8.2. Model: The model rocket will be treated as a particle. Kinematic equations in two dimensions apply. Air
resistance is neglected.
Visualize:
The horizontal velocity of the rocket is equal to the speed of the car, which is 3.0 m/s.
Solve: For the rocket, Newton’s second law along the y-direction is:
(Fnet ) y  FR  mg  maR  ay 



 



1 
8.0 N  0.5 kg 9.8 m/s2   6.2 m /s 2
0.5 kg 

2
Thus using y1  y0  (v0 ) y t1  t0  12 a y t1  t0 ,
20 m 0 m  0 m  6.2 m/s t
2
1
2
1R
0 s
  20 m  3.1 m/s t
2
2
2
1
 t1  2.54 s
Since t1 is also the time for the rocket to move horizontally up to the hoop,



x1  x0  (v0 ) x t1  t0  12 ax t1  t0
  0 m  3.0 m/s2.54 s  0 s 0 m  7.6 m
2
Assess: In view of the rocket’s horizontal speed of 3.0 m/s and its vertical thrust of 8.0 N, the above-obtained value for
the horizontal distance is reasonable.
8.6. Model: Treat the block as a particle attached to a massless string that is swinging in a circle on a frictionless table.
Visualize:
Solve: (a) The angular velocity and speed are
rev2 rad

  75471.2
 rad/min

min1 rev
1min
vrt    0.50 m471.2
radmin3.93

 m/s 
60 s
The tangential velocity is 3.9 m/s.
(b) The radial component of Newton’s second law is
 FT
r
Thus

mv2
r
3.93 m/s 
T  0.20 kg6.2
 N
0.50 m
2
8.10. Model: Model the ball as a particle in uniform circular motion. Rolling friction is ignored.
Visualize:
Solve:
The track exerts both an upward normal force and an inward normal force. From Newton’s second law,
Fnmr
net2

60 rev2 rad1min

 2  0.030 kg0.20

 m0.24
 N 
 min1 rev60 s
2

 

8.14. Model: The passengers are particles in circular motion.
Visualize:
Solve: The center of the circle of motion of the passengers is directly above them. There must be a net force pointing up
that provides the needed centripetal acceleration. The normal force on the passengers is their weight. Ordinarily their
weight is FG , so if their weight increases by 50%, nF 1.5 .G Newton’s second law at the bottom of the dip is
 GG 1.510.5
  
 FnFFmg
r 

mv2
r
2
 vgr
 0.50.59.8

m/s30
 m12.1
 m/s 
Assess: A speed of 12.1 m/s is 27 mph, which seems very reasonable.
8.18. Model: Use the particle model for the car, which is undergoing nonuniform circular motion.
Visualize:
Solve:
The car is in circular motion with radius r 
d
 100 m. We require
2
22
1.5 m/s1.5
m/s



r
100 m
The definition of the angular velocity can be used to determine the time t using the angular acceleration
a 1.5 m/s2
22 
 t 
 1.510
 s.
r
100 m
arr   221  1.5 m/s0.122
  s
  i  t
 t 
11
  i 0.122 s0
s


0.015 s2

 8.2 s
8.22. Model: Treat Sam as a particle.
Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xy-coordinates for the
second. Sam’s final velocity at the top of the slope is his initial velocity as he becomes airborne.
Solve: Sam’s acceleration up the slope is given by Newton’s second law:
()sin10
FFmgma

net0 s 
ag
0 

F
200 N
22
 sin10(9.8
  m/s)sin100.965

m/s  
m
75 kg
The length of the slope is s1  (50 m)/sin10288
  m.
His velocity at the top of the slope is
222
vvassasv
  2()22(0.965

 m/s)(288
 m)23.6

m/s
10010011

sin10410
  m/s.
This is Sam’s initial speed into the air, giving him velocity components vv
cos1023.2

m/s and vv
11
y 
11
x 
This is not projectile motion because Sam experiences both the force of gravity and the thrust of his skis. Newton’s second law for
Sam’s acceleration is
a1x 
a1y 
()Fnet
m
y

()(200
Fnet x N)cos10

m
75 kg

 2.63 m/s2
(200 N)sin10(75
  kg)(9.80 m/s)2
 9.34 m/s2
75 kg
The y-equation of motion allows us to find out how long it takes Sam to reach the ground:
222
1
yyvtattt
 0 m50
 m(4.10
 yy m/s)(4.67
 m/s)

21121222
2

This quadratic equation has roots t2  2.86 s (unphysical) and t2  3.74 s.The x-equation of motion—this time with an
acceleration—is
xxvtattt
 
21121222
xx
 11
22
Sam lands 105 m from the base of the cliff.
 0 m(23.2

m/s)(2.63
 m/s)105 m
222

8.26. Model: The model rocket is treated as a particle and its motion is determined by constant-acceleration kinematic
equations.
Visualize:
Solve: As the rocket is accidentally bumped v0x  0.5 m/s and v0 y  0 m/s. On the other hand, when the engine is fired
Fmaa

xxx
 
(a) Using yyvttatt
1001010

yy

  12 



Fx
20 N

 40 m/s2
m 0.500 kg
,
2
0 m40
 m0 m9.8

m/s2.857
 12 
s
tt 
22
11

The distance from the base of the wall is
xxvttatt
 
1001010

xx

  12 


2
2
 0 m0.5
  m/s2.85740
s  12  s165
 m/s2.857

m

2

(b) The x- and y-equations are
yyvttatt
 0000

xxvttatt
 0000

2
Except for a brief interval near ttt 0, 200.5.


  12 


  12 
yy
xx



2

2
 404.9
 t2
 0.520
tt
2
Thus xt 20,2 or tx2  /20. Substituting this into the y-equation gives
yx 400.245

