Unit 8. Center of Mass A point that represents the average location for the total mass of a system...

Unit 8

Center of Mass• A point that represents the average location

for the total mass of a system• For symmetric objects, made from uniformly

distributed material• Center of mass = Geometric center

21

2211

mm

xmxmxcm

Center of Mass & Motion

• Center of mass follows a projectile path

Rotational Motion

• Focuses on pure rotational motion• Motion that consists of rotation

about a fixed axis• Points on a rigid object move in

circular paths around an axis of rotation

• Examples:• Ferris Wheel; CD

Angular Position

• In the study of rotational motion, position is described using angles (θ)

• Angular Position = The amount of rotation from a reference point• Counterclockwise rotation = positive

angle• Clockwise rotation = negative angle

• Units = radians

Angular Position

r

s

Radius

length Arcradians)(in

For a full revolution:

360rad 2 rad 22

r

r

Angular Displacement

• In the study of rotational motion, angular displacement is also described using angles (θ)

• Change in angular position• Δθ = θf − θi

• Δθ = angular displacement• θf = final angular position• θi = initial angular position• Units = radians (rad)

• Counterclockwise rotation = Positive• Clockwise rotation = Negative

Angular Displacement

The angle through which the object rotates is called theangular displacement.

o

PROBLEM: Adjacent Synchronous Satellites

Synchronous satellites are put into an orbit whose radius is 4.23×107 m.

If the angular separation of the twosatellites is 2 degrees, find the arc length that separates them.

rad 0349.0deg360

rad 2deg00.2

€

s = rθ = 4.23×107m( ) 0.0349 rad( )

=1.48 ×106m (920 miles)

r

s

Radius

length Arcradians)(in

ANSWER

o

How we describe the rate at which angular displacement is changing…

Angular Velocity

timeElapsed

ntdisplacemeAngular locity angular ve Average

ttt o

o

SI Unit of Angular Velocity: radian per second (rad/s)

Average Angular Velocity

Gymnast on a High Bar

A gymnast on a high bar swings throughtwo revolutions in a time of 1.90 s.

Find the average angular velocityof the gymnast.

Problem

€

=−2.00 rev2π rad

1 rev

⎛

⎝ ⎜

⎞

⎠ ⎟= −12.6 rad

srad63.6s 90.1

rad 6.12

Answer

Problem

It takes a motorcycle rider 2 seconds to make a counterclockwise lap around a track. What is his average angular velocity?

ttt o

o

€

2π rad

2 s= π rad s

Changing angular velocity means that an angular acceleration is occurring.

DEFINITION OF AVERAGE ANGULAR ACCELERATION

ttt o

o

timeElapsed

locityangular vein Change on acceleratiangular Average

SI Unit of Angular acceleration: radian per second squared (rad/s2)

Angular Acceleration

A Jet Revving Its Engines

As seen from the front of the engine, the fan blades are rotating with an angular speed of -110 rad/s. As theplane takes off, the angularvelocity of the blades reaches-330 rad/s in a time of 14 s.

Find the angular acceleration, assuming it tobe constant.

Problem

2srad16s 14

srad110srad330

Answer

ttt o

o

Problem

Over the course of 1 hour, what is (a) The angular displacement(b) The angular velocity and (c) The angular acceleration of the

minute hand?

AnswerVARIABLES:Elapsed Time Δt = 1 h 3600 sAngular Displacement ΔθAngular Velocity ωAngular Acceleration α

STEP-BY-STEP SOLUTION* Angular Displacement

Δθ = -2πrad (Minute hand travels clockwise one revolution)

Δt = 1 h 3600 s

* Angular Velocityω = Δθ/Δt = -2πrad/3600s = -((2)(3.14))/3600 = -1.75 x 10-3

rad/s

* Since the angular velocity is constant, angular acceleration is zero

Problem

At a particular instant, a potter's wheel rotates clockwise at 12.0 rad/s; 2.50 seconds later, it rotates at 8.50 rad/s clockwise. Find its average angular acceleration during the elapsed time.

