Chapter 7 Rotational Motion. Slide 7-3 © 2015 Pearson Education, Inc. Curvilinear coordinates

Chapter 7

• Rotational Motion

Slide 7-3

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𝑠=𝑟 𝜃

Curvilinear coordinates

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∆ 𝑥=𝑣𝑖 𝑡+12𝑎𝑡 2

∆ 𝜃=𝜔 𝑖 𝑡+12𝛼𝑡 2 𝜔 𝑓

2 =𝜔𝑖2+2𝛼 ∆𝜃

𝑣 𝑓2 =𝑣𝑖

2+2𝑎∆ 𝑥

Kinematic equations are the same, just different variables

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Angular Velocity

𝑣=𝑟 𝜔 𝑣

“omega”

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Ball rolling across frictionless floor

This slope is

𝜔=∆𝜃∆ 𝑡

Checking UnderstandingTwo coins rotate on a turntable. Coin B is twice as far from the axis as coin A.

A. The angular velocity of A is twice that of B.

B. The angular velocity of A equals that of B.

C. The angular velocity of A is half that of B.

Slide 7-13

Answer

All points on the turntable rotate through the same angle in the same time. All points have the same period.

Two coins rotate on a turntable. Coin B is twice as far from the axis as coin A.

A. The angular velocity of A is twice that of B.

B. The angular velocity of A equals that of B.

C. The angular velocity of A is half that of B.

Slide 7-14

Angular acceleration α measures how rapidly the angular velocity is changing:

Slide 7-17

𝛼=𝑎t

𝑟

Angular Acceleration

𝑎t

Tangential acceleration

Slide 7-18

Linear and Circular motion compared

Two coins rotate on a turntable. Coin B is twice as far from the axis as coin A.

A. The speed of A is twice that of B.

B. The speed of A equals that of B.

C. The speed of A is half that of B.

Checking Understanding

Slide 7-15

Answer

Twice the radius means twice the speed

Two coins rotate on a turntable. Coin B is twice as far from the axis as coin A.

A. The speed of A is twice that of B.

B. The speed of A equals that of B.

C. The speed of A is half that of B.

Slide 7-16

v r

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3 types of related motion

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Tangential velocity of circular motion can become linear motion

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Tangential velocity of circular motion can become linear motion

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Tangential velocity of circular motion can become linear motion

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Combination motion

𝑣𝜔

Center of mass follows original trajectory

Slide 7-19

The equations have the same form

Example ProblemA high-speed drill rotating CCW takes 2.5 s to speed up to 2400 rpm.

A. What is the drill’s angular acceleration?B. How many revolutions does it make as it reaches top speed?

Slide 7-21

∆ 𝜃=𝜔 𝑖 𝑡+12𝛼𝑡 2 𝜔 𝑓

2 =𝜔𝑖2+2𝛼 ∆𝜃

2 ∆𝜃𝑡 2 =𝛼

𝜔 𝑓2

2𝛼=∆𝜃

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Comparing rotational motion plots

Slide 7-22

Centripetal and Tangential acceleration

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The difference of nonuniform circular

motion

The speed is changing

Center of Gravity

=

Slide 7-29

Calculating the Center-of-Gravity Position

Slide 7-30

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Center of mass practice

𝑥cg=𝑚1 𝑥1+𝑚2 𝑥2

𝑚1+𝑚2

=5 kg ∙ m15 kg

=13

m

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𝜏=𝑟𝐹 sin 𝜃 Angle between lever arm and force

Torque –when force causes circular motion

Checking UnderstandingWhich point could be the center of gravity of this L-shaped piece?

Slide 7-32

Which point could be the center of gravity of this L-shaped piece?

Answer

(a)

Slide 7-33

Interpreting Torque

rF rF sin

Torque is due to the component of the force perpendicular to the radial line.

Slide 7-25

Signs and Strengths of the Torque

Slide 7-27

The four forces below are equal in magnitude. Which force would be most effective in opening the door?

A. B. C. D. E. Either or

$$ 1

$$ 2

$$ 3

$$ 4

Either

$$ 1 o

r $$

3

98%

0% 0%2%0%

Example torque Problem Revolutionaries attempt to pull down a statue of the Great Leader by pulling on a rope tied to the top of his head. The statue is 17 m tall, and they pull with a force of 4200 N at an angle of 65° to the horizontal. What is the torque they exert on the statue? If they are standing to the right of the statue, is the torque positive or negative?

Slide 7-28

65°

17 m

F = 4200 N

𝜏=17 m ∙ 4200 N ∙ sin 𝜃

𝜃≠65°

𝜏=17 m ∙ 4200 N ∙ sin 25°

𝜏=𝑟𝐹 sin 𝜃

pivot

Negative torque, but why?

Rotating it in the CW directionr = 17m

Which force vector on point P would keep the wheel from spinning?

A. AB. CC. DD. E

A C D E

98%

0%2%0%

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Torque practice

𝜏=𝑟𝐹 sin𝜗  

𝜏=𝑟𝑚𝑔 sin𝜗

𝜏= (1.6 m ) (3.2 kg )(− 9.8m

s2 )sin ( 6 5° )

𝜗

Which torques are equal?

A. B = C = D = E only

B. A = B and C = D = E

C. None are equal

D. B = E and C = D

B = C = D =

E only

A = B and C = D = E

None are e

qual

B = E a

nd C = D

0%

98%

0%2%

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Torque practice with centers of mass

What is the Net Torque is exerted by the gymnast about an axis through the rings?