This is the equation of a straight line, so the rocket follows a linear trajectory to the ground.
8.30. Model: Model the ball as a particle which is in a vertical circular motion.
Visualize:
Solve: At the bottom of the circle,
F
mv 2
 0.500 kg  v  v  5.5 m/s
 15 N    0.500 kg   9.8 m/s 2  
r
1.5 m 
2
r
 T  FG 
8.34. Model: Use the particle model for the car, which is in uniform circular motion.
Visualize:
Solve: Newton’s second law is
 sin20 
 FTma
mv2
r

rr

 FTF
z
cos200

G

N
These equations can be written as
mv2
Tmg
cos20 
r
T sin20 
Dividing these two equations gives
22
tan20tan204.55
  vrgvrg
 m9.8
 m/stan204.03
   m/s 


8.38. Model: Use the particle model for the car and the model of kinetic friction.
Visualize:
Solve: We will apply Newton’s second law to all three cars.
Car A:
    
 FnfFfma
xx
kGkxx
x
  
 Fnfynmg
yy
y
k
y

 0 N0 N 
  0 N0 N 
, we have fmg
 . Since fn
The y-component equation means nmg
kk  
kk  
 fmg

kk

 k g9.8
  m/s
mm
Car B: Car B is in circular motion with the center of the circle above the car.
ax 

. From the x-component equation,
2
    
 FnfFnmgma
rr
kG rr
r
  0 N


        0 N0 N 
 FnfFfma
tt

kGktt
t
mv2
r
From the r-equation
22
mvv
nmgfnmg



rr

kkk

 





Substituting back into the t-equation,
ag
t 
fmv

kk

mmr

 

2
2


   
22 25 m/s




9.8
m/s12.9

m/s

k

200 m 


Car C: Car C is in circular motion with the center of the circle below the car.
    
 FnfFnmgma
rr
r
kG rr
   0 N


        0 N0 N 
 FnfFfma
ttt
  
From the r-equation nmgvr
2
mv2
r

kGktt
. Substituting this into the t-equation yields
 fn

kk
agvr

t 
mm
  k  
22
 6.7 m/s
8.42. Model: Model the chair and the rider as a particle in uniform circular motion.
Visualize:
Solve: Newton’s second law along the r-axis is
   
 FTFma
rrr
G r
 Tmr
sin0
 N

Since rL sin, this equation becomes
2rad

TmL
  2  150 kg9.0
 m3330
 N  
 4.0s 
2
Thus, the 3000 N chain is not strong enough for the ride.
2
8.46. Model: Masses m1 and m2 are considered particles. The string is assumed to be massless.
Visualize:
Solve: The tension in the string causes the centripetal acceleration of the circular motion. If the hole is smooth, it acts
like a pulley. Thus tension forces T1 and T2 act as if they were an action/reaction pair. Mass m1 is in circular motion of
radius r, so Newton’s second law for m1 is
 FTr 
1 
2
mv
1
r
Mass m2 is at rest, so the y-equation of Newton’s second law is
 
 FTmgTmg
y
Newton’s third law tells us that TT
12
2222
 0 N

. Equating the two expressions for these quantities:
2
mvmrg
12
 mgv

2 
rm
1
8.50. Model: Model the car as a particle on a circular track.
Visualize:
Solve: (a) Newton’s second law along the t-axis is
 
 FFma
 1000 N1500
  kg23aa
m/s
tt 
ttt

2
With this tangential acceleration, the car’s tangential velocity after 10 s will be
vvatt
1010
ttt 



2
 m/s10 s0
  0 ms23
s203m/s 
The radial acceleration at this instant is
v12t 203 m/s 16


m/s2
r
25 m9
2
ar 
The car’s acceleration at 10 s has magnitude
aaa
1

22222
tr

m/s
23 m/s169
  m/s1.90


22
a
ar
11 t
  tantan21


 23 


 169 

where the angle is measured from the r-axis.
(b) The car will begin to slide out of the circle when the static friction reaches its maximum possible value
 fnssmax  
.
That is,
mv22t
2
 vrg
 25 m9.8
 m/s15.7
  m/s
2t 
r

In the above equation, nmg
follows from Newton’s second law along the z-axis. The time when the car begins to slide
can now be obtained as follows:
 
 Ffnmg
r
vvatt
2020
ttt 
sssmax






 ms23m/s0
  15.7 m/s0

2
t
2

  t2  24 s
8.54. Solve: (a) At what speed does a 1500 kg car going over a hill with a radius of 200 m have a weight of 11,760 N?
(b) The weight is the normal force.
1500kg v2
2940N19.8
 m/s
v
200 m
8.58. Model: Use the particle model for the ball, which is in uniform circular motion.
Visualize:
Solve:
From Newton’s second law along r and z directions,
 Fn
r
cos 
mv2
r
 sin0sin

 Fnmgnmg
z

 
Dividing the two force equations gives
tan 
From the geometry of the cone, tan.  ry
gr
v2
Thus
rgr

yv
 vgy

2
8.62. Model: Use the particle model for the ball.
Visualize:
Solve: (a) Newton’s second law along the r- and z-directions is
 cos 
 Fnmr
2
r
Using Fmg
G 
 FnF
z
sin0
 N
G

and dividing these equations yields:
tan 

where you can see from the figure that tan.  Ryr

gRy

rr 2

Thus  
g
.
Ry
(b)  will be minimum when (y)
R is maximum or when y  0 m. Then min  gR
/.
(c) Substituting into the above expression,

g
Ry

2
9.8 m/srad60
s1 rev
 9.9 95 rpm
0.20 m0.10

ms1 min2 rad


