classical mechanics lecture 15 - astronomy · classical mechanics lecture 15 ... the contribution...

Classical Mechanics Lecture 15

Today’s Concepts:a) Parallel Axis Theoremb) Torque & Angular Acceleration

Mechanics Lecture 15, Slide 1

Unit 14 Main Points


Clicker Question A.

B.

C.

D.

0% 0%0%0%

A)

B)

C)

D)

A mass M is uniformly distributed over the length L of a thin rod. The mass inside a short element dx is given by:

dxM

Mdx

dxLM

dxML

dx

x

L

M


Clicker Question A.

B.

C.

D.

0% 0%0%0%

A)

B)

C)

dxLMx2

dxLM

x21

2dxLM

A mass M is uniformly distributed over the length L of a thin rod. The contribution to the rod’s moment of inertia provided by element dx is given by:

dx

x

L

M


In both cases shown below a hula hoop with mass M and radius Ris spun with the same angular velocity about a vertical axis through its center. In Case 1 the plane of the hoop is parallel to the floor and in Case 2 it is perpendicular.

In which case does the spinning hoop have the most kinetic energy?A) Case 1 B) Case 2 C) Same

Everyone got this right too!!!…

Clicker Question

ω

R R

ω

Case 2Case 1


In which case does the spinning hoop have the most kinetic energy?A) Case 1 B) Case 2 C) Same

A) In case one, more mass is located away from its axis, so it has larger moment of inertia. Therefore it has more kinetic energy.

ω

R R

ω

Case 2Case 1

2

21 ωIK =

Clicker Question


A wheel which is initially at rest starts to turn with a constant angular acceleration. After 4 seconds it has made 4 complete revolutions.

How many revolutions has it made after 8 seconds?A) 8 B) 12 C) 16

CheckPoint

30.3% got this right on first attempt

α


After 4 seconds it has made 4 complete revolutions.

How many revolutions has it made after 8 seconds?A) 8 B) 12 C) 16

CheckPoint Response

C) The number of revolutions is proportional to time squared.

α2

00 21)( ttt αωθθ ++=

t0ωωα −

=

0;0 00 == ωθ

2

1

2

21

22

1

2

2121

)()(

==

tt

t

t

tt

α

α

θθ


A triangular shape is made from identical balls and identical rigid, massless rods as shown. The moment of inertia about the a, b, and c axes is Ia, Ib, and Icrespectively.

Which of the following orderings is correct?

59% got this right on first attempt..

CheckPoint

a

b

c

A) Ia > Ib > Ic

B) Ia > Ic > Ib

C) Ib > Ia > Ic


Which of the following orderings is correct?CheckPoint Response

B) Ia = 8mr^2 Ib = 3mr^2 Ic = 4mr^2

a

b

c

A) Ia > Ib > Ic

B) Ia > Ic > Ib

C) Ib > Ia > Ic

( ) ( ) 2222 8)0(22 mRmRmRmIa =++=

( ) ( ) ( ) 2222 3mRRmRmRmIb =++=

( ) ( ) 2222 4)0(02 mRmmRmIc =++=


Clicker Question A.

B.

C.

0% 0%0%

r

dr

A)

B)

C)

A disk has a radius R. The area of a thin ring inside the disk with radius r and thickness dr is:

drr2π

rdrπ2

drr34π

Review geometry….


Clicker Question A.

B.

C.

0% 0%0%

A disk spins at 2 revolutions/sec.

What is its period?

A) T = 2 sec

B) T = 2π sec

C) T = ½ sec


Clicker Question A.

B.

C.

0% 0%0%

rad/sec

rad/sec

rad/sec

A disk spins at 2 revolutions/sec.

What is its angular velocity?

A)

B)

C)

πω 2=

2πω =

πω 4=


Moment of Inertia of Sphere

http://hyperphysics.phy-astr.gsu.edu/hbase/isph.html#sph4

Must integrate from z-axis not center of sphere!


http://hyperphysics.phy-astr.gsu.edu/hbase/isph.html%23sph4

Volume of Sphere

33

3

03

0

2

00

2

0

0 0

22

0

2

34

34

3|

31

2)11(|cossin

2

sin

sin

RRV

Rrdrr

d

d

drrddI

ddrdrdV

dVV

sphere

Rr

r

sphere

sphere

ππ

θθθ

πϕ

θθϕ

ϕθθ

ππ

π

ππ

==

==

=−−−=−=

=

=

=

=

∫

∫

∫

∫ ∫∫

∫


(i)

(ii)

(iii)

200 2

1 tt αωθθ ++=

tαωω += 0

θαωω ∆+= 220

2

Using (ii)t

0ωωα −=

Using (i) 2

21 tαθ =


Use (iv)

Use (v)

(iv) 2

21 MRIDISK =

(v) 2

21 ωIK =


Use (viii)

Use (ix)

(vi) θRd =

(vii) ωRv =

(viii) αRaT =

(ix) RRvac

22

ω==


Use (vii)

Use (vi)

(vi) θRd =

(vii) ωRv =

(viii) αRaT =

(ix) RRvac

22

ω==


rodI

2

31 MLIrod =

2

21 MRIdisk =

22

35

215 MLMRIII roddiskfan +=+=


( )2

0

20

200

221

21

t

t

tt

θθα

αθθ

αωθθ

−=

=−

++=

tf αωω += 0

2

21

ffanf IK ω=


if

if

iiff IKIK

ωω

ωω

ωω

2121

41

21

21

22

22

=

=

===

t

t

f

f

0

0

ωωα

αωω

−=

+=


Main Points


Parallel Axis Theorem


Parallel Axis Theorem

Smallest when D = 0


Clicker Question A.

