8.2 Torque – 8.4 Equilibrium Revisited – 8.8 Angular …

PHYSICS 149: Lecture 21• Chapter 8: Torque and Angular Momentum

– 8.2 Torque

– 8.4 Equilibrium Revisited

– 8.8 Angular Momentum

Lecture 21 Purdue University, Physics 149 1

Midterm Exam 2• Wednesday, April 6, 6:30 PM – 7:30 PM• Place: PHY 333• Chapters 5 - 8• The exam is closed book.• The exam is a multiple-choice test.• There will be ~15 multiple-choice problems.

– Each problem is worth 10 points.• Note that total possible score for the course is 1,000 points (see the

course syllabus)

• The difficulty level is about the same as the level of textbook problems.

• You may make a single crib sheet– you may write on both sides of an 8.5” × 11.0” sheet

Lecture 20 2Purdue University, Physics 149


Rotational Kinetic Energy • Consider a mass M on the end of a string being spun

around in a circle with radius r and angular frequency ω– Mass has speed v = ω r– Mass has kinetic energy K = ½ M v2 = ½ M ω2 r2

• Consider a disk with radius R and mass M, spinning with angular frequency ω– Each “piece” of disk has speed vi = ωri– Each “piece” has kinetic energy Ki = ½ mi v2 = ½ mi ω2 ri

2

– Combine all the piecesΣKi = Σ ½ mi ω2 r2

= ½ (Σ mi ri2) ω2

= ½ I ω2

• Rotational InertiaI = Σ miri

2 (units kg m2)

M

ri


Rotational Inertia TableFor objects with finite number of masses, use I = Σ m r2. For “continuous” objects, use table below.


Torque• Rotational effect of force. Tells how effective

force is at twisting or rotating an object.τ = ± r Fperpendicular = r F sin θ

τ = ±rperpendicular F rperpendicular = lever arm– Units N m– Sign, CCW rotation is positive

F⊥

Torque• Torque measures the effectiveness of a force for

twisting or turning an object.– Magnitude:

– Sign:• If a torque causes a CCW rotation, assign “+” sign.• If a torque causes a CW rotation, assign “-” sign.

– Direction:• If a torque causes a CCW rotation, the torque’s direction is “out

of” the plane and normal to the plane.• If a torque causes a CW rotation, the torque’s direction is “into”

the plane and normal to the plane.• Torque is a vector quantity.• Units: N⋅m

– But, torque is not a form of energy.• Torque is denoted by τ

beyond the scopeof this class

lever arm



Torque• Rotational effect of force. Tells how effective force is at

twisting or rotating an object.• τ = ± r Fperpendicular = r F sin θ

– Units N m– Sign, CCW rotation is positive

• Work done by torque– Recall W = F d cos θ– For a wheel

W = Ftangential d= Ftangential 2 π r θ / (2 π) (θ in radians)= Ftangential r θ= τ θ

P = W/t = τ θ/t = τ ω


The picture below shows three different ways of using a wrench to loosen a stuck nut. Assume the applied force F is the same in each case.

In which of the cases is the torque on the nut the biggest?

A) Case 1 B) Case 2 C) Case 3

ILQ: Torque

In which of the cases is the torque on the nut the smallest? A) Case 1 B) Case 2 C) Case 3


ILQ 1Two forces produce the same torque. Does it follow that they have the same magnitude?

A) Yes

B) No

ILQ 2Which of the forces in the figure produces anegative (clockwise) torque about the rotation axis indicated?

A) 1 and 2B) 1, 2, and 4C) 3 onlyD) 4 only


ILQ• The figure shows an overhead view of a meter stick that

can pivot about the dot shown at the position marked 20 (for 20 cm). All five horizontal forces on the stick have the same magnitude. Rank those forces according to the “magnitude” of the torque they produce, greatest first.

a) 2 and 3 tie, 1, then 1 and 5 tieb) 1 and 3 tie, 4, then 2 and 5 tiec) All tied) 1, 4, then 2 and 3 tie, 5


ILQ• A heavy box is resting on the floor. You would like to

push the box to tip it over on its side, using the minimumforce possible.

