

y

x

m1 = 5 kg at (0, 0)

m2 = 4 kg at (0, 3m)

m3 = 3 kg at (4m, 0)

ANSWER:

A system of three masses, m1, m2 andm3 are arranged as shown. Calculatethe coordinates of the center of massof this system.

y

x

m1 = 5 kg at (0, 0)

m2 = 4 kg at (0, 3m)

m3 = 3 kg at (4m, 0)

1 1 2 2 3 3

1 2 3

(5 )(0 ) (4 )(0 ) (3 )(4 )

5 4 3

1

cg

cg

cg

m x m x m xx

m m m

kg m kg m kg mx

kg kg kg

x m

1 1 2 2 3 3

1 2 3

(5 )(0 ) (4 )(3 ) (3 )(0 )

5 4 3

1

cg

cg

cg

m y m y m yy

m m m

kg m kg m kg my

kg kg kg

y m

MOMENT OF INERTIA:

Objects have an intrinsic resistance to changes in their rate ofrotation. This is referred to as rotational inertia and is quantified bymoment of inertia, I.

Moment of inertia depends on two factors: 1. the object’s mass, and 2. how that mass is distribute relative to the axis of rotation.

2i iI m r

For many solid objects there are simple formulas for their momentsof inertia.

225

212

sp

s

here

di k

I

I R

MR

M

Any axis through center.

Axis through center andperpendicular to disk.

EXAMPLE: y

x

m1 = 5 kg at (0, 0)

m2 = 4 kg at (0, 3m)

m3 = 3 kg at (4m, 0)r1

r2

r3

Suppose the three masses in the diagramare fixed together by massless rods and arefree to rotate in the xy plane about theircenter of gravity. The coordinates of theircenter of gravity are (1m, 1m) as calculatedin the previous example.

Calculate the moment of inertia of thesemasses about this axis.

2 2

1

2 2

2

2 2

3

1 0 1 0 1.41

1 0 1 3 2.24

1 4 1 0 3.16

r m m m m m

r m m m m m

r m m m m m

2 2 21 1 2

2

2 3 3

2

2

2

(5 )(1.41 )

60.00

(4 )(2.24 )

(3 )(3.16 )

I m r m r m r

I kg m

k

I kg m

g m

kg m

ANSWER: y

x

m1 = 5 kg at (0, 0)

m2 = 4 kg at (0, 3m)

m3 = 3 kg at (4m, 0)r1

r2

r3

Suppose the three masses in the diagramare fixed together by massless rods and arefree to rotate in the xy plane about theircenter of gravity. The coordinates of theircenter of gravity are (1m, 1m) as calculatedin the previous example.

Calculate the moment of inertia of thesemasses about this axis.

ROTATIONAL DYNAMICS I:

If the net torque acting on an object is not zero, the object willexperience an angular acceleration, a. a is related to torque byNewton’s second law of motion for rotation:

I EXAMPLE:

An object has a moment of inertia of 60 kgm2 and is acted upon by atorque of 20Nm. What is the magnitude of the resulting angularacceleration of the object?

ANSWER:

An object has a moment of inertia of 60 kgm2 and is acted upon by atorque of 20Nm. What is the magnitude of the resulting angularacceleration of the object?

2

2

20

60

0.33 rads

I

N m

I kg m

ROTATIONAL DYNAMICS II:

A rotating object has kinetic energy due to its rotational motion inaddition to whatever kinetic energy it has due to the linear motion ofits center of gravity.

212rotKE I

The work-energy theorem applies in the following way:

lin rot NCKE KE PE W

An object’s total kinetic energy is given by:

tot lin rotKE KE KE

EXAMPLE: A solid ball rolls down a 40o incline from a height of 4m without slipping. The radius of the ball is 20cm and it begins from rest. What is the linear speed of the ball at the bottom of the incline?

ANSWER: A solid ball rolls down a 40o incline from a height of 4m without slipping. The radius of the ball is 20cm and it begins from rest. What is the linear speed of the ball at the bottom of the incline?

, ,

21, 2

21, 2

22 21 2 1

2 5 52

0

0

0

Lin Rot

Lin i Rot i

Lin f

Rot f

i f

KE KE PE

KE KE

KE mv

KE I

vmr mv

r

PE mgh PE

Basic Equation

Values and expressions for initial and final quantities

Continued on next slide

2

2 21 12 5

2710

0

10 9.8 410

7 7

7.48 m

m

s

s

mv mv mgh

v gh

v

mghv

Substitutions made to produce working equation

Simplify and solve for v

Continues from previous slide

The speed of the ball does not depend on its mass nor on its radius. It even does not depend on the angle of the incline, just the height from which it starts.

