Ch. 8Ch. 8

Rotational Motion and Rotational Motion and EquilibriumEquilibrium

Ch. 8 OverviewCh. 8 Overview

Rolling MotionRolling Motion TorqueTorque Rotational EquilibriumRotational Equilibrium Rotational DynamicsRotational Dynamics Rotational Kinetic EnergyRotational Kinetic Energy Conservation of Angular MomentumConservation of Angular Momentum

Tangential VariablesTangential Variables

To describe the connection between To describe the connection between linear and rotational motion, we use linear and rotational motion, we use tangential variables.tangential variables.

Arc lengthArc length Tangential speedTangential speed Tangential accelerationTangential acceleration

Arc LengthArc Length

ΔΔs = rs = rΔθΔθ

Δθ

r

Δs

Tangential SpeedTangential Speed

t

rvt

t

svt

Δθ

r

Δs

vt

but ω=Δθ/Δt so rvt

t

rvt

Ex. The turntable turns at 33 1/3 Ex. The turntable turns at 33 1/3 rpm. Find the angular velocity of the rpm. Find the angular velocity of the motor if the radius of the small pulley motor if the radius of the small pulley is 1.27 cm. is 1.27 cm.

Gear RatioGear Ratio

v is the same for both wheelsv is the same for both wheels rr11ωω11 = r = r22ωω22

ωω22 = r = r11/r/r22 ωω11

γγ = r = r11/r/r22 is called the gear ratio is called the gear ratio

r1 r2

v

A ladybug sits at the outer edge of a merry go-A ladybug sits at the outer edge of a merry go-round, and a gentleman bug sits halfway round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-between her and the axis of rotation. The merry-go-round makes a complete revolution once each go-round makes a complete revolution once each second. The gentleman bug’s angular speed issecond. The gentleman bug’s angular speed is

1.1. ½ the ladybug’s½ the ladybug’s

2.2. twice the ladybug’stwice the ladybug’s

3.3. equal to the equal to the ladybug’sladybug’s

4.4. cannot be cannot be determineddetermined

1 2 3 4

0% 0%0%0%

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A ladybug sits at the outer edge of a merry go-A ladybug sits at the outer edge of a merry go-round, and a gentleman bug sits halfway round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-between her and the axis of rotation. The merry-go-round makes a complete revolution once each go-round makes a complete revolution once each second. The gentleman bug’s tangential speed issecond. The gentleman bug’s tangential speed is

1.1. ½ the ladybug’s½ the ladybug’s

2.2. twice the ladybug’stwice the ladybug’s

3.3. equal to the equal to the ladybug’sladybug’s

4.4. cannot be cannot be determineddetermined

1 2 3 4

0% 0%0%0%

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Tangential AccelerationTangential Acceleration

If the tangential speed If the tangential speed changes, we can define a changes, we can define a tangential accelerationtangential acceleration

r

t

r

t

va tt

Rolling MotionRolling Motion

If a car rolls without If a car rolls without slipping, then the slipping, then the distance the car distance the car travels = the arc travels = the arc length turned by length turned by the wheelthe wheel

d = s = rd = s = rθθ The condition for The condition for

rolling w/o slipping rolling w/o slipping can then be can then be expressed as v = rexpressed as v = rωω

How does the instantaneous How does the instantaneous velocity of the two points velocity of the two points compare?compare?

1.1. vvAA = v = vBB

2.2. vvAA > v > vBB

3.3. vvAA < v < vBB

4.4. Cannot be Cannot be determineddetermined

B

1 2 3 4

0% 0%0%0%

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Rolling w/o SlippingRolling w/o Slipping

The instantaneous The instantaneous velocity at the velocity at the point of contact is point of contact is 00

A perfect wheel A perfect wheel has no kinetic has no kinetic frictionfriction v = rω

v = 2 rω

v = 0

Ex. A car starts from rest and Ex. A car starts from rest and accelerates to 15 m/s in a time of accelerates to 15 m/s in a time of 5.0 s. a) Sketch the situation. 5.0 s. a) Sketch the situation. b) Find the car’s average b) Find the car’s average acceleration during the 5.0 s. acceleration during the 5.0 s. c) Find the angular velocity of the c) Find the angular velocity of the tires at t = 5.0 s if their diameter is tires at t = 5.0 s if their diameter is 16”. d) Find the angular 16”. d) Find the angular acceleration of the wheel during the acceleration of the wheel during the 5.0 s.5.0 s.