B.

C.

D.

0% 0%0%0%

A solid ball of mass M and radius is connected to a rod of mass m and length L as shown. What is the moment of inertia of this system about an axis perpendicular to the other end of the rod?

ML

mR

axis

222

31

52 MLmLMRI ++=

222

31

52 mLMLMRI ++=

22

31

52 mLMRI +=

22

31 mLMLI +=

A)

B)

C)

D)


A ball of mass 3M at x = 0 is connected to a ball of mass M at x = L by a massless rod. Consider the three rotation axes A, B, and C as shown, all parallel to the y axis.

For which rotation axis is the moment of inertia of the object smallest? (It may help you to figure out where the center of mass of the object is.)

B

3M M

CA

L/2L/40

x

y

L

100% got this right !!!

CheckPoint


Right Hand Rule for finding Directions

Why do the angular velocity and acceleration point perpendicular to the plane of rotation?


Clicker Question A.

B.

C.

D.

0% 0%0%0%

A ball rolls across the floor, and then starts up a ramp as shown below. In what direction does the angular velocity vector point when the ball is rolling up the ramp?

A) Into the page

B) Out of the page

C) Up

D) Down


Enter Question TextA.

B.

C.

D.

0% 0%0%0%

A ball rolls across the floor, and then starts up a ramp as shown below. In what direction does the angular acceleration vector point when the ball is rolling up the ramp?

A) Into the page

B) Out of the page


Enter Question TextA.

B.

C.

D.

0% 0%0%0%

A ball rolls across the floor, and then starts up a ramp as shown below. In what direction does the angular acceleration vector point when the ball is rolling back down the ramp?

A) into the page

B) out of the page


Torque

τ = rF sin(θ )


In Case 1, a force F is pushing perpendicular on an object a distance L/2 from the rotation axis. In Case 2 the same force is pushing at an angle of 30 degrees a distance L from the axis.

In which case is the torque due to the force about the rotation axis biggest?

A) Case 1 B) Case 2 C) Same

FL/2 90o

Case 1

axis

L F30o

Case 2

axis

100% got this right

CheckPoint


In which case is the torque due to the force about the rotation axis biggest?

A) Case 1 B) Case 2 C) Same

A) Perpendicular force means more torque.

B) F*L = torque. L is bigger in Case 2 and the force is the same.

C) Fsin30 is F/2 and its radius is L so it is FL/2 which is the same as the other one as it is FL/2.

FL/2 90o

Case 1

axis

L F30o

Case 2

axis

CheckPoint


Torque and AccelerationRotational “2nd law”


Similarity to 1D motion


Summary : Torque and Rotational “2nd law”


Clicker Question A.

B.

C.

0% 0%0%

Strings are wrapped around the circumference of two solid disks and pulled with identical forces. Disk 1 has a bigger radius, but both have the same moment of inertia.

Which disk has the biggest angular acceleration?

A) Disk 1

B) Disk 2

C) same FF

ω1ω2


Clicker/CheckpointA.

B.

C.

0% 0%0%

Two hoops can rotate freely about fixed axles through their centers. The hoops have the same mass, but one has twice the radius of the other. Forces F1 and F2 are applied as shown.

How are the magnitudes of the two forces related if the angular acceleration of the two hoops is the same?

A) F2 = F1

B) F2 = 2F1

C) F2 = 4F1

F1

F2

Case 1 Case 2


CheckPoint

How are the magnitudes of the two forces related if the angular acceleration of the two hoops is the same?

A) F2 = F1

B) F2 = 2F1

C) F2 = 4F1

B) twice the radius means 4 times the moment of inertia, thus 4 times the torque required. But twice the radius=twice the torque for same force. 4t = 2F x 2R

M, R M, 2R

F1

F2

Case 1 Case 2


θ

θτ sinRF=

090=θ

00=θ

543690 =−=θMechanics Lecture 15, Slide 50

Direction is perpendicular to both R and F, given by the right hand rule

θτ sinRF=

0=xτ

0=yτ

321 FFFz ττττ ++=


Use (i) & (ii)

Use (iii)

(ii) ατ I=

(i) 2

21 MRIDISK =

(iii) 2

21 ωIK =


Moment of Inertia

2

2

222

222

)()15726(

)()2552

1516(

)5(52)4(

151

)(52

31

spherespheretotal

spherespheretotal

spherespherespherespherespherespheretotal

sphererodspherespheresphererodrodtotal

sphererodtotal

RmI

RmI

RmRmRmI

RLmRmLmI

III

=

++=

++=

+++=

+=


Moment of Inertia

2)()15726(

2

90sin2

sin

spheresphere

rod

rod

Rm

FL

I

I

FLrF

==

=

==

τα

ατ

θτ


Moment of Inertia

2

2222

2222

,,

)(41

52

45

6016

)21(

52)

25(

51)4(

601

)21(

52)

25(

121

5.4

spherespheretotal

spherespherespherespherespherespherespherespheretotal

spherespherespherespheresphererodrodrodtotal

sphererodtotal

spheresphererod

sphereCMsphererodCMrodCM

RmI

RmRmRmRmI

RmRmRmLmI

III

Rmm

RmRmR

+++=

++

+=

++

+=

+=

=+

+=


Moment of Inertia

2)(41

52

45

6016

0

0sin25sin

spheresphere

sphere

RmI

I

FRrF

+++

==

=

==

τα

ατ

θτ


Moment of Inertia

++

+=

+=

2222 )(52)4(

121

spherespherespherespheresphererodrodrodtotal

sphererodtotal

RmRmRmLmI

III


classical mechanics lecture 15 - astronomy · classical mechanics lecture 15 ... the contribution...

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