Answer

Average Angular Acceleration:α = Δω / Δt = (-8.5 rad/s) – (-12 rad/s) / 2.5 s

= 1.4 rad/s2

Did NOT do slides beyond this point

Angular Velocity Vector

Angular Velocity Vector:* Parallel to the axis of rotation* Magnitude of the angular velocity vector is

proportional to the angular speed- Faster the object rotates longer the vector

* Direction of the angular velocity vector determined by the right-hand rule

Right-Hand Rule

Right-hand Rule:* Used to determine the direction of the angular

velocity vector* Fingers of the right hand curl in the direction of

rotation* Thumb then points in the direction of the

angular velocity vector

Right-Hand Rule: Grasp the axis of rotation with your right hand, so that your fingers circle the axisin the same sense as the rotation.

Your extended thumb points along the axis in thedirection of the angular velocity.

Angular Velocity Vector

Angular Acceleration Vector

Angular Acceleration Vector:* If the object is speeding up its angular

velocity vector is increasing in magnitude Angular acceleration vector points in the same direction as the angular velocity vector

* If the object is slowing down its angular velocity vector is decreasing in magnitude Angular acceleration vector points in the opposite direction as the angular velocity vector

Preview Kinetic Books: 10.15

atvv o

tvvx o 21

axvv o 222

221 attvx o

Five kinematic variables:

1. displacement, x

2. acceleration (constant), a

3. final velocity (at time t), v

4. initial velocity, vo

5. elapsed time, t

Recall… The Kinematic Equations For Constant Acceleration

to

to 21

222 o

221 tto

ANGULAR DISPLACEMENT

ANGULAR VELOCITY

ANGULAR ACCELERATION

TIME

Rotational Kinematic

Equations: For Constant Angular

Acceleration

Rotational Kinematic

Equations

1. Make a drawing.2. Decide which directions are to be called positive (+) and negative (-).3. Write down the values that are given for any of the fivekinematic variables.4. Verify that the information contains values for at least threeof the five kinematic variables. Select the appropriate equation.5. When the motion is divided into segments, remember thatthe final angular velocity of one segment is the initial velocity for the next.6. Keep in mind that there may be two possible answers to a kinematics problem.

Reasoning Strategy

PROBLEM: Blending with a Blender

The blades are whirling with an angular velocity of +375 rad/s whenthe “puree” button is pushed in.

When the “blend” button is pushed,the blades accelerate and reach agreater angular velocity after the blades have rotated through anangular displacement of +44.0 rad.

The angular acceleration has a constant value of +1740 rad/s2.

Find the final angular velocity of the blades.

θ α ω ωo t

+44.0 rad +1740 rad/s2 ? +375 rad/s

222 o

srad542rad0.44srad17402srad375

2

22

2

o

Tangential Velocity

Tangential Velocity:* Linear velocity at an instant

- Magnitude = magnitude of linear velocity- Direction: Tangent to circle

* vT = rωvT = tangential speedr = distance to axisω = angular velocityDirection: tangent to circle

Preview Kinetic Books: 10.11

velocityl tangentiaTv

Tangential Velocity

t

rt

r

t

svT

t

rad/s)in ( rvT

Tangential Velocity

Tangential Acceleration

Tangential Acceleration:- Rate of change of tangential speed- Increases with distance from center- Direction of vector is tangent to circle- aT = rα

aT = tangential accelerationr = distance to axisα = angular acceleration

Preview Kinetic Books: 10.12

t

rt

rr

t

vva ooToTT

to

)rad/sin ( 2raT

Tangential Acceleration

PROBLEM: A Helicopter Blade

A helicopter blade has an angular speed of 6.50 rev/s and anangular acceleration of 1.30 rev/s2.For point 1 on the blade, findthe magnitude of

(a) The tangential speed

(b) The tangential acceleration.

srad 8.40rev 1

rad 2

s

rev 50.6

3.00 m 40.8rad s 122m sTv r

22 sm5.24srad17.8m 3.00 raT

22

srad 17.8rev 1

rad 2

s

rev 30.1

rad/s)in ( 2

22

r

r

r

r

va Tc

Centripetal Acceleration

Problem: A Discus Thrower

Starting from rest, the throweraccelerates the discus to a finalangular speed of +15.0 rad/s ina time of 0.270 s before releasing it.During the acceleration, the discusmoves in a circular arc of radius0.810 m.

Find the magnitude of the totalacceleration.

