Fisika Dasar - Minggu 08 - Torsi

Review Problem: A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her?

Sample Review ProblemSample Review Problem

An engineer wishes to design a curved exit ramp for a toll road in such a way that a car will not have to rely on friction to round the curve without skidding. She does so by banking the road in such a way that the force causing the centripetal acceleration will be supplied by the component of the normal force toward the center of the circular path. Find the angle at which the curve should be banked if a typical car rounds it at a 50.0-m radius and a speed of 13.4 m/s.

Rotational Equilibrium Rotational Equilibrium andand

Rotational DynamicsRotational Dynamics

TorqueTorque► Consider force Consider force

required to open door. required to open door. Is it easier to open the Is it easier to open the door by door by pushing/pulling pushing/pulling awayaway from hinge or from hinge or closeclose to to hinge?hinge?

closeclose to hinge to hinge

awayaway from from hingehinge

Farther Farther from hinge, from hinge, larger larger rotational rotational effect!effect!

Physics concept: Physics concept: torquetorque

TorqueTorque

► TorqueTorque, , is the , , is the tendency of a forcetendency of a force to rotate an object about some axisto rotate an object about some axis

is the torqueis the torque dd is the is the leverlever arm arm (or (or momentmoment arm) arm) FF is the force is the force

Fd

Door example:

Lever ArmLever Arm

► The lever arm, d, The lever arm, d, is the is the shortest shortest (perpendicular)(perpendicular) distance from the axis distance from the axis of rotationof rotation to a line to a line drawn along the the drawn along the the direction of the forcedirection of the force

d = L sin Φd = L sin Φ

► It is It is not necessarilynot necessarily the the distance between the distance between the axis of rotation and axis of rotation and point where the force is point where the force is appliedapplied

Direction of TorqueDirection of Torque

► Torque is a vector quantityTorque is a vector quantity The direction is The direction is perpendicularperpendicular

to the to the planeplane determined by determined by the the lever armlever arm and the and the forceforce

Direction and sign: Direction and sign: If the turning tendency of the If the turning tendency of the

force is force is counterclockwisecounterclockwise, the , the torque will be positivetorque will be positive

If the turning tendency is If the turning tendency is clockwiseclockwise, the torque will be , the torque will be negativenegative

Direction of torque: out of page

UnitsUnits

SISI Newton meter (Nm)Newton meter (Nm)

US CustomaryUS Customary Foot pound (ft lb)Foot pound (ft lb)

An Alternative Look at An Alternative Look at TorqueTorque

► The force could also be The force could also be resolved into its resolved into its x- and x- and y-componentsy-components The x-component, The x-component, F cos F cos

ΦΦ, produces , produces 0 torque0 torque The y-component, The y-component, F sin F sin

ΦΦ, produces a non-zero , produces a non-zero torquetorque sinLF

F is the forceF is the force

L is the distance along the objectL is the distance along the object

Φ is the angle between force and objectΦ is the angle between force and object

LL

Question 1Question 1

You are trying to open a door that is stuck by pulling on the doorknob in a direction perpendicular to the door. If you instead tie a rope to the doorknob and then pull with the same force, is the torque you exert increased? Will it be easier to open the door?

1. No2. Yes


You are using a wrench and trying to loosen a rusty nut. Which of the arrangements shown is most effective in loosening the nut? List in order of descending efficiency the following arrangements:

2, 1, 4, 3 or2, 4, 1, 3

What if two or more different forces act on lever arm?

Net TorqueNet Torque

►The The net torquenet torque is the sum of all the is the sum of all the torques produced by all the forcestorques produced by all the forces

Remember to account for the direction of Remember to account for the direction of the tendency for rotationthe tendency for rotation►CounterclockwiseCounterclockwise torques are positive torques are positive►ClockwiseClockwise torques are negative torques are negative

Example 1:Example 1:

Given:

weights: w1= 500 Nw2 = 800 N

lever arms: d1=4 m d2=2 m

Find:

= ?

1. Draw all applicable forces

(500 )(4 ) ( )(800 )(2 )

2000 1600

400

N m N m

N m N m

N m

2. Consider CCW rotation to be positive

500 N 800 N

4 m 2 m

Rotation would be CCW

N

Determine the net torque:

Where would the 500 N person have to be relative to fulcrum for zero torque?

