•  Main points of today’s lecture: •  Analogies between rotational and

translational motion •  Solutions involving:

•  Rolling motion:

ω

ω

ατ

IL

IKE

II

rot

=

=

=

2

21

22

21

21 ωIMvKEtotal +=

Physics 231 Lecture 20

Example •  A uniform steel beam of length 5.00 m has a weight of 4.5x103 N. One end of the beam is

bolted to a vertical wall. The beam is held in a horizontal position by a cable attached between the other end of the beam and a point on the wall. The cable makes an angle of 25o above the horizontal. A load whose weight is 12.0x103 N is hung from the beam at a point that is 3.5 m from the wall. Find (a) the magnitude of the tension in the supporting cable and (b) the magnitude of the force exerted on the end of the beam by the bolt that attaches to the wall.

25o

load

Choose pivot to be at hinge: (cancels out hinge force)

τ i = 0 = T(5m)sin 25o( )i∑ − Wload 3.5m( )− Wbeam 2.5m( )

T!

R!

loadW!

beamW!

( )( )

( )( )load beamo o

3.5m 2.5mT W W

(5m)sin 25 (5m)sin 25= ⋅ + ⋅

4T 2.54x10 N⇒ =

Hinge forces:

!Fi = 0; x-comp: Rx -Tcos 25o( ) = 0

i∑

x comp: R x = Tcos 25o( ) = 2.29x104 N

oy load beam y-comp: R W W Tsin(25 ) 0− − + =

o 3 3 3y load beamR W W Tsin(25 ) 12.0x10 N 4.5x10 N 10.7x10 N= + − = + −

3 2 2 3y x y R 5.8x10 N R R R 23.6x10 N= ⇒ = + =

τ = Frsin θ( ); counterclockwise is positive, clockwise negative

quiz

•  A 1-kg rock is suspended by a massless string from one end of a 1-m measuring stick.What is the mass of the measuring stick if it is balanced by placing the fulcrum at the 0.25-m mark? –  a) 0.25 kg –  b) 0.5 kg –  c) 1 kg –  d) 2 kg –  e) none of the above

Calculate the torques choosing the pivot to be at the fulcrum:i stick

i0 0.25m 1kg g 0.25m M g 0τ = = ⋅ ⋅ − ⋅ ⋅ =∑

stickM 1kg=

!

WW=mg

Quiz

•  A uniform beam of length 5 m and mass 100kg is supported at the middle by a pivot and is clamped at the right end as shown below. A 80 kg man stands on the beam 0.5 m from the left end. What is the force exerted by the clamp on the beam? –  a) 1760 N up –  b) 627 N down –  c) 1411 N up –  d) 830 N down

clamp 0.5 m

Put axis at center to remove torque from pivot

clampF must be down to counteract torque from man

( )i man clampi

0 W (2.5m 0.5m) F 2.5mτ = = ⋅ − −∑

( ) ( )2clamp man

(2.0m)F W 80kg 9.8m / s (.8) 627N down2.5m

⇒ = ⋅ = =

Conceptual example

•  A box, with its center-of-mass off-center as indicated by the dot, is placed on an inclined plane. In which of the four orientations shown, if any, does the box tip over?

If the box is about to tip over, it’s weight will be resting mainly on its leftmost corner. Assume that is the starting point. Make that the pivot, then the only torque comes from gravity on the box. In cases A,B and D, the gravitation torque makes it rotate clockwise, which increases its stability. In the case of C the torque is counter clockwise, which makes it tip over.

Conceptual question

•  The center of gravity of an object must be physically located somewhere within the material that makes up the object. –  a) true –  b) false

Moment of inertia for collections of masses •  Let’s consider the tangential acceleration of a

single mass m1 on attached to a massless rod of length r1 and fixed to rotate about axis O: m1

r1

Fig 8.14, p.229

Slide 25

F1t

m1 r1 r2

m2

m3 ( ) αατα 121111111111 IrmrFrmamF ttt ===⇒==

•  If we attach a second mass m2 on attached to a massless rod of length r2 which is rigidly attached to the pivot so its angular acceleration is the same:

( )( ) αααατττ

αταIrmrmrmrm

rmrFrmamF

net

ttt

=+=+=+=

==⇒==222

211

222

21121

22222222222

•  Similarly for three particles:

( ) ααττττ Irmrmrmnet =++=++= 233

222

211321

•  In general:

∑≡++++=i

iirmrmrmrmrmI 2244

233

222

211 ....