TorqueTorque A torque produces an angular A torque produces an angular

accelerationacceleration Torque = Torque = ττ = rF sin = rF sinθθ = F = Fperpendicularperpendicular ∙r ∙r If the torque produces a ccw rotation If the torque produces a ccw rotation

+, +,

cw -cw -r

F

θ

Ex. A force of 10.0 N is applied Ex. A force of 10.0 N is applied along the tangent of a wheel of along the tangent of a wheel of radius .50 m. a) sketch the radius .50 m. a) sketch the situation. b) Find the net situation. b) Find the net torque on the wheel.torque on the wheel.

Ex. Find the net force and torque Ex. Find the net force and torque for each situation shown below.for each situation shown below.

r = .25 mr = .25 m

5.0 N

5.0 N

5.0 N

5.0 N

How do the magnitudes of the torques How do the magnitudes of the torques produced by identical forces for the two produced by identical forces for the two situations shown compare?situations shown compare?

A B

1 2 3 4

0% 0%0%0%

1.1. TTAA > T > TBB

2.2. TTAA = T = TBB

3.3. TTAA < T < TBB

4.4. Cannot be Cannot be determineddetermined

11 22 33 44 55

How do the magnitudes of the torques How do the magnitudes of the torques produced by identical forces for the two produced by identical forces for the two situations shown compare?situations shown compare?

AB

1 2 3 4

0% 0%0%0%

1.1. TTAA > T > TBB

2.2. TTAA = T = TBB

3.3. TTAA < T < TBB

4.4. Cannot be Cannot be determineddetermined

11 22 33 44 55

Torque Applied to a Rigid BodyTorque Applied to a Rigid Body

A A rigid bodyrigid body is a system where the is a system where the particles are strongly bound to each particles are strongly bound to each other and maintain their relative other and maintain their relative orientation during motionorientation during motion

The net torque is the sum of all the The net torque is the sum of all the indivisual torques applied to the indivisual torques applied to the systemsystem

ττnetnet = = ΣτΣτii = = ΣΣrriiFFii sin sinθθii

Newton’s Second Law Applied to Newton’s Second Law Applied to Rotating ObjectsRotating Objects

ττnetnet = = ΣΣrrii m miiaaii sin sin θθii = = ΣΣrrii m miiaatt

aatt = r = rαα

ττnetnet = = ΣΣrrii m mi i rrii αα = = ΣΣmmi i rrii22 αα

ττnetnet = I = Iαα

Moment of InertiaMoment of Inertia

Moment of Inertia is defined as I Moment of Inertia is defined as I = = ΣΣmmi i

rrii22

Moment of Inertia resists changes in Moment of Inertia resists changes in rotational motionrotational motion

It depends not just on the mass but how It depends not just on the mass but how far the mass is located from the axis of far the mass is located from the axis of rotation rotation

Ex. Find the moment of inertia for Ex. Find the moment of inertia for the following system of objects the following system of objects connected by light rods of the given connected by light rods of the given length and rotating about the given length and rotating about the given axis.axis.

1.0 kg

.50 kg

.25 kg

.75 kg.20 m

.10 m

.075 m

.40 m42°

A hoop and a disk (made of different A hoop and a disk (made of different materials) have identical masses and materials) have identical masses and radii. Which has a greater moment of radii. Which has a greater moment of inertia?inertia?

1 2 3 4

0% 0%0%0%

1.1. HoopHoop

2.2. DiskDisk

3.3. They are the They are the samesame

4.4. Cannot be Cannot be determineddetermined

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Moments of Inertia of Extended Moments of Inertia of Extended ObjectsObjects

Different shaped Different shaped objects have objects have different moments different moments of inertiaof inertia

In general, the In general, the further the mass is further the mass is from the axis, the from the axis, the greater the greater the moment of inertiamoment of inertia

Identical torques are applied to a hoop Identical torques are applied to a hoop and a disk of the same mass and and a disk of the same mass and radius. How do the resulting angular radius. How do the resulting angular accelerations compare?accelerations compare?