Example 2:Example 2:

Given:

weights: w1= 500 Nw2 = 800 N

lever arms: d1=4 m = 0

Find:

d2 = ?

1. Draw all applicable forces and moment arms

2

2 2

(800 )(2 )

(500 )( )

800 2 [ ] 500 [ ] 0 3.2

RHS

LHS

N m

N d m

N m d N m d m

500 N 800 N

2 md2 m

According to our understanding of torque there would be no rotation and no motion!

N’ y

What does it say about acceleration and force?

Thus, according to 2nd Newton’s law F=0 and a=0! NN

NNNFi

1300'

0)800(')500(

Torque and EquilibriumTorque and Equilibrium

► First Condition of EquilibriumFirst Condition of Equilibrium► The The net external net external forceforce must be must be zerozero

This is a necessary, but not sufficient, This is a necessary, but not sufficient, condition to ensure that an object is in condition to ensure that an object is in complete mechanical equilibriumcomplete mechanical equilibrium

This is a statement of This is a statement of translational translational equilibriumequilibrium

► Second Condition of EquilibriumSecond Condition of Equilibrium► The The net external net external torquetorque must be must be zerozero

► This is a statement of This is a statement of rotational equilibriumrotational equilibrium

0

0 and 0x y

F

F F

BBBBBBBBBBBBBB

0

Axis of RotationAxis of Rotation

► So far we have chosen obvious axis of rotationSo far we have chosen obvious axis of rotation► If the object is If the object is in equilibriumin equilibrium, it does not , it does not

matter where you put the axis of matter where you put the axis of rotation for calculating the net torquerotation for calculating the net torque The location of the axis of rotation is The location of the axis of rotation is

completely arbitrarycompletely arbitrary Often the nature of the problem will suggest Often the nature of the problem will suggest

a convenient location for the axisa convenient location for the axis When solving a problem, you When solving a problem, you mustmust specify specify

an axis of rotationan axis of rotation►Once you have chosen an axis, you must maintain Once you have chosen an axis, you must maintain

that choice consistently throughout the problemthat choice consistently throughout the problem

Center of GravityCenter of Gravity

►The force of gravity acting on an The force of gravity acting on an object must be consideredobject must be considered

► In finding the torque produced by the In finding the torque produced by the force of gravity, all of the weight of the force of gravity, all of the weight of the object can be considered to be object can be considered to be concentrated at concentrated at oneone point point

Calculating the Center of Calculating the Center of GravityGravity

1.1. The object is divided up into a large The object is divided up into a large number of very small particles of number of very small particles of weight (mg)weight (mg)

2.2. Each particle will have a set of Each particle will have a set of coordinates indicating its location (x,y)coordinates indicating its location (x,y)

3.3. The The torquetorque produced by produced by each particleeach particle about the axis of rotation is about the axis of rotation is equalequal to its to its weight times its lever armweight times its lever arm

4.4. We wish to locate the point of We wish to locate the point of application of the application of the single forcesingle force , whose , whose magnitude is equal to the weight of the magnitude is equal to the weight of the object, and whose effect on the rotation object, and whose effect on the rotation is the same as all the individual is the same as all the individual particles.particles.

► This point is called the This point is called the center of gravitycenter of gravity of the objectof the object

Coordinates of the Center of Coordinates of the Center of GravityGravity

► The coordinates of the center of gravity can be The coordinates of the center of gravity can be found from the sum of the torques acting on found from the sum of the torques acting on the individual particles being set equal to the the individual particles being set equal to the torque produced by the weight of the objecttorque produced by the weight of the object

► The center of gravity of a homogenous, The center of gravity of a homogenous, symmetric body must lie on the axis of symmetric body must lie on the axis of symmetry.symmetry.