Moments of inertia of extended objects

•  To get the moments of inertia, one calculates I=Σmiri

2 as well. Some values are given in the table to the right.

•  Last row is an illustration of the “parallel axis theorem”:

2CMCM MRII +=

•  Here ICM is the moment of inertial about an axis going through the center of mass. I is the moment of inertia about another axis which is parellel to and displaced a distance RCM from the center of mass.

What is the moment of inertia of a cylinder with M=2 kg and and R=0.5 m?

a) 0.25 kg�m2

b) 0.5 kg�m2

c) 1.0 kg�m2 d) 0.125 kg�m2

I = 12

MR2

= 12⋅2kg ⋅ 0.5m( )2

= 0.25kg ⋅m2

Analogies between straight line and rotational motion •  We already introduced the following analogous equations and quantities:

•  New analogies:

( )

( )

θαωω

αωθ

ωωθ

ωωω

αωωωθ

Δ+=

+=Δ

+=Δ

+=

+==Δ

2

21

21

21

20

2

20

0

0

0

tt

t

tt

( )

( )

xavv

attvx

tvvx

vvv

atvvtvx

o

o

Δ=−

+=Δ

+=Δ

+=

+==Δ

221

2121

20

2

20

0

mvp

mvKE

maFm

trans

=

=

=

2

21

ω

ω

ατ

IL

IKE

II

rot

=

=

=

2

21

rotational kinetic energy

angular momentum

Example

•  A turntable platter has a radius of 0.150 m and is rotating at 3.49 rad/s. When the power is shut off, the platter slows down and comes to rest in 15.0 s, due to a net retarding torque of 6.2x10-3 N�m. Assume the platter to be a uniform solid disk. Determine the mass of the platter. (Hint: calculate α from the change in ω, plug into Newton’s second law to find M.) –  a) 1.37 kg –  b) 2.37 kg –  c) 3.37 kg –  d) 4.37 kg

21I MR2

=

200

0 3.49rad / st .233rad / st 15s

ω ωω ω α α − −= + ⇒ = = = −

21I MR2

τ α α= =

( ) ( )3

22 2

6.2x10 NmM 2.37kg1 1R 0.15m 0.233rad / s2 2

τ

α

−−⇒ = = =−

R 0.15m

ω0 3.49 rad/s

ω 0

t 15 s

τ -6.2x10-3Nm

M ?

Reading Quiz 2.  An object will be stable if

A.  its center of gravity is below its highest point. B.  its center of gravity lies over its base of support. C.  its center of gravity lies outside its base of support. D.  the height of its center of gravity is less than 1/2 its total

height.

Slide 8-7

Reading Quiz 4.  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

Reading Quiz 3.  Which statement about an object’s center of gravity is not true

A.  If an object is free to rotate about a pivot, the center of gravity will come to rest below the pivot.

B.  The center of gravity coincides with the geometric center of the object.

C.  The torque due to gravity can be calculated by considering the object’s weight as acting at the center of gravity.

D.  For objects small compared to the earth, the center of gravity and the center of mass are essentially the same.

Slide 7-9

Example •  A cylindrical 5.00-kg reel with a radius of 0.600 m

and a frictionless axle, starts from rest and speeds up uniformly as a 3.00-kg bucket falls into a well, making a light rope unwind from the reel (a) What is the linear acceleration of the falling bucket? (b) What is the angular acceleration of the reel?

bucket bucketa) For bucket: T m g m a− =

reelFor reel: TR Iτ α= =

also: a / Rα = −

T − mbucketg = −

IreelaR2 − mbucketg

bucket reel bucket1m a M a m g2

+ = −

22a 5.3m / sb) - 8.9rad / s

R 0.6mα = = =

2 2bucket

bucket reel

m 3kga g 9.8m / s 5.3m / s1 5.5kgm M2

−⇒ = = − = −

+

bucketm a=

reelITRα

⇒ =

reel2

I aTR

⇒ = − 2reel reel

1and I M R2

=

/

/

T

Mreel

R

mbucket

= −

12

MreelR2a

R2 − mbucketg

Energy •  A car is designed to get its energy from a rotating flywheel with a

radius of 2.00 m and a mass of 500 kg. Before a trip, the flywheel is attached to an electric motor, which brings the flywheel’s rotational speed up to 5000 rev/min. (a) Find the kinetic energy stored in the flywheel. (b) If the flywheel is to supply energy to the car as would a 10.0-hp motor, find the length of time the car could run before the flywheel would have to be brought back up to speed.