1 2 3 4

0% 0%0%0%

1.1. ααhoophoop > α > αdiskdisk

2.2. ααhoophoop = α = αdiskdisk

3.3. ααhoophoop < α < αdiskdisk

4.4. Cannot be Cannot be determineddetermined

11 22 33 44 55

Four engines oriented 90Four engines oriented 90° apart ° apart fire cw fire cw along the tangent of a 1200 kg cylindrical along the tangent of a 1200 kg cylindrical satellite of radius 2.5 m initially at rest. The satellite of radius 2.5 m initially at rest. The engines each exert a thrust of 250 N. engines each exert a thrust of 250 N. a) Sketch the situation. b) Draw a free a) Sketch the situation. b) Draw a free body diagram for the satellite. c) What is body diagram for the satellite. c) What is the net force on the satellite? d) What is the net force on the satellite? d) What is the net torque on the satellite? e) Find the the net torque on the satellite? e) Find the angular acceleration of the satellite. f) If the angular acceleration of the satellite. f) If the rockets fire for 10.0 s, what is the angular rockets fire for 10.0 s, what is the angular velocity of the satellite?velocity of the satellite?

Center of GravityCenter of Gravity The weight of an extended object can The weight of an extended object can

exert a torque on the object called the exert a torque on the object called the gravitational torquegravitational torque

ττgravgrav = = ΣΣrrii m miig = Mg = Mtottotrrcgcg g g

We can calculate the torque as though it We can calculate the torque as though it is due to all the weight Mis due to all the weight Mtottotg acting at a g acting at a single point rsingle point rcgcg called the center of gravity called the center of gravity

Center of GravityCenter of Gravity

The center of gravity of an object is The center of gravity of an object is the balance point. If the object is the balance point. If the object is supported at the center of gravity, supported at the center of gravity, the gravitational torque on it is zero.the gravitational torque on it is zero.

For a uniform object, the center of For a uniform object, the center of gravity coincides with the center of gravity coincides with the center of the objectthe object

Static EquilibriumStatic Equilibrium

An object in An object in static equilibriumstatic equilibrium has a has a velocity of 0 and an angular velocity velocity of 0 and an angular velocity of 0of 0

Which of the following is true Which of the following is true about an object in static about an object in static equilibrium? equilibrium? (TPS)(TPS)

1 2 3 4 5

0% 0%0%

0%0%

1.1. FFnetnet = 0 = 0

2.2. ττnetnet = 0 = 0

3.3. I = 0I = 0

4.4. 1 and 21 and 2

5.5. 1, 2, and 31, 2, and 3

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Which of the following is true Which of the following is true about an object in static about an object in static equilibrium? equilibrium? (TPS)(TPS)

1 2 3 4 5

0% 0% 0%0%0%

1.1. FFnetnet = 0 = 0

2.2. ττnetnet = 0 = 0

3.3. I = 0I = 0

4.4. 1 and 21 and 2

5.5. 1, 2, and 31, 2, and 3

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Conditions for Static EquilibriumConditions for Static Equilibrium

Translational Equilibrium – Translational Equilibrium – FFnetnet = 0 = 0 Rotational Equilibrium – Rotational Equilibrium – ττnetnet = 0 = 0

If an object is in static equilibrium, If an object is in static equilibrium, then its net torque is 0 regardless then its net torque is 0 regardless about which axis the torque is about which axis the torque is calculatedcalculated

Ex. A beam of negligible mass is Ex. A beam of negligible mass is supported at its center by a fulcrum. supported at its center by a fulcrum. A .50 kg mass hangs .30 m from one A .50 kg mass hangs .30 m from one side of the support. And a .40 kg mass side of the support. And a .40 kg mass hangs from the other side of the beam. hangs from the other side of the beam. a) Sketch the situation. b) Draw a free a) Sketch the situation. b) Draw a free body diagram for the beam. c) Find body diagram for the beam. c) Find the distance form the center of the .40 the distance form the center of the .40 kg beam so that equilibrium will result.kg beam so that equilibrium will result.

Ex. A uniform board of length 2.0 Ex. A uniform board of length 2.0 m and mass 5.0 kg is supported by m and mass 5.0 kg is supported by a fulcrum at .35 m from one end. a fulcrum at .35 m from one end. a) Sketch the situation. b) On a) Sketch the situation. b) On which end must a mass be placed which end must a mass be placed so that equilibrium results. c) How so that equilibrium results. c) How much mass must be placed at that much mass must be placed at that end of the board so that equilibrium end of the board so that equilibrium results?results?

