today: chapter 9, rotational energy homework #9 - due...

April 9, Week 12

Today: Chapter 9, Rotational Energy

Homework #9 - Due April 16 at 11:59pmMastering Physics: 7 questions from chapter 9. WrittenQuestion: 10.80

Rotational Energy – p. 1/10

Review

The rate at which an objects spins is given by its angularvelocity, −→ω , and angular acceleration −→

α .

ω = dθ

dt, α = dω

dt

b


Review


α .

ω = dθ

dt, α = dω

dt

RHR I - Curl the fingers of your righthand in the “sense" of the rotation.Your extended thumb, points in directionof −→ω .

b


Review


α .

ω = dθ

dt, α = dω

dt


−→ω

⊗

b


Review


α .

ω = dθ

dt, α = dω

dt


−→ω

⊗

b

b


Review


α .

ω = dθ

dt, α = dω

dt


−→ω

⊗

b

−→r

b


Review


α .

ω = dθ

dt, α = dω

dt


−→ω

⊗

b−→v −→

r

b


Review


α .

ω = dθ

dt, α = dω

dt


−→ω

⊗

b−→v −→

r

−→v = −→ω ×

−→r

b


Review


α .

ω = dθ

dt, α = dω

dt


−→ω

⊗

b−→v −→

r

−→v = −→ω ×

−→r

Take the fingers of the right hand and “sweep"−→

A into−→

B ,extended thumb points in direction of cross product


Connected Rotating Objects

r1

r2



r1

r2

−→v

−→v

When connected by a non-slippingchain or belt, the two rotating objectsmust have the same linear velocity



r1

r2

−→v

−→v


v1 = v2



r1

r2

−→v

−→v


v1 = v2

ω1r1 = ω2r2



r1

r2

−→v

−→v


v1 = v2

ω1r1 = ω2r2

Different Angular Velocities


Linear Accelerations

Every point on a rotating object has two accelerationcomponents.

b




b

b




−→ω⊙

b

b




−→ω⊙

b

−→r

b




−→ω⊙

b

−→r

−→arad - changes in direction

b




−→ω⊙

b

−→r −→

arad


b




−→ω⊙

b

−→r −→

arad


arad = v2

r= ω2r

b




−→ω⊙

b

−→r −→

arad


arad = v2

r= ω2r

−→arad = −ω2−→

r

b




−→ω⊙

b

−→r −→

arad


−→atan - changes in speed

arad = v2

r= ω2r

−→arad = −ω2−→

r

b




−→ω⊙

b

−→r −→

arad

−→atan



arad = v2

r= ω2r

−→arad = −ω2−→

r

b




−→ω⊙

b

−→r −→

arad

−→atan



arad = v2

r= ω2r

−→arad = −ω2−→

r

atan = αr

b




−→ω⊙

b

−→r −→

arad

−→atan



arad = v2

r= ω2r

−→arad = −ω2−→

r

atan = αr −→atan = −→

α ×−→r

b




−→ω⊙

b

−→r −→

arad

−→atan



arad = v2

r= ω2r

−→arad = −ω2−→

r


α ×−→r

−→a = −→

arad +−→atan

b




−→ω⊙

b

−→r

−→atan

−→arad



arad = v2

r= ω2r

−→arad = −ω2−→

r


α ×−→r

−→a = −→

arad +−→atan a =

√

a2rad

+ a2tan


Non-Circular Objects

Putting the origin of the coordinate system at the axis ofrotation allows us to use all of the equations for circularobjects.

b

b




br

b




br

r = distance from axis of rotation


Clicker Quiz

The following object is rotated about one end as shown.What is the linear speed of the point A?

b A

b


Clicker Quiz


d d/2b A

b


Clicker Quiz


d

d/2 b A

b


Clicker Quiz


dd/2

b A

b


Clicker Quiz


d

d/2

bA

b


Clicker Quiz


dd/2

bA

b


Clicker Quiz


dd/2bA

b


Clicker Quiz


d

d/2bA

b


Clicker Quiz


dd/2

bA

b


Clicker Quiz


d

d/2bA

b


Clicker Quiz


d d/2b A

b


Clicker Quiz


d d/2b A

(a) dω

b


Clicker Quiz


d d/2b A

(a) dω

(b) d

2ω

b


Clicker Quiz


d d/2b A

(a) dω

(b) d

2ω

(c) 3d

2ω

b


Clicker Quiz


d d/2b A

(a) dω

(b) d

2ω

(c) 3d

2ω

(d) 2dω

b


Clicker Quiz


d d/2b A

(a) dω

(b) d

2ω

(c) 3d

2ω

(d) 2dω


Rotational Kinetic Energy

Any rotating object has a kinetic energy due to its motion.




We have to imagine splitting the rotating object up intomany small pieces.





Look at the i-th piece






It has kinetic energy, Ki =1

2Miv

2

i







2Miv

2

i

ri







2Miv

2

i

ri

vi = riω







2Miv

2

i

ri

vi = riω All pieces have same ω







2Miv

2

i

ri


Ki =1

2Mi (riω)

2 = 1

2Mir

2

iω2







2Miv

2

i


ri

Ki =1

2Mi (riω)

2 = 1

2Mir

2

iω2

K ≈

∑

i

Ki =∑

i

1

2Mir

2

iω2


Rotational Kinetic Energy II



ri

K ≈1

2

(

∑

i

Mir2

i

)

ω2





ri

K ≈1

2

(

∑

i

Mir2

i

)

ω2

This expression becomes exactin the limit as the number of piecesapproaches infinity (and so the sizeof each piece approaches zero)





r

K ≈1

2

(

∑

i

Mir2

i

)

ω2

This expression becomes exactin the limit as the number of piecesapproaches infinity (and so the sizeof each piece approaches zero)


Moment of Inertia

In the limit, the sum becomes the Moment of Inertia, I, forthe rotating object.


Moment of Inertia


ri

I ≈

∑

i

Mir2

i


Moment of Inertia


ri

I ≈

∑

i

Mir2

i

I = limi→∞

∑

i

Mir2

i


Moment of Inertia


r

I ≈

∑

i

Mir2

i

I = limi→∞

∑

i

Mir2

i

I =∫

r2dM =∫

r2ρ dV

(ρ = density)


Moment of Inertia


r

I ≈

∑

i

Mir2

i

I = limi→∞

∑

i

Mir2

i

I =∫

r2dM =∫

r2ρ dV

(ρ = density)

K = 1

2Iω2


Moment of Inertia II

The moment of inertia, I, is the rotational counterpart tomass, i.e., it plays the same role in rotation as mass does inlinear motion.




The moment of inertia tells us how “hard" it is to make anobject rotate.





The moment of inertia depends on:






(a) The object’s shape.






(a) The object’s shape.(b) The axis of rotation.






(a) The object’s shape.(b) The axis of rotation.(c) The total mass of the object.


today: chapter 9, rotational energy homework #9 - due...

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