# Physics 231 Lecture 19 Torques and rotational dynamicslynch/PHY231/post_files/lecture_19.pdf ·...

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Archimedes: “GIVE ME A PLACE TO STAND AND I WILL MOVE THE EARTH”

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Physics 231 Lecture 19 Torques and rotational dynamics

• Angular kinematics:

• Torques: • Equilibrium:

• Dynamics:

( ) ( )0

0 0

2 2 20 0

t t1 1 t2 2

1t t 22

θ ω ω ω α

ω ω ω θ ω ω

θ ω α ω ω α θ

Δ = = +

= + Δ = +

Δ = + = + Δ

τ ≡ Frsinθ

τ i = 0i∑ ,

!Fi

i∑ = 0 ,

"Rcm =

mi

"rii∑ = 0

mtotal

I = miri

2

i∑ ; τ = Iα

Analogy between linear and angular kinematics for constant acceleration

• If the acceleration is constant, we obtained a set of equation on the left side of the table relating Δx, v, a and t. If the angular acceleration is constant and we go through the same steps, we obtain the analogous set of equation on the right side of the table relating Δθ, ω, α, and t.

( )

( )

ave

0

0

0

20

2 20

t; t t

1 21 t2

1t t 2

2

θ ω θ ω ωω ω α

ω ω ω

θ ω ω

θ ω α

ω ω α θ

Δ = Δ Δ = ≡= +

= +

Δ = +

Δ = +

= + Δ

( )

( )

ave

0

o

o

20

2 20

x vt; x t v vv v at

1v v v2

1x v v t2

1x v t at2

v v 2a x

Δ = Δ Δ = ≡= +

= +

Δ = +

Δ = +

− = Δ

setting t=tΔ

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Example

• After 10 s, a spinning roulette wheel has slowed to an angular speed of 1.88 rad/s. During this time, the wheel rotates through an angle of 44.0 rad. Determine the angular acceleration of the wheel.

o ot tω ω α ω ω α= + ⇒ = −

( ) ( )

θαωωαωθ

ωωθωωω

αωωωθ

Δ+=+=Δ

+=Δ+=

+==Δ

2 21

21

21

20

220

00

0

tt

t

tt ω 1.88 rad/s

t 10 s

Δθ 44.0 rad

( )2 2o

1 1t t t t t2 2

θ ω α ω α αΔ = + = − +

21t t2

θ ω α⇒ Δ = −

( )2

22

t 1.88rad / s 10s 44rad 0.5rad / s1 0.5 10st2

ω θα − Δ ⋅ −⇒ = = = −

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Example • A spinning roulette wheel has slowed to rest from an initial angular

speed of 1.88 rad/s. During this time, the wheel rotates through an angle of 44.0 rad. Determine the angular acceleration of the wheel.

Δθ =ωt ω =ω0 +αt

ω = 12ω +ω0( ) Δθ = 1

2ω +ω0( )t

Δθ =ω0t +12αt2 ω 2 =ω0

2 + 2αΔθ

ω0 1.88 rad/s

ω 0

t 10 s

Δθ 44.0 rad

15

0 =ω 2 =ω02 + 2αΔθ −ω0

2 = 2αΔθ

α = −ω02 / (2Δθ )

α = − 1.88( )2

/ (2 ⋅44)rad / s2

α = −0.0402rad / s2

Quiz

• A centrifuge decelerates to rest over a time interval of 10 seconds from an initial angular velocity of 1000 rad/s. Over this time interval, the angular displacement is about: – a) 5 rad – b) 5000 rad – c) 10 rad – d) 10000 rad

2oo

1000rad / st 100rad / st 10s

ω ωω ω α α − −= + ⇒ = = = −

( ) ( ) ( ) ( )22 2o

1 1t t 1000rad / s 10s 100rad / s 10s2 2

θ ω αΔ = + = + −

5000radθΔ =

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Δθ =ωt ω =ω0 +αt

ω = 12ω +ω0( ) Δθ = 1

2ω +ω0( )t

Δθ =ω0t +12αt2 ω 2 =ω0

2 + 2αΔθ

could also use:

Δθ = 12ω +ω0( )t = 1

20+1000( )10rad = 5000rad

Rolling motion

!!

θrsd ==

• Similarly there are corresponding relationships between v, a and ω, α..

αωraarvv

t

t

====

• Consider a wheel of radius r rolling on the ground. The relationship between the distance traveled and the angle is obtained by considering the arc length s:

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Torques • The direction that you make an object turn

about an axis depends not only on the force, but the point at which you apply the force.

• The relevant quantity that governs rotation is not the force, but is the torque instead:

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• The direction that you make an object turn about an axis depends not only on the force, but the point at which you apply the force.

• The relevant quantity that governs rotation is not the force, but is the torque instead:

Torques

( )1 1 1 1 1 1Fr sin Fd counterclockwiseτ θ= =

• To calculate the torque about an axis: 1. Chose your axis 2. Draw your force. 3. Draw the line, which is perpendicular to the force and goes from the line

of force to the axis. 4. This line is your lever arm d. 5. Torque has the units N�m or Joules.

Fig 8.3, p.222

Slide 3

r1 r2

d1

d2

Fig 8.3, p.222

Slide 3

r1 r2

d1

d2 θ2

θ1

21

( )2 2 2 2 2 2F r sin F d clockwiseτ θ= − =Here, I assume positive torque is in the counterclockwise direction.

