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9.19 Emilie's potter's wheel rotates with a constant 2.25 rad/s2angular acceleration. After

4.0 s, the wheel has rotated through an angle of 60.0 rad. What was the angular elocit!

of the wheel at the "eginning of the 4.0 s interal#

We choose the direction of rotation of the fl!wheel to "e positie.

Then: (-o) = ot +(1/2)t2

gieso = (-o)/t -(1/2)t = (60rad)/4s (1/2)(2.25rad/s2)(4s) = 10.5 rad/s

9.26 A fl!wheel with a radius of 0.$00 m starts from rest and accelerates with a constant

angular acceleration of 0.600 rad/s2. %ompute the magnitude of the tangential

acceleration, the radial acceleration, and the resultant acceleration of a& a point on its rim

at the start "& a point on its rim after it has turned through 60.0o c& a point on its rimafter it has turned through (20o.

We have: arad= r

2

and rada depending on the rotation rate, it changes as the flywheelaccelerates. atan = rand it is constant. Since rada and tana are in perpendicular

directions, the resultant acceleration is 2 2tan rad .a a a= +

Solve: (a) = 0 so rad 0.a = atan = r = (0.3m)(0.6rad/s2) = 0.18m/s2

Hence: a = 0.18m/s2

(b) for =( /3) : 2= 0+ 2 ( -0); since0= 0; then 2= 2 ( -0); with arad= r

2

arad= r(2(-0)); arad= 0.377m/s2

( ) ( )2 22 2 20.()0 m/s 0.$** m/s 0.4() m/sa = + =

(c) for =(2 /3) : arad= r(2(-0)); arad= 0.754m/s2

( ) ( )2 2

2 2 20.()0 m/s 0.*54 m/s 0.**5 m/sa = + =

9.28Electric tooth"rushes can "e effectie in remoing dental pla+ue. ne modelconsists of a head (.(0 cm in diameter that rotates "ac- and forth through a *0.0oangle

*600 times/minute. he rim of the head contains a thin row of "ristles a& What is the

aerage angular speed in each direction of the rotating head, in rad/s# "& What is theaerage linear speed in each direction of the "ristles against the teeth# c& sing !our own

o"serations, what is the approimate speed of the "ristles against !our teeth when !ou

"rush "! hand with an ordinarytooth"rush#

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In one back and forth cycle the head turns through 140o= 2.443 rad.

(a) In ( min 60 s,= the head turns through ( ) 4*600 2.44$ rad (.)5* (0 rad.= 3

= ( /t) = (1.8573 10 4rad)/(60s) = 310/rad/s

(b) = ( /t) = 5( /2)/(1s) = 7.9 rad/s

() ! = r = (0.0055")(#.9rad/s) = 0.04 "/s

9.$$1our small 0.200 -g spheres, each of which !ou can regard as a point mass, are

arranged in a s+uare 0.400 m on a side and connected "! light rods. 3ee the figure &.

1ind the moment of inertia of the s!stem a"out a& an ais through the center of thes+uare, perpendicular to its plane at point O "& a"out an ais along the line AB c& a"out

an ais along the line CD.

(a) Each mass is a distance( )

22 0.400 m 0.400 m

2 2= from the axis.

mr2 40.2-g&0.4m/ 2 &2 0.064 -g.m2

(%) Each mass is a distance 0.2m from the ais

mr2 40.2-g&0.2m&2 0.0$2-g.m2

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() two masses are at a distance 0.4m/ 2 & and two masses are on the ais7

mr2 20.2-g&0.4m/ 2 &2 0.0$2 -g.m2

9.$6A wagon wheel is constructed as shown in the figure. he radius of the wheel is0.$00 m, and the rim has a mass of (.*0 -g. Each of the wheel's eight spo-es, whichcome out from the center and are 0.$00 m long, has a mass of 0.220 -g. What is the

moment of inertia of the wheel a"out an ais through its center and perpendicular to the

plane of the wheel#

The rim is a thin-walled hollow cylinder with2

rimI m R= and each of the 8 spokes can be

treated as a slender rod with the axis at one end, so for each spoke2(

spo-e$.I m R=

& = ("r'"2)) + 8(1/$)"s*oe

2, = 0.19$."2

9.$8A grinding wheel in the shape of a solid dis- is 0.200 m in diameter and has a mass

of $.0 -g. he wheel is rotating at 2200 rpm a"out an ais through its center a& What is

its -inetic energ!# "& 8ow far would it hae to drop in free fall to ac+uire the sameamount of -inetic energ!#

For a solid disk and an axis through its center,2(

2.I MR= K = (1/2)I2cwhere must be

in rad/s.

We also have 1rpm = (2 rad)(60s)

(a) K = (1/2)[(1/2)(3.kg)(0.1m)2][(2200)( (2 rad)(60s) = 397 J

(b) In free fall, gravitational potential energy mgyis converted to kinetic energy.

K mgy= and

( )( )2$9* :

($.5 m$.00 -g 9.)0 m/s

Ky

mg= = =

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9.4$ A solid uniform $.25 -g c!linder, 65.0 cm in diameter and (2.4 cm long, is

connected to a (.50 -g weight oer two massless frictionless pulle!s as shown in thefigure. he c!linder is free to rotate a"out an ale through its center perpendicular to its

circular faces, and the s!stem is released from rest a& 8ow far must the (.50 -g weight

fall "efore it reaches a speed of 2.50 m/s# "& 8ow fast is the c!linder turning at thisinstant#

9.46 A "ic!cle racer is going downhill at ((.0 m/s when, to his horror, one of his 2.25 -g

wheels comes off when he is *5.0 m a"oe the foot of the hill. We can model the wheel

as a thin;walled c!linder )5.0 cm in diameter and neglect the small mass of the spo-esa& 8ow fast is the wheel moing when it reaches the foot of the hill if it rolled without

slipping all the wa! down# "& 8ow much total -inetic energ! does the wheel hae when

it reaches the "ottom of the hill#

The wheel has 2 ,I MR= with 2.25 -gM = and 0.425 m.R = Rolling without slipping

means vcm= R ; = (v cm/R) for the wheel. Initially the wheel has vcm,i = 11m/s . Use

coordinates where +y is upward and 0y = at the bottom of the hill, so i *5.0 my = and

f 0.y =

Solve: (a) Conservation of energy gives i i f f .K U K U+ = +

( )2

2 2 2 2 2cm( ( ( (cm cm cm cm2 2 2 2

.K m I m mR mR

= + = + =

yy v y y

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K = (1/2)mvcm2 + (1/2)Icm

2

K = mvcm2

Ki= mvcm,i2 ; Kf= mvcm,f

2 ; Ui= mgyi ; Uf = mgyf

2i cm, i ,K m= y 2f cm,f .K m= y i i ,U mgy= f f 0U mgy= = so 2 2i cm,i cm ,f .mgy m m+ =y y

hence: vcm,f2= vcm,i

2+ gyi vcm,f= 29.3m/s

(b) Kf= mvcm,f2 = 1.93 x 103J

9.50A solid uniform mar"le and a "loc- of ice, each with the same mass, start from restat the same height 8 a"oe the "ottom of a hill and moe down it. he mar"le rolls

without slipping, "ut the ice slides without friction a& 1ind the speed of each of these

o"