mi vi

ri

2i i i1KE m v2

=

2a) 5000 / 60 2 rad / s 5.23x10 rad / sω π= ⋅ =

( )2i i1 m r2

ω= 2 2i i

1 m r2

ω= 2i

1 I2

ω=

2total i i

i

1KE m v2

=∑ 2i

i

1 I2

ω=∑ 2total

1 I2

ω=

( )22 21 1I MR 500kg 2m 1000kg m2 2

= = ⋅ ⋅ = ⋅

( ) ( )22 2 21 1KE I 1000kg m 5.23x10 rad / s2 2

ω= = ⋅8KE 1.37x10 J=

( )( )8

4KE 1.37x10 Jb) KE P t t 1.8x10 s 5hP 10hp 746W / hp

= Δ ⇒ Δ = = = ≈

R 2m M 500kg ω 5000

rev/min P 10 hp t ?

one mass:

all masses:

Rolling: Energy of the center of Mass Energy in the center of Mass

•  If you have an object rolling with velocity v, the total kinetic energy of the object is equal to the sum of the translational kinetic energy of the center of mass:

plus the rotational kinetic energy of the object about its center of mass: •  Often vcm is obvious if the center of mass is obvious. If you only have

the masses and velocities of the particles that make up the object, you can compute vcm from

2trans cm

1KE Mv2

=

2rot

1KE I2

ω=

!vcm = 1

Mmi!vi ≡!Ptot

Mi∑

Which one wins?

•  Which object will be fastest? –  a) hoop –  b) cylinder –  c) sphere

Rolling down the hill

•  Starting from rest, an object with moment of mass M, moment of inertia I, and radius R rolls down an incline of height h. What is the velocity at the bottom? (Hint use conservation of energy.)

h 22

21

21 ωIMvKEKEKE rottranstotal +=+=

Conservation of energy:

ghv

ghvc

ghvb

ghva

2 d)7

10 )

34 )

)

=

=

=

=

•  What is v for a hoop? (I=MR2)

•  What is v for a cylinder? (I=1/2MR2)

•  What is v for a sphere? (I=2/5MR2)

•  What do you use if the object slides without rolling?

KEtotal,f =

12

Mv2 + 12

Iω 2 = MghvRolling motion: v RR

ω ω= ⇒ =

22 2

2

1 1 v 1 1 IMgh Mv I M v2 2 R 2 2 R

⎛ ⎞ ⎡ ⎤= + = +⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

22

Mgh 2ghv I1 1 I 1MMR2 2 R

= =⎡ ⎤ ++⎢ ⎥⎣ ⎦

Rolling down the hill

•  Starting from rest, an object with moment of mass M, moment of inertia I, and radius R rolls down an incline of height h. What is the velocity at the bottom? (Hint use conservation of energy.)

h 22

21

21 ωIMvKEKEKE rottranstotal +=+=

Conservation of energy:

ghv

ghvc

ghvb

ghva

2 d)7

10 )

34 )

)

=

=

=

=

•  What is v for a hoop? (I=MR2)

•  What is v for a cylinder? (I=1/2MR2)

•  What is v for a sphere? (I=2/5MR2)

KEtotal,f =

12

Mv2 + 12

Iω 2 = MghvRolling motion: v RR

ω ω= ⇒ =

22 2

2

1 1 v 1 1 IMgh Mv I M v2 2 R 2 2 R

⎛ ⎞ ⎡ ⎤= + = +⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

22

Mgh 2ghv I1 1 I 1MMR2 2 R

= =⎡ ⎤ ++⎢ ⎥⎣ ⎦

Rolling down the hill

•  Starting from rest, an object with moment of mass M, moment of inertia I, and radius R rolls down an incline of height h. What is the velocity at the bottom? (Hint use conservation of energy.)