1 2 1 2d and d are the lever arms for and .τ τI is the rotational analog of F maτ α= =∑ ∑

Example

• Find the net torque (magnitude and direction) produced by the forces F1 and F2 about the rotational axis shown in the drawing. The forces are acting on a thin rigid rod, and the axis is perpendicular to the page.

τ1 = −F1d1 d1 = r1 ⋅sin 900( )d1 = 0.5m ⋅sin 900( ) = 0.50m

( ) ( )1 20N 0.5m 10Nmτ⇒ = − = −

τ 2 = F2d2 d2 = r 2⋅sin 600( ) = 1.10m ⋅sin 600( ) = 0.95m

( ) ( )2 35N 0.95m 33Nmτ⇒ = =

net 10Nm 33Nm 23Nm 23Nm counterclockwiseτ = − + = =

net 1 2τ τ τ= +

300

600 d2

22

τ1 = −F1r1 sin 900( ) ; clockwise (neg.)

τ 2 = F2r2 sin 600( ) ;counterclockwise; pos.

r

r1 r2

Conceptual quiz

• You are using a wrench and trying to loosen a rusty nut. Which of the arrangements shown is least effective in loosening the nut?

a) b)

d) c)

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c) has the smallest lever arm

Reading Quiz 1. An object is in equilibrium if

A. . B. . C. either A or B. D. both A and B.

Slide 8-5

!Fnet = 0

!τ net = 0

Reading Quiz 2. Which factor does the torque on an object not depend on?

A. The magnitude of the applied force. B. The object’s angular velocity. C. The angle at which the force is applied. D. The distance from the axis to the point at which the force is

applied.

Slide 7-7

Conceptual problem

The same magnitude of force is being applied at the same point, only the angle is not specified. The maximum torque is when the force is perpendicular to the door, which it already was

Example

• One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. One force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The other force has a magnitude of 6.00 N and acts at a 300 angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end that is pinned.

2.00 N

6.00 N

30o

h

ii

constant 0ω τ= ⇒ =∑d

0 2.0N 1m 6.0N d= ⋅ − ⋅

16.0N d 2.0N m d m3

⋅ = ⋅ ⇒ =

( ) ( )o

o

d 2d h sin 30 h m3sin 30

= ⇒ = =

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• But is the object necessarily in rotational equilibrium? Consider the case of a merry-go-round sitting on a frictionless surface. The only forces parallel to the surface are shown in the Figure. Obviously, the net horizontal force vanishes. What can you say about the angular acceleration around the vertical axis shown? – a) It is zero. The two torques cancel out. – b) It is non-zero and counterclockwise. – c) It is non-zero and clockwise

Rotational Equilibrium I

• Newton’s second law for linear motion states that the net force on an object is equal to its mass times its acceleration.

∑ =i

i amF !!

100 N

100 N

• If the net force is zero, the acceleration of the object is zero. It could be at rest or moving with constant non-zero velocity.

Rotational equilibrium II: • Newton’s 2nd law for rotational motion states that the angular

acceleration is given by the net torque. It states that the net torque, i.e. the sum of all torques about an axis, on a body is equal to the moment of inertial I times the angular acceleration α about that axis:

∑ =i

i Iατ

• Here, I is the moment of inertia. It plays the same role as the mass in Newton’s law for linear motion. We discuss I later.

• We define equilibrium to be the situation when both the linear acceleration and angular acceleration vanish. This occurs when both the net force and the net torque vanish. In other words:

0 and 0 == ∑∑i

ii

i F!

τ

• Here, torques must vanish around any axis. Axes can be chosen so as to make the solutions of problems easier. If you are not in equilibrium then the location has to be what object rotates around.

Choosing the Pivot Point

Slide 8-12

This is true BECAUSE the net force is zero in equilibrium. If the net force is not zero, then we are not in equilibrium, and the torque will depend on where we place the pivot.

Example • A 71 kg boulder is placed on a 2.00 m long board at a point that is 1.4

m from one end. Cliff and Will support the board at each end so that it is horizontal. Cliff is nearest the boulder. If the weight of the board is negligible, what force does each apply to the board?

2.00m

CliffF!

WillF!1.4m

Place pivot at Will⇒ lever arm for Will vanishes

gMboulder

i boulder Cliff0 M g 0.6m F 2m 0τ = ⇒ ⋅ − ⋅ =∑2

Cliff boulder0.6 0.6F M g 71kg(9.8m / s ) 208.7N2.0 2.0

⇒ = ⋅ = ⋅ =

i Cliff Will boulderF 0 F F M g 0= ⇒ + − =∑Will boulder CliffF M g F 487.1N⇒ = − =

What if the board has mass? Center of mass

• What if the weight of the board, or any other extended object, is not zero? How do we calculate the torque due to the weight?

Fig 8.7, p.224

Slide 10

• We can break up the object into little mass elements mi. When these mass elements are added, Σi mi = m total, where mtotal is the total mass of the object.

• When this is done, the torque due to gravity around axis O can be calculated as follows:

• Thus, we put the entire weight of the object at . For a uniform rod or some other uniform object, is at the center of the object. In general, however, may not even be within the object itself.

!rcm = (xcm ,ycm )

cmr!

cmr!

τ i = Fyixii∑

i∑ = migxi

i∑

= mgmixi

i∑

mtotal

= mgxcm; where, xcm =mixi

i∑mtotal

• Main points of today’s lecture: • Analogies between rotational and

translational motion • Solutions involving:

• Rolling motion:

ω

ω

ατ

IL

IKE

II

rot

=

=

=

2

21

22

21

21 ωIMvKEtotal +=

Physics 231 Lecture 20