Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley

PowerPoint® Lectures for University Physics, Twelfth Edition – Hugh D. Young and Roger A. Freedman

Lectures by James Pazun

Chapter 9

Rotation of Rigid Bodies

Rotational Motion Chapters 9 and 10 are about rotation •  Start with Fixed Axis motion •  Rotational kinematics: just like in Ch 1-3 •  The relationship between linear and

angular variables •  Definition of “angular mass” •  Rotating and translating at the same time •  Eventually move from kinematics to

dynamics

Rotational Motion •  Real life problems often have something to

do with rotation of objects – Examples: spinning wheel, moving parts inside

your car engine, rotors of all kinds etc. •  Methods that we have developed so far

dealt mostly with translational motion (brick sliding on an incline) – We want to have simpler means and tools

tailored to this new special case – Old methods still work, but often hard to apply or

we need to do the same tricks every time •  Makes sense to do them once and then just

remember our lessons and use when applicable

Some Jargon •  Fixed axis: i.e, an object spins about

a fixed place… – Example: pulley on a well, merry go

round • Rigid body: i.e, the objects don’t

change as they rotate. – Example: a bicycle wheel – Examples of Non-rigid bodies?

Overview: Rotational Motion •  Take our results from “linear” physics and do the same

for “angular” physics •  Analogue of

– Position ← – Velocity ← – Acceleration ← – Force ← – Mass ← – Momentum ← – Energy ←

Cha

pter

s 1-3

C

hapt

ers 4

-8

Angular motions in revolutions, degrees, and radians

•  One complete cycle of 360° is one revolution.

•  One complete revolution is 2π radians.

•  Relating the two, 360° = 2 π radians or 1 radian = 57.3°.

€

s = Rθ

Velocity and Acceleration

Angular velocity is a vector •  You can visualize the position of the vector by sweeping

out the angle with the fingers of your right hand. The position of your thumb will be the position of the angular velocity vector. This is called the “right-hand rule.”

Motion on a Wheel: relation between angular velocity and linear velocity

What is the linear tangential speed of a point rotating around in a circle with angular speed ω, and constant radius R?

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Δl = RΔθ = RωΔt⇒ vtan =ΔlΔt

= Rω

In the same way we can relate the angular acceleration and the linear tangential acceleration

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atan =Δ vtanΔt

= Rα

For the RADIAL acceleration we have (no matter if v or ω are constant)

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arad =v2tanR

= Rω 2

Kinematic equations for angular kinematics

Motion of a hard drive A typical hard drive nowadays spins at a rate of 10,000 rpm and have a radius of about 8 cm. The hard this gets to this speed in about 10 ms. What is the angular acceleration to reach the full reading speed of the hard drive? What is the speed at which the edge of the hard drive is moving?

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α = ?θ =

ω 0 = 0ω =10000 rpm =1047 rad/st = 0.01s€

ω =ω 0 +αt

α =ω −ω 0

t=10470.1

=1.047 ×104 rads2

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v = Rω = (0.08)(1.047 ×103) = 83.8 m/s =187 mph

Gives a new meaning to low flying!!

An athlete throwing the discus A discus thrower moves the discus in a circle of radius 80.0 cm. At a certain instant the thrower is spinning at an angular speed of 10.0 rad/s and the angular speed is increasing at 50.0 rad/s2. At this instant find the acceleration of the discus and its magnitude

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atan = rα = (0.800 m)(50.0 rad/s2) = 40.0 m/s2

arad = rω 2 = (10.0 rad/s)2(0.800 m) = 80.0 m/s2

a = atan2 + arad

2 = 89.4 m/s2

Kinetic energy of an object rotating about a fixed axis defining moment of inertia

Kinetic energy of a small piece of the object rotation at an angular velocity ω

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12mi vi

2 =12mi(riω)

2 =12miri

2ω 2

But a solid object is made of many of those pieces

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KErot =12m1r1

2ω 2 +12m2r2

2ω 2 + =12miri

2ω 2 =12i

∑ miri2

i∑⎛

⎝ ⎜

⎞

⎠ ⎟ ω 2

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KErot =12Iω 2

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I = miri2

i∑ Moment of inertia

or “angular mass”

Rotational energy changes if parts shift and I changes

Moment of inertia about axis passing through A

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IA = mArA2 +mBrB

2 +mCrC2

= (0.30 kg)(0.0 m)2 + (0.10 kg)(0.5 m)2 +(0.20 kg)(0.40 m)2

= 0.057 kg m2

Moment of inertia about axis passing through BC

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IBC = mArA2 +mBrB

2 +mCrC2

= (0.30 kg)(0.4 m)2 + (0.10 kg)(0.0 m)2 +(0.20 kg)(0.0 m)2

= 0.048 kg m2

Finding the moment of inertia for common shapes

Calculating rotational energyThe cable is wrapped around a cylinder. If it unwinds 2.0 m by pulling it with a force of 9.0 N and it starts at rest, what is its final angular velocity and velocity of the cable? (use work energy theorem)

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Wtotal = KE f −KEi

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FΔx =12Iω 2 − 0

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ω =2FΔxI

=4FΔxmR2 = 20 rad/s

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v = Rω = (20 rad/s)(0.060 m) =1.2 m/s

What is the velocity of the block when it hits the ground?

The work done by the cable is zero since the two tension forces cancel each other out so energy is conserved

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KEi + PEi = KE f + PE f

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0 +mgh =12mv2 +

12Iω 2 + 0

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0 +mgh =12mv2 +

1212MR2

⎛

⎝ ⎜

⎞

⎠ ⎟ vR⎛

⎝ ⎜

⎞

⎠ ⎟ 2

+ 0

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v = 2mgh(m +M/2)

Parallel axis theorem: important for next chapter and to find complicated moments of inertia

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IP = Icm + Md2

The moment of inertia about an axis going through point P a distance d away from the center of mass is equal to the moment of inertia about the center of mass plus Md2