Rotational Work and EnergyRotational Work and Energy

A spool with a string can be wound A spool with a string can be wound up to lift a weightup to lift a weight

Rotating objects can do workRotating objects can do work

Rotational Kinetic EnergyRotational Kinetic Energy

In a rotating object each atom In a rotating object each atom instantaneously moves along the instantaneously moves along the tangenttangent

Consequently, objects have kinetic Consequently, objects have kinetic energy when they rotateenergy when they rotate

Rotational Kinetic EnergyRotational Kinetic Energy The total kinetic energy of a rotating object The total kinetic energy of a rotating object

is KEis KErotrot = = ΣΣ1/2 m1/2 miivvii22

But we can relate the tangential speed to But we can relate the tangential speed to the angular velocity vthe angular velocity vtt = = ωω r r

KEKErotrot = = ΣΣ1/2 m1/2 miiωω22rrii22 = ( = (ΣΣ1/2 m1/2 miirrii

22) ) ωω22

KEKErotrot =1/2 I =1/2 Iωω22

Rotational Work Energy TheoremRotational Work Energy Theorem

Rotational work changes rotational Rotational work changes rotational KEKE

τθτθ = = ΔΔ1/2 I 1/2 I ωω22

FPE - Ex. A hoop of mass 0.5 kg FPE - Ex. A hoop of mass 0.5 kg and radius .25 m is initially rotating and radius .25 m is initially rotating at 20.0 sat 20.0 s-1-1. Friction applies a torque . Friction applies a torque of -2.0 Nm bringing the hoop to rest. of -2.0 Nm bringing the hoop to rest. a) Sketch the situation. b) Find a) Sketch the situation. b) Find the amount of work done by the the amount of work done by the friction. c) Through what angular friction. c) Through what angular displacement does the hoop turn?displacement does the hoop turn?

Total Kinetic Energy of a Rolling Total Kinetic Energy of a Rolling ObjectObject

If an object slides it has KE = ½ mvIf an object slides it has KE = ½ mv22

If an object rotates it has KE = ½ IIf an object rotates it has KE = ½ Iωω22

If an object is rolling it has both If an object is rolling it has both translational (sliding) and rotational translational (sliding) and rotational kinetic energykinetic energy

KEKErollingrolling = ½ mv = ½ mv22 + ½ I + ½ Iωω22

A hoop and disk of the same mass and A hoop and disk of the same mass and radius are released from rest at the top of radius are released from rest at the top of a ramp. If both roll without slipping, which a ramp. If both roll without slipping, which will reach the bottom of the ramp first? will reach the bottom of the ramp first? (TPS)(TPS)

1 2 3 4

0% 0%0%0%

1.1. HoopHoop

2.2. DiskDisk

3.3. TieTie

4.4. Cannot be Cannot be determineddetermined

11 22 33 44 55

A hoop and disk of the same mass and A hoop and disk of the same mass and radius are released from rest at the top of radius are released from rest at the top of a ramp. If both roll without slipping, which a ramp. If both roll without slipping, which will reach the bottom of the ramp first? will reach the bottom of the ramp first? (TPS)(TPS)

1 2 3 4

0% 0%0%0%

1.1. HoopHoop

2.2. DiskDisk

3.3. TieTie

4.4. Cannot be Cannot be determineddetermined

11 22 33 44 55

A hoop and solid sphere of the same mass A hoop and solid sphere of the same mass and radius are released from rest at the and radius are released from rest at the top of a ramp. If both roll without slipping, top of a ramp. If both roll without slipping, which will reach the bottom of the ramp which will reach the bottom of the ramp first? first? (TPS)(TPS)

1 2 3 4

0% 0%0%0%

1.1. SphereSphere

2.2. HoopHoop

3.3. TieTie

4.4. Cannot be Cannot be determineddetermined

11 22 33 44 55

A hoop and solid sphere of the same mass A hoop and solid sphere of the same mass and radius are released from rest at the and radius are released from rest at the top of a ramp. If both roll without slipping, top of a ramp. If both roll without slipping, which will reach the bottom of the ramp which will reach the bottom of the ramp first? first? (TPS)(TPS)