Which of the force vectors in the diagram shows the correct location and direction of the force? Assume enough friction so that the box does not slide; instead it rotates about point P.

a) ab) bc) c

r


Example• The pull cord of a lawnmower engine is wound

around a drum of radius of 6.00 cm. While the cord is pulled with a force of 75 N to start the engine, what magnitude torque does the cord apply to the drum?

– F = 75.0 N– r = 6.00 cm = 0.0600 m– θ = 90°

– τ = F⋅r⋅sinθ = 4.5 N⋅m

75 N

6 cm



Torque ExampleA person raises one leg to an angle of 30 degrees. An ankle weight (89 N) is attached a distance of 0.84 m from her hip. What is the torque due to this weight?

1) Draw Diagram2) τ = F r sin θ

= F r sin(90 – 30)

If she raises her leg higher, the torque due to the weight will

A) IncreaseB) SameC) Decrease

30 F=89 N= 65 N m


Equilibrium• Conditions for Equilibrium

Σ F = 0 Translational EQ (Center of Mass)Σ τ = 0 Rotational EQ

• Can choose any axis of rotation…. Choose Wisely!• A meter stick is suspended at the center. If a 1 kg weight is

placed at x=0. Where do you need to place a 2 kg weight to balance it?A) x = 25 B) x=50 C) x=75 D) x=100 E) Impossible

Σ τ = 09.8 (0.5) – (19.6)d = 0d = 259.8 N 19.6 N

50 cm d

x

y

pivot


Static Equilibrium and Center of Mass

Center of masspivot

d

W=mg

• Gravitational Force Weight = mg– Acts as force at center of mass– Torque about pivot due to gravity τ = mgd – Object not in static equilibrium


A method to find center of mass of an irregular object

Center of mass

pivot d

W=mg

Torque about pivot ≠ 0

Not in equilibrium

Center of mass

pivot

Torque about pivot = 0

Equilibrium

Static Equilibrium


Equilibrium Acts• A rod is lying on a table and has two equal but

opposite forces acting on it. What is the net force on the rod?A) Up B) Down C) Zero

• Will the rod move? A) Yes B) No F

F

y

x

y-direction: Σ Fy = may

+F – F = 0

Yes, it rotates!

Example 8.6• The beam’s weight is 425 N. For equilibrium, what should be the

magnitudes of the forces?

The entire gravitational forceacts at the center of mass(recall in Ch 7).

Choose a rotation axis.

For Στ = 0 (rotational equilibrium), calculate and add all the torques.* torque due to gravity: τg = (lever arm)⋅(force) = +(2.44m / 2)⋅(425N)

= +518.5 N⋅m(We assign “+” sign, because this torque causes a CCW rotation.)

* torque due to F1: τF1 = (lever arm)⋅(force) = –(2.44m–1.00m)⋅(F1) = –(1.44m)⋅(F1)

(We assign “–” sign, because this torque causes a CW rotation.)Στ = τg + τF1 + τF2 = +518.5 N⋅m – (1.44m)⋅(F1) + 0 = 0 F1 = 360 N

For ΣF = 0 (translational equilibrium), calculate and add all the forces.ΣFy = F1 + F2 – mg = 360 N + F2 – 425 N = 0 F2 = 65 N


Example 8.6: Different Axis• The beam’s weight is 425 N. For equilibrium, what should be the

magnitudes of the forces?

The entire gravitational forceacts at the center of mass(recall in Ch 7).

Choose a rotation axis.

For Στ = 0 (rotational equilibrium), calculate and add all the torques.* torque due to gravity: τg = (lever arm)⋅(force) = –(2.44m / 2 –

1.00m)⋅(425N) = – 93.5 N⋅m

(We assign “–” sign, because this torque causes a CW rotation.)* torque due to F2: τF2 = (lever arm)⋅(force) = +(2.44m–1.00m)⋅(F2)

= +(1.44m)⋅(F2)(We assign “+” sign, because this torque causes a CCW rotation.)

Στ = τg + τF1 + τF2 = –93.5 N⋅m + 0 + (1.44m)⋅(F2) = 0 F2 = 65 N

For ΣF = 0 (translational equilibrium), calculate and add all the forces.ΣFy = F1 + F2 – mg = F1 + 65 N – 425 N = 0 F1 = 360 NLecture 21 20Purdue University, Physics 149

ILQ• A 1 kg ball is hung at the end of a rod 1 m long.

