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Copyright © 2010 Pearson Education, Inc. Chapter 10 Rotational Kinematics and Energy
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Chapter 10 Rotational Kinematics and Energy

Feb 03, 2022

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Page 1: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

Chapter 10

Rotational Kinematics and Energy

Page 2: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

10-1 Angular Position, Velocity, and Acceleration

Page 3: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

10-1 Angular Position, Velocity, and Acceleration

Degrees and revolutions:

Page 4: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

10-1 Angular Position, Velocity, and Acceleration

Arc length s, measured in radians:

Page 5: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

10-1 Angular Position, Velocity, and Acceleration

Page 6: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

10-1 Angular Position, Velocity, and Acceleration

Page 7: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

10-1 Angular Position, Velocity, and Acceleration

Page 8: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

10-1 Angular Position, Velocity, and Acceleration

Page 9: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

10-2 Rotational Kinematics

If the angular acceleration is constant:

Page 10: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

10-2 Rotational Kinematics

Analogies between linear and rotational kinematics:

Page 11: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

Example: A high speed dental drill is rotating at 3.14×104 rads/sec. Through how many degrees does the drill rotate in 1.00 sec?

Given: ω = 3.14×104 rads/sec; Δt = 1 sec; α = 0

Want Δθ.

( )( )degrees 101.80rads1014.3

sec 0.1rads/sec1014.3

21

64

40

00

200

×=×=

×=Δ=Δ

Δ+=

Δ+Δ+=

tt

tt

ωθ

ωθθ

αωθθ

Page 12: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

10-3 Connections Between Linear and Rotational Quantities

Page 13: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

10-3 Connections Between Linear and Rotational Quantities

Page 14: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

10-3 Connections Between Linear and Rotational Quantities

Page 15: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

10-3 Connections Between Linear and Rotational Quantities

This merry-go-round has both tangential and centripetal acceleration.

Page 16: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

10-4 Rolling Motion If a round object rolls without slipping, there is a fixed relationship between the translational and rotational speeds:

Page 17: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

10-4 Rolling Motion We may also consider rolling motion to be a combination of pure rotational and pure translational motion:

Page 18: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

10-5 Rotational Kinetic Energy and the Moment of Inertia

For this mass,

Page 19: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

10-5 Rotational Kinetic Energy and the Moment of Inertia

We can also write the kinetic energy as

Where I, the moment of inertia, is given by

Page 20: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

Example: (a) Find the moment of inertia of the system below. The masses are m1 and m2 and they are separated by a distance r. Assume the rod connecting the masses is massless.

ω

r1 r2 m1 m2

r1 and r2 are the distances between mass 1 and the rotation axis and mass 2 and the rotation axis (the dashed, vertical line) respectively.

Page 21: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

(b) What is the moment of inertia if the axis is moved so that is passes through m1?

Example continued:

( )( ) ( )( )2

22

222

211

2

1

2

m kg 00.1m 00.1kg 00.1m 00.0kg 00.2

=

+=

+==∑=

rmrmrmIi

ii

Take m1 = 2.00 kg, m2 = 1.00 kg, r1= 0.33 m , and r2 = 0.67 m.

( )( ) ( )( )2

22

222

211

2

1

2

m kg 67.0m 67.0kg 00.1m 33.0kg 00.2

=

+=

+==∑=

rmrmrmIi

ii

Page 22: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

10-5 Rotational Kinetic Energy and the Moment of Inertia

Moments of inertia of various regular objects can be calculated:

Page 23: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

Example: What is the rotational inertia of a solid iron disk of mass 49.0 kg with a thickness of 5.00 cm and a radius of 20.0 cm, about an axis through its center and perpendicular to it?

From the previous slide:

( )( ) 222 m kg 98.0m 2.0kg 0.4921

21

=== MRI

Page 24: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

10-6 Conservation of Energy The total kinetic energy of a rolling object is the sum of its linear and rotational kinetic energies:

The second equation makes it clear that the kinetic energy of a rolling object is a multiple of the kinetic energy of translation.

Page 25: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

10-6 Conservation of Energy If these two objects, of the same mass and radius, are released simultaneously, the disk will reach the bottom first – more of its gravitational potential energy becomes translational kinetic energy, and less rotational.

Page 26: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

Example: Two objects (a solid disk and a solid sphere) are rolling down a ramp. Both objects start from rest and from the same height. Which object reaches the bottom of the ramp first?

h

θ

The object with the largest linear velocity (v) at the bottom of the ramp will win the race.

Page 27: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

!"

#$%

& +=

!"

#$%

& +=

!"

#$%

&+=++=+

+=+

=

2

22

2222

2

21

21

21

21

2100

RIm

mghv

vRImmgh

RvImvImvmgh

KUKUEE

ffii

fi

ω

Apply conservation of mechanical energy:

Example continued:

Solving for v:

Page 28: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

2sphere

2disk

5221

mRI

mRI

=

=

For the disk:

For the sphere:

ghv34

disk =

ghv710

sphere =

Since Vsphere> Vdisk the sphere wins the race.

ghv 2box =Compare these to a box sliding down the ramp.

Example continued:

The moments of inertia are:

Page 29: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

Summary of Chapter 10

•  Period:

•  Counterclockwise rotations are positive, clockwise negative

•  Linear and angular quantities:

Page 30: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

Summary of Chapter 10

•  Linear and angular equations of motion:

Tangential speed:

Centripetal acceleration:

Tangential acceleration:

Page 31: Chapter 10 Rotational Kinematics and Energy

Copyright © 2010 Pearson Education, Inc.

Summary of Chapter 10

• Rolling motion:

• Kinetic energy of rotation:

• Moment of inertia:

• Kinetic energy of an object rolling without slipping:

• When solving problems involving conservation of energy, both the rotational and linear kinetic energy must be taken into account.