► Often, the center of gravity of such an object Often, the center of gravity of such an object is the is the geometricgeometric center of the object. center of the object.

i

iicg

i

iicg m

ymyand

m

xmx

Example:Example:

Given:

masses: m1= 5.00 kgm2 = 2.00 kgm3 = 4.00 kg

lever arms: d1=0.500 m d2=1.00

m

Find:

Center of gravity

Find center of gravity of the following system:

m

kg

mkgmkgmkg

mmm

xmxmxm

m

xmx

i

iicg

136.0

0.11

)00.1(00.4)0(00.2)500.0(00.5

321

332211

Experimentally Determining Experimentally Determining the Center of Gravitythe Center of Gravity

► The wrench is hung freely The wrench is hung freely from two different pivotsfrom two different pivots

► The intersection of the lines The intersection of the lines indicates the center of indicates the center of gravitygravity

► A rigid object can be A rigid object can be balanced by a single force balanced by a single force equal in magnitude to its equal in magnitude to its weight as long as the force is weight as long as the force is acting upward through the acting upward through the object’s center of gravityobject’s center of gravity

Equilibrium, once againEquilibrium, once again

►A zero net torque does not mean the A zero net torque does not mean the absence of rotational motionabsence of rotational motion An object that rotates at An object that rotates at uniform angular uniform angular

velocityvelocity can be under the influence of a can be under the influence of a zero net torquezero net torque►This is analogous to the translational situation This is analogous to the translational situation

where a zero net force does not mean the where a zero net force does not mean the object is not in motionobject is not in motion

Example of aExample of aFree Body DiagramFree Body Diagram

► Isolate the object to Isolate the object to be analyzedbe analyzed

►Draw the free body Draw the free body diagram for that diagram for that objectobject Include all the Include all the

external forces external forces acting on the objectacting on the object

Suppose that you placed a 10 m ladder (which weights 100 N) against the wall at the angle of 30°. What are the forces acting on it and when would it be in equilibrium?

ExampleExample

Example:Example:

Given:

weights: w1= 100 Nlength: l=10 mangle: =30° = 0

Find:

f = ?n=?P=?


N.P

P.N

sinPLcosL

mg

6862

118660

2

11000

030302

Torques: Forces:

Nn

mgnF

Nf

PfF

y

x

100

0

6.86

0

2. Choose axis of rotation at bottom corner ( of f and n are 0!)

Note: f = s n, so 866.0100

6.86

N

N

n

fs

mg

So far: net torque was zero.What if it is not?

Torque and Angular Torque and Angular AccelerationAcceleration

► When a rigid object is When a rigid object is subject to a net torque subject to a net torque (≠0),(≠0), it undergoes it undergoes an an angular accelerationangular acceleration

► The angular The angular acceleration is directly acceleration is directly proportional to the net proportional to the net torquetorque The relationship is The relationship is

analogous to ∑F = maanalogous to ∑F = ma►Newton’s Second LawNewton’s Second Law

Torque and Angular Torque and Angular AccelerationAcceleration

so,

:onacceleratitangential

bymultiply,

ra

rmarF

rmaF

t

tt

tt

2mrrFt

torque dependent upon object and axis of rotation. Called moment of inertia I. Units: kg mkg m22

2iirmI

IThe angular acceleration is inversely The angular acceleration is inversely proportional proportional to the analogy of the mass in a rotating to the analogy of the mass in a rotating systemsystem

Example: Moment of Inertia Example: Moment of Inertia of a Uniform Ringof a Uniform Ring

► Image the hoop is Image the hoop is divided into a number of divided into a number of small segments, msmall segments, m11 … …

► These segments are These segments are equidistant from the axisequidistant from the axis

22 MRrmI ii

Other Moments of InertiaOther Moments of Inertia

Newton’s Second Law for a Newton’s Second Law for a Rotating ObjectRotating Object

► The The angular accelerationangular acceleration is is directlydirectly proportional to the proportional to the net net torquetorque

► The The angular accelerationangular acceleration is is inverselyinversely proportional to the proportional to the moment of inertia of the objectmoment of inertia of the object

There is a There is a major differencemajor difference between between moment of inertiamoment of inertia and and massmass: :

► The moment of inertia depends on the quantity of matter The moment of inertia depends on the quantity of matter and and its distributionits distribution in the rigid object. in the rigid object.

► The moment of inertia also depends upon the location of The moment of inertia also depends upon the location of the axis of rotationthe axis of rotation

I

Example:Example:

Consider a flywheel (cylinder pulley) of mass M=5 kg and radius R=0.2 m and weight of 9.8 N hanging from rope wrapped around flywheel.

What are forces acting on flywheel and weight? Find acceleration of the weight.

mg

Example:Example:

Given:

masses: M = 5 kgweight: w = 9.8 Nradius: R=0.2 m

Find:

Forces=?