Centripetal Acceleration &

Tangential Acceleration

2

22

sm182

srad0.15m 810.0

rac

2sm0.45

s 0.270

srad0.15m 810.0

t

ω-ωrra o

T

2 22 2 2 2 2182m s 45.0m s 187 m sT ca a a

Rotational Motion Review

• UNIFORM CIRCULAR MOTION:• Example- Spinning Ferris wheel or an orbiting satellite• Object moves in a circular path and at a constant speed

• The object is accelerating, however, because the direction of the object’s velocity is constantly changing

• Centripetal acceleration Directed toward the center of the circle

• Net force causing the acceleration is a centripetal force Also directed toward the center of the circle

• In this section, we will examine a related type of motion The motion of a ROTATING RIGID OBJECT

The Motion of a Rotating Rigid Object• A football spins as it flies through the

air• If gravity is the only force acting on the

football Football spins around a point called its center of mass• As the football moves through the air,

its center of mass follows a parabolic path

Rotational Dynamics

• Study of Rotational Kinematics• Focuses on analyzing the motion of a

rotating object• Determining such properties as its angular

displacement, angular velocity or angular acceleration

• Study of Rotational Dynamics• Explores the origins of rotational motion

* Preview Kinetic Books 11.1

Torque• Every time you open a door, turn of a water faucet, or

tighten a nut with a wrench you are exerting a TURNING FORCE• Turning force produces a TORQUE

• Review…• If you want to make an object move Apply a force

• Forces make things accelerate

• If you want to make an object turn or rotate Apply a torque• Torques produce rotation

Torque

• Torque • A force that causes or opposes rotation• The amount of torque depends on…

• The amount of force• When more force is applied there is

more torque

• The distance from the axis of rotation to the point of force

The amount of torque depends on where and in what direction the force is applied, as well as the location of the axis of rotation.

Torque• For a force applied perpendicularly

r F ZZZZZZZZZZZZZ Z

τ = magnitude of torque (Greek letter tau)r = distance from axis to forceF = force•Direction:

•Torque resulting from a force is + if the rotation is counterclockwise and – if the rotation is clockwise

•Counterclockwise +, clockwise −•S.I. Units: newton-meters (N∙m)

Torque

• Imagine a dog-flap door• Rotates on a hinge• Allowing pets to enter and leave a house at

will

Axis of Rotation

Force: * If the dog pushed on the door with the same force but at a point closer to the hinge, the door would be more difficult to rotate

Lever Arm: Perpendicular distance from the axis of rotation to a line draw along the direction of the force

Lever A

rm

Torque, angle & lever arm

• * Preview Kinetic Books 11.2•Forces do not have to be perpendicular to an object to cause the object to rotate

τ = r × Fτ = rF sin θ

τ = Torque (N x m) Vector quantityr = Position vector (m) (AKA: Displacement)F = Force (N) Vector quantityθ = angle between r and F

Calculating Torque using the Lever Arm

• Torque can also be calculated using the concept of a lever arm• Lever arm is the perpendicular distance

from the axis of rotation to the line containing the force vector• “Line of action”

τ = F (r sin θ)* r sin θ = lever arm

Mathematically:τ = F (r sin θ) is the same as τ = rF

sin θ

Torque

Problem

• A basketball is being pushed by two players during a tip-off. One player exerts an upward force of 15 N at a perpendicular distance of 14 cm from the axis of rotation. The second player applies a downward force of 11 N at a perpendicular distance of 7 cm from the axis of rotation. Find the net torque acting on the ball about its center of mass.

Answer

• Given:• F1 = 15N r1 = 0.14m

• F2 = 11N r2 = 0.07m

• τnet = ?

τ = F1r1 + F2r2

(** The factor sin Θ is not included because each distance is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force)

Answer Continued

• τ = F1r1 + F2r2

• * Use the standard convention of signs…• τ = (15N)(-0.14m) + (-11N)(0.070m)

• τ = -2.9 Nxm

• Net torque is negative, so the ball rotates in a clockwise direction

The Achilles Tendon

The tendon exerts a force of magnitude 790 N. Determine the torque (magnitude and direction) of this force about the ankle joint.

790 N

23.6 10 m 790 sin145

16.3 N m

N

Answer

τ = rF sin θ