If the system balances at a point on the rod one third of the distance from the end holding the mass, what is the mass of the rod?

a) 0.25 kgb) 0.50 kgc) 1.0 kgd) 2.0 kg


ILQ• Find the mass of the tomatoes which make this

mobile a "balanced meal." Assume the rods are massless with the lengths indicated.

a) 0.5 kgb) 1.0 kgc) 1.5 kgd) 2.0 kg


Equilibrium Example• A 50 kg diver stands at the end of a 4.6 m diving board.

Neglecting the weight of the board, what are the forces acted by the supports?

Draw a FBD.Choose a rotation axis.Στ = 0 (rotational equilibrium)

Στ = +(1.2m)⋅(F1) – (4.6m)⋅(50kg⋅9.8m/s2) = 0 F1 = 1880 N

ΣF = 0 (translational equilibrium)ΣFy = F1 – F2 – mg

= 1880 N – F2 – (50kg⋅9.8m/s2) = 0 F2 = 1390 N

mg

F1

F2


ILQ• A child's pull-toy has two wheels attached to an

axle, which has a string wrapped around it. A cut-away view is shown. Assume that friction exists between the wheels and floor. When the string is pulled to the right, the wheels will roll to the

a) right, winding up the string.b) left, unwinding the string.


Example: Dumbbell• Consider a rotational motion around a point P.

– For normal force and F3, θ = 0°.So τ=0 no rotation

– For gravity, θ = 180°.So τ=0 no rotation

– F1 and F2 cause CCW rotations.

– F4 causes a CW rotation.



The HammerYou want to balance a hammer on the tip of your finger, which way is easier

A) Head upB) Head downC) Same

mg

R

Angular acceleration decreases with R!, so large R is easier to balance.

τ = I α

m g R sin(θ) = mR2 αTorque increases with R

Inertia increases as R2

g sin(θ) / R = α


Energy Conservation• Friction causes an object to roll, but if it rolls w/o

slipping friction does NO work!– W = F d cos θ θ is zero for point in contact

• No dissipated work, energy is conserved

• Need to include both translation and rotation kinetic energy. – K = ½ m v2 + ½ I ω2


Translational + Rotational KE• Consider a cylinder with radius R and mass M, rolling w/o

slipping down a ramp. Determine the ratio of the translational to rotational KE.

H

use and

Translational: KT = ½ M v2

Rotational: KR = ½ I ω2

KR = ½ (½ M R2) (V/R)2

= ¼ M v2

= ½ KT

Energy conservation: Ki + Ui = Kf + Uf


Rolling Act• Two uniform cylinders are machined out of

solid aluminum. One has twice the radius of the other.– If both are placed at the top of the same ramp

and released, which is moving faster at the bottom?

A) bigger one B) smaller one C) same

Ki + Ui = Kf + Uf


Define Angular Momentum

Momentum Angular Momentump = mv L = Iω

conserved if ΣFext = 0 conserved if Στext = 0

Vector Vector

units: kg-m/s units: kg-m2/s

For a particle:


Linear and Angular

Linear Angular

Displacement x θ

Velocity v ω

Acceleration a α

Inertia m I

Kinetic Energy ½ m v2 ½ I ω2

Newton’s 2nd

Law F = ma τ = Iα

Momentum p = mv L = Iω


Act: Two Disks• A disk of mass M and radius R rotates around the z

axis with angular velocity ωi. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity ωf.

A) ωf = ωi B) ωf = ½ ωi C) ωf = ¼ ωi

ωi

z

ωf

z


Act: Two Disks• First realize that there are no external torques

acting on the two-disk system.– Angular momentum will be conserved!

ω0

z2

1

ωf

z


DemoYou are sitting on a freely rotating bar-stool with your arms stretched out and a heavy glass mug in each hand. Your friend gives you a twist and you start rotating around a vertical axis though the center of the stool. You can assume that the bearing the stool turns on is frictionless, and that there is no net external torque present once you have started spinning. You now pull your arms and hands (and mugs) close to your body.

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8.2 Torque – 8.4 Equilibrium Revisited – 8.8 Angular …

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