I

RT

IRT

Forces: Torques:

!Tneed

maTmgF

Tangential acceleration at the edge of flywheel (a=at):

ttt

t

akgam

mkga

I

TRRa

5.22.0

10.0

R

IT

or

aor

2

2

2

2

t

N

T

T

Mg

mg

28.2

5.3

8.9

5.2

5.2

smkg

N

kgm

mga

maakgmg

maTmgF

t

22 10.02

1mkgMRI


A force F is applied to a dumbbell for a time interval t, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater center-of-mass speed?

1. (a)2. (b)3. no difference4. The answer depends on the

rotational inertia of the dumbbell.


A force F is applied to a dumbbell for a time interval t, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater center-of-mass speed?

Force acts the same time: change of momentum is the same. Thus CM speed is the same as well.



Return to our example:Return to our example:

Consider a flywheel (cylinder pulley) of mass M=5 kg and radius R=0.2 m with weight of 9.8 N hanging from rope wrapped around flywheel. What are forces acting on flywheel and weight? Find acceleration of the weight.

mg

If flywheel initially at rest and then begins to rotate, a torque must be present:

tt

I

I

00 since,

Define physical quantity:

t

L

intervaltime

momentumang.inchange LI momentumangular

Angular MomentumAngular Momentum

► Similarly to the relationship between force and Similarly to the relationship between force and momentum in a linear system, we can show the momentum in a linear system, we can show the relationship between torque and angular momentumrelationship between torque and angular momentum

► Angular momentumAngular momentum is defined as L = I ω is defined as L = I ω

► If the net torque is zero, the angular momentum If the net torque is zero, the angular momentum remains constantremains constant

► Conservation of Linear MomentumConservation of Linear Momentum states: states: The angular The angular momentum of a system momentum of a system is conservedis conserved when the net when the net external torque acting on the systems is zero.external torque acting on the systems is zero.

That is, whenThat is, when

t

L

ffiifi IIorLL ,0

t

pF

(compare to )

Return to our example once Return to our example once again:again:

Consider a flywheel (cylinder pulley) of mass M=5 kg and radius R=0.2 m with weight of 9.8 N hanging from rope wrapped around flywheel. What are forces acting on flywheel and weight? Find acceleration of the weight.

mg

Each small part of the flywheel is moving with some velocity. Therefore, each part and the flywheel as a whole have kinetic energy!

2

2

massipulley

ipulleyipulley vKE 2

2

1 IKE pulley

Thus, total KE of the system: 22

2

1

2

1mvIKE pulleytot

Total Energy of Rotating Total Energy of Rotating SystemSystem

► An object rotating about some axis with an An object rotating about some axis with an angular speed, ω, has rotational kinetic energy angular speed, ω, has rotational kinetic energy ½Iω½Iω22

► Energy concepts can be useful for simplifying the Energy concepts can be useful for simplifying the analysis of rotational motionanalysis of rotational motion

► Conservation of Mechanical EnergyConservation of Mechanical Energy

Remember, this is for conservative forces, no dissipative Remember, this is for conservative forces, no dissipative forces such as friction can be presentforces such as friction can be present

fgrtigrt PEKEKEPEKEKE )()(


A force F is applied to a dumbbell for a time interval t, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater energy?

Please fill your answer as question 36 of General Purpose Answer Sheet




A force F is applied to a dumbbell for a time interval t, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater energy?

Since CM speeds are the same, translational kinetic energies are the same. But (b) also rotates, so it also has rotational energy.



Note on problem solving:Note on problem solving:

► The same basic techniques that were used in The same basic techniques that were used in linear motion can be applied to rotational motion.linear motion can be applied to rotational motion. Analogies: Analogies: FF becomes , becomes , mm becomes becomes II and a becomes and a becomes

, v becomes ω and x becomes θ, v becomes ω and x becomes θ

► Techniques for conservation of energy are the Techniques for conservation of energy are the same as for linear systems, as long as you same as for linear systems, as long as you include the rotational kinetic energyinclude the rotational kinetic energy

► Problems involving angular momentum are Problems involving angular momentum are essentially the same technique as those with essentially the same technique as those with linear momentumlinear momentum The moment of inertia may change, leading to a change The moment of inertia may change, leading to a change

in angular momentumin angular momentum

Fisika Dasar - Minggu 08 - Torsi

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