Do the same calculation with a cylinder and determine which will reach the bottom of the incline first, the sphere or the cylinder if released together.

ANGULAR MOMENTUM:

A rotating object has rotational momentum called angular momentum,L.

L I

Angular momentum and linear momentum are different physicalquantities and thus do not add.

Whenever a system’s rotational motions are governed exclusively byinternal torques, the system’s total angular momentum will beconstant. Under such circumstances angular momentum is conserved. Li = Lf.

EXAMPLE: A merry-go-round on a playground has a radius or 1.5 m and a mass of 225 kg. One child is sitting on its outer edge as it rotates 1 rev/s. If the mass of the child is 50 kg, what will the new rotation rate be when the child crawls half way to the center of the merry-go-round?

ANSWER: A merry-go-round on a playground has a radius or 1.5 m and a mass of 225 kg. One child is sitting on its outer edge as it rotates 1 rev/s. If the mass of the child is 50 kg, what will the new rotation rate be when the child crawls half way to the center of the merry-go-round?

, ,

2 2 2 21 1, ,2 2

2 21,2

2 21,2

2 212

2 212

225 1.5 50 1.52

225 1.5 50

8.17 1.3

.75

or

i f

system i i system f f

mgr mgr c c i i mgr mgr c c f f

mgr mgr c c if i

mgr mgr

rad re

c c f

radf s

vf s s

L L

I I

m r m r m r m r

m r m r

m r m r

kg m kg m

kg m kg

f Hz

m

EQUILIBRIUM:

The forces acting on an object in equilibrium must satisfy both thefirst and second conditions of equilibrium.

0 0

0

x y

All

F F

An object in equilibrium has no linear and no angular accelerations. That does not mean it isn’t moving, just not accelerating. When an object is stationary it is said to be in static equilibrium. Many of the equilibrium problems here will be static equilibrium problems.

THE SECOND CONDITION OF EQUILIBRIUM:

An object in rotational equilibrium has uniform angular velocity, =constant. This means its angular acceleration, , is zero. In order for to be zero, the net or total torque acting on the object must be zero.

0

i i iAll CCW CW

iAll

i iCCW CW

Solution on next several slides

EXAMPLE:

30o

cable

wall

hinge

45o

rod

A uniform rod of length l = 4mand mass m = 75kg is hinged toa wall at the left and supportedat the right by a cable. Thecable is attached to the rod l/4from its right edge. The rodmakes a 30o angle with thehorizontal and the cable makesa 45o angle with the rod. Calculate the tension in thecable and the force exerted onthe rod by the hinge.

Add Force Vectors: T = Tension W = Weight R = Reaction

T

W

R

STEP 1:

30o

cable

wall

hinge

45o

rod

Add Force Vectors: T = Tension W = Weight R = ReactionAdd Needed Angles

STEP 1:

210O

165O

pivot30o

cable

wall

hinge

45orod

T

W

R

30o

cable

wall

hinge

45o

3l/4= 3m

l/2 = 2m

roddT

dW

pivot

Add Force Vectors: T = Tension W = Weight R = ReactionAdd Needed AnglesAdd Pivot and Moment Arms: dT = Moment Arm For Tension

dW = Moment Arm For Weight

T

W

R

STEP 1:

210O

165O

STEP 2: Force analysis

Force X Y

T Tcos 165o Tsin 165o

W 0 -735N

R Rx Ry

cos165 0

735 sin165 0

ox x

oy y

F R T

F R N T

First Condition of Equilibrium:On the next slide these force equations will be used to solve for Rx and Ry.

STEP 3: Torque Analysis

Force d Torque Direction

T (3m)sin45o T(2.1213m) CCW 2.1213m

W (2m)cos30o 1273.06Nm CW 1.7321m

R 0 0 -----

Second Condition of Equilibrium:

2.1213 12

600.13

73.06

CCW CW

T m Nm

T N

From force equations on previousslide and using T=600.13N:

Rx=579.68N Ry=579.67N

Then using the PythagoreanTheorem and tan-1:

R = 819.8N = 45o ANSWERANSWER