h 22

21

21 ωIMvKEKEKE rottranstotal +=+=

Conservation of energy:

ghv

ghvc

ghvb

ghva

2 d)7

10 )

34 )

)

=

=

=

=

•  What is v for a hoop? (I=MR2)

•  What is v for a cylinder? (I=1/2MR2)

KEtotal,f =

12

Mv2 + 12

Iω 2 = MghvRolling motion: v RR

ω ω= ⇒ =

22 2

2

1 1 v 1 1 IMgh Mv I M v2 2 R 2 2 R

⎛ ⎞ ⎡ ⎤= + = +⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

22

Mgh 2ghv I1 1 I 1MMR2 2 R

= =⎡ ⎤ ++⎢ ⎥⎣ ⎦

Rolling down the hill

•  Starting from rest, an object with moment of mass M, moment of inertia I, and radius R rolls down an incline of height h. What is the velocity at the bottom? (Hint use conservation of energy.)

h 22

21

21 ωIMvKEKEKE rottranstotal +=+=

Conservation of energy:

ghv

ghvc

ghvb

ghva

2 d)7

10 )

34 )

)

=

=

=

=

•  What is v for a hoop? (I=MR2)

KEtotal,f =

12

Mv2 + 12

Iω 2 = MghvRolling motion: v RR

ω ω= ⇒ =

22 2

2

1 1 v 1 1 IMgh Mv I M v2 2 R 2 2 R

⎛ ⎞ ⎡ ⎤= + = +⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

22

Mgh 2ghv I1 1 I 1MMR2 2 R

= =⎡ ⎤ ++⎢ ⎥⎣ ⎦

Conceptual question

•  An external force F is applied to a dumbbell for a time interval !t, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater center-of-mass speed?

–  a) (a) –  b) (b) –  c) no difference –  d) The answer depends on the rotational inertia of the

dumbbell.

recall that

ext tot cmF t P M vΔ = Δ = Δv v v

Angular momentum

•  In analogy to the linear momentum we have the angular momentum . For a single particle moving a circle, the magnitude of L is:

vmp !! =ωIL !!

=

tmrvωmrIωL === 2

m1 r1

Fig 8.14, p.229

Slide 25

m r

vt

•  Recall that if there are no external forces, the linear momentum will be conserved. Similarly, if there no external torques, the angular momentum will be conserved.

•  For example, this means that an isolated object will keep the same value of the angular momentum throughout a problem. –  In linear motion, conservation of for an isolated object meant that

the velocity must remain constant. –  In angular motion, conservation of for an isolated object does not

mean must remain constant, because I can change!

vmp !! =v!

ωIL !!=

ω!

0LLf = 00ωω II ff =⇒ 00 ωωf

f II=⇒

Example

•  A woman stands at the center of platform. The woman and the platform rotate with an angular speed of 5.00 rad/s. Friction is negligible. Her arms are outstretched and she is holding a dumbbell in each hand. In this position, the total moment of inertia of the rotating system (platform, woman and dumbbells) is 5.4 kg�m2. By pulling in her arms, the moment of inertia is reduced to 3.8 kg�m2. Find her new angular speed. –  a) 3.1 rad/s –  b) 7.1 rad/s –  c) 13.8 rad/s –  d) 23.4 rad/s

0f 0

f

II

ω ω=

f5.4 5rad / s 7.1rad / s3.8

ω = =

ω0 5 rad/s I0 5.4 kg�m2

If 3.8 kg�m2

ωf ?

0LLf =

00ωω II ff =⇒

quiz

•  A diver leaps from the 10 m platform and executes a triple forward somersault dive. While in the tuck, the diver has a moment of inertia of about 3.3 kg�m2 and during the final layout, the diver stretches her body and increases her moment of inertia to 10 kg�m2. If the diver achieves an angular velocity of 2 rev/s while in the tuck, what is the angular velocity during the layout? –  a) 6 rev/s –  b) 66 rev/s –  c) 0.66 rev/s –  d 0.17 rev/s

0f 0

f

II

ω ω=

ω0 2 rad/s I0 3.3 kg�m2

If 10 kg�m2

ωf ?

f3.3 2rad / s 0.66rad / s10

ω = =

0LLf =

00ωω II ff =⇒