1 2 3 4

0% 0%0%0%

1.1. SphereSphere

2.2. HoopHoop

3.3. TieTie

4.4. Cannot be Cannot be determineddetermined

11 22 33 44 55

A solid sphere and disk of the same mass A solid sphere and disk of the same mass and radius are released from rest at the and radius are released from rest at the top of a ramp. If both roll without slipping, top of a ramp. If both roll without slipping, which will reach the bottom of the ramp which will reach the bottom of the ramp first? first? (TPS)(TPS)

1 2 3 4

0% 0%0%0%

1.1. SphereSphere

2.2. DiskDisk

3.3. TieTie

4.4. Cannot be Cannot be determineddetermined

11 22 33 44 55

A solid sphere and disk of the same mass A solid sphere and disk of the same mass and radius are released from rest at the and radius are released from rest at the top of a ramp. If both roll without slipping, top of a ramp. If both roll without slipping, which will reach the bottom of the ramp which will reach the bottom of the ramp first? first? (TPS)(TPS)

1 2 3 4

0% 0%0%0%

1.1. SphereSphere

2.2. DiskDisk

3.3. TieTie

4.4. Cannot be Cannot be determineddetermined

11 22 33 44 55

Angular MomentumAngular Momentum

The angular moment of a rigid body The angular moment of a rigid body rotating about an axis isrotating about an axis is

L = IL = Iωω

SI Units?SI Units?

Ex. A thin rectangular sheet of Ex. A thin rectangular sheet of mass .50 kg and width mass .50 kg and width perpendicular to the axis of .20 m perpendicular to the axis of .20 m rotates about an axis through its rotates about an axis through its center at 15 scenter at 15 s-1-1. a) Sketch the . a) Sketch the situation. b) Find the angular situation. b) Find the angular momentum of the board. c) What momentum of the board. c) What would we need to do to change would we need to do to change the angular momentum of the the angular momentum of the board?board?

Angular Momentum of a System of Angular Momentum of a System of ObjectsObjects

If we have a system of objects thanIf we have a system of objects than

LLtotaltotal = = ΣΣLLii

To change the angular momentum of To change the angular momentum of the system we exert a torque the system we exert a torque ΔΔL/L/ΔΔt = t = ττ

ττtotaltotal = = Σ τΣ τii = = Σ τΣ τi, exti, ext + + Σ τΣ τi, inti, int

But But Σ τΣ τi, inti, int = 0 so = 0 so ττtotaltotal = = Σ τΣ τi, exti, ext = = ττnet, extnet, ext

Conservation of Angular Conservation of Angular MomentumMomentum

ΔΔLLtotaltotal//ΔΔt = t = ττnet, extnet, ext

If If ττnet, extnet, ext = 0 then = 0 then ΔΔLLtotaltotal//ΔΔt = 0t = 0

If the net external torque on a If the net external torque on a system is 0, then the total angular system is 0, then the total angular momentum of the system is momentum of the system is conservedconserved

Ex. A figure skater is spinning on ice. a) Ex. A figure skater is spinning on ice. a) Describe the system. b) Are there any Describe the system. b) Are there any external torques acting on the system? c) If external torques acting on the system? c) If she brings her arms in, is she exerting an she brings her arms in, is she exerting an internal or external torque. d) What internal or external torque. d) What happens to her angular velocity when she happens to her angular velocity when she brings her arms in? e) Explain in terms of brings her arms in? e) Explain in terms of conservation of angular momentum.conservation of angular momentum.

Ex. A disk of mass .45 kg and radius .20 m Ex. A disk of mass .45 kg and radius .20 m is initially spinning at .40 sis initially spinning at .40 s-1-1 about an axis about an axis through the center of the circular face. A through the center of the circular face. A hoop of the same mass and radius is hoop of the same mass and radius is dropped from rest concentrically onto the dropped from rest concentrically onto the disk so that the two stick together. a) disk so that the two stick together. a) Sketch the situation. b) Is the torque Sketch the situation. b) Is the torque between the hoop and disk an external or between the hoop and disk an external or internal torque to the system of hoop and internal torque to the system of hoop and disk? c) Is angular momentum conserved? disk? c) Is angular momentum conserved? d) Find the final angular velocity of the d) Find the final angular velocity of the system.system.