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𝑟1

𝐹 1

𝑟2

𝐹 2

∑𝜏=0Torque equilibrium

Reading Quiz2. 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

Answer2. 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 7-8

Example ProblemAn object consists of the three balls shown, connected by massless rods. Find the x- and y-positions of the object’s center of gravity.

Slide 7-31

𝑥cg=𝑚1 𝑥1+𝑚2 𝑥2+𝑚3𝑥3

𝑚1+𝑚2+𝑚3

𝑦 cg=𝑚1𝑦1+𝑚2 𝑦2+𝑚3 𝑦3

𝑚1+𝑚2+𝑚3

An object consists of the three balls shown, connected by massless rods. Find the x- and y-positions of the object’s center of gravity.

Slide 7-31

𝑥cg=𝑚1 (0 )+𝑚2 ( 0 )+𝑚3 (1 m )

𝑚1+𝑚2+𝑚3

=2 kg ∙ m4 kg

=12

m

𝑦 cg=𝑚1 (1 m )+𝑚2 (0 )+𝑚3 (0 )

𝑚1+𝑚2+𝑚3

=1 kg ∙ m4 kg

=14

m

The center of mass for these 3 bodies

The moment of Inertia the rotational equivalent of mass

Newton’s Second Law for Rotation

/ II = moment of inertia. Objects with larger moments of inertia are harder to get rotating.

I miri2

Slide 7-34

Rotational and Linear Dynamics Compared

Slide 7-36

Which moment of inertia is greatest?

A. AB. BC. CD. D

𝑀𝑚

A B C D

25% 25%25%25%

Which force vector applied to point P will stop this rolling ball?

A. AB. BC. CD. DE. E

A B C D E

20% 20%20%20%20%

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Formulas more common moments of inertia 𝐼

Which red vector is your best bet for getting this bolt as tight as possible?

A. AB. BC. CD. D

A B C D

25% 25%25%25%

Reading Quiz1. Moment of inertia is

A. the rotational equivalent of mass.

B. the point at which all forces appear to act.

C. the time at which inertia occurs.

D. an alternative term for moment arm.

Slide 7-5

Answer1. Moment of inertia is

A. the rotational equivalent of mass.

B. the point at which all forces appear to act.

C. the time at which inertia occurs.

D. an alternative term for moment arm.

Slide 7-6

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Torque causing rotation

∑𝜏 ≠ 0

∑𝜏=¿ 𝐼 𝛼 ¿What happens to these masses when you let go?

What happens to this pulley system?

A. It does not moveB. The 10N force accelerates

the mass upwardC. The force of gravity on the

mass results in a net force upward

D. The mass moves upward at a constant speed

It does

not move

The 1

0N force

accelera

tes t.

.

The f

orce of g

ravit

y on th

e ...

The m

ass m

oves u

pward at a

...

25% 25%25%25%

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Do we always have to use torque?

∑ 𝐹1=¿𝐹𝑔 1+𝑇1=𝑚1𝑎1¿∑ 𝐹2=¿𝐹𝑔 2+𝑇 2=𝑚2𝑎2¿

𝑇 1=−𝑇2

Newton’s Third

𝑇 1

𝑇 2

𝑇 1

𝐹 𝑔

𝐹 𝑔

Starting from rest, how long does it take to hit the ground?

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Use forces to solve

𝑚1𝑔+𝑇1=𝑚1𝑎

𝑚2|𝑔|−𝑇 1=𝑚2𝑎

𝑎2=𝑎1=𝑎

𝑇 1

𝑇 2𝐹 𝑔

𝐹 𝑔

𝑎1 𝑎2

Get rid of by solving for it and substituting into the other equation

𝑎=− 3.27m

s2

Starting from rest, how long does it take to hit the ground?

∆ 𝑦=𝑣 𝑖𝑦𝑡+12𝑎𝑡 2

𝑡=√ 2 ∆ 𝑦𝑎

𝑡=.78 s

Reading Quiz4. A net torque applied to an object causes

A. a linear acceleration of the object.

B. the object to rotate at a constant rate.

C. the angular velocity of the object to change.

D. the moment of inertia of the object to change.

Slide 7-11

Answer4. A net torque applied to an object causes

A. a linear acceleration of the object.

B. the object to rotate at a constant rate.

C. the angular velocity of the object to change.

D. the moment of inertia of the object to change.

Slide 7-12

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Friction in rotational motionDraw the normal force for the wheel against the break

Draw the frictional force from the break

𝐹 𝑓

𝑁 𝑏

Is this beam balanced?A. YesB. No, it will spin CWC. No, it will spin CCWD. Not enough information

Yes

No, it w

ill sp

in CW

No, it w

ill sp

in CCW

Not enough

inform

ation

25% 25%25%25%

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Now the pulley has friction𝐼𝑝=

12𝑚𝑝𝑟

2

∑𝜏=¿𝜏1+𝜏2= 𝐼𝑝𝛼 ¿

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Now the pulley has friction and mass𝐼𝑝=

12𝑚𝑝𝑟

2

.02m ∙ 20N − .02m ∙ 30 N=(.0008 kg ∙m 2)𝛼

4 kg

Additional Example ProblemA baseball bat has a mass of 0.82 kg and is 0.86 m long. It’s held vertically and then allowed to fall. What is the bat’s angular acceleration when it has reached 20° from the vertical? (Model the bat as a uniform cylinder).

Slide 7-43