Examples in Chapter 9
9.25

A flywheel with a radius of 0.3 m starts from
rest and accelerates with a constant angular
acceleration of 0.6 rad/s2 . Compute the
magnitude of



a)
b)
c)
the angular accleration
the radial acceleration
The resultant acceleration of a point on its rim
At the start
After it has turned through 600
After it has turned through 1200
Starting conditions
a = 0.3 rad/s
 w0 =0
 r= 0.3 m

Equations to think about
v  rw
atan  ra
aradial  w 2 r
a a
2
2
tan
a
r  0.3
a  0.6
2
radial
At the start
v  rw
atan  ra
aradial  w 2 r
2
2
a 2  atan
 aradial
v  0.3*0  0
atan  0.3*0.6  0.18 m/s 2
aradial  w 2 r  0
2
2
a 2  atan
 aradial
a  0.18 m/s 2
r  0.3
a  0.6
At 600
v  rw
w 2  w02  2a    0 
atan  ra
w 2  02  2*0.6*
aradial  w r
2
aradial  1.2562 *.3  .3769
a 2  .182  .37692
a  0.417 m/s
r  0.3
a  0.6
w0  0

1800
3
 w  1.256
atan  .18
2
2
a 2  atan
 aradial
  0  600 *



3
At 1200
v  rw
w 2  w02  2a    0 
atan  ra
w 2  02  2*0.6*
aradial  w r
2
aradial  1.2562 *.3  .3769
a 2  .182  .37692
a  0.417 m/s
r  0.3
a  0.6
w0  0

1800
3
 w  1.256
atan  .18
2
2
a 2  atan
 aradial
   0  1200 *


2
3
9.46

A light flexible rope is wrapped several times
around a hollow cylinder with a weight of 40 N
and a radius of 0.25 m that rotates without
friction about a fixed horizontal axis. The
cylinder is attached is attached to the axle by
spokes of a negligible moment of inertia. The
cylinder is initially at rest. The free end of the
rope is pulled with a constant force P for a
distance of 5 m at which the end of the rope is
moving at 6 m/s. If the rope does not slip on
the cylinder, what is the value of P?
Some figurin’

W=DR

Where R= ½ I w2
v=rw or v/r=w
 Initially, w0 =0 so DR= ½ I (v/r)2
 From I=m*r2=(40 N/9.8)*(.25)2=0.255
 W=F*d or P*(5 m)
 W=5P
 P= (1/2)*0.255*(6/.25)2/5=14.7 N

9.71

a)
b)
A vacuum cleaner belt is
looped over a shaft of
radius 0.45 cm and a
wheel of radius 2.0 cm .
The arrangement is shown
below. The motor turns the
shaft at 60 rev/s and the
shaft is connected to a
beater bar which sweeps
the carpet. Assume that
the belt doesn’t slip.
What is the speed of a
point on the belt?
What is the angular
velocity of the belt?
Some more figurin’

Part a) v=r*w where
r=0.45 cm
 w= 60 rev/s *(2* radians/rev)=377 rad/s
 v=0.45*377=169 cm/s or 1.69 m/s


Part b) w2= 169 cm/s / 2 cm =84.8 rad/s
Hint on 9.72

The wheels are coupled so that there is
the tangential velocity is constant so that
w1/w2 = r2/r1