Sample Problem 3/4

c03.qxd
6/15/06
130
12:31 PM
Chapter 3
Page 130
Kinetics of Particles
Sample Problem 3/4
8
We approximate the resistance-velocity relation by R kv2 and
find k by substituting R 8 N and v 2 m/s into the equation, which gives
k 8/22 2 N s2/m2. Thus, R 2v2.
The only horizontal force on the model is R, so that
Solution.
[ΣFx max]
R max
or
2v2
6
R, N
The design model for a new ship has a mass of 10 kg and is tested in an experimental towing tank to determine its resistance to motion through the water at
various speeds. The test results are plotted on the accompanying graph, and the
resistance R may be closely approximated by the dashed parabolic curve shown. If
the model is released when it has a speed of 2 m/s, determine the time t required
for it to reduce its speed to 1 m/s and the corresponding travel distance x.
4
2
0
0
1
2
v, m /s
v0 = 2 m/s
v
dv
10
dt
x
W
We separate the variables and integrate to obtain
dt 5 t
0
v
2
dv
v2
t5
1v 12 s
1
1
Thus, when v v0/2 1 m/s, the time is t 5(1 2 ) 2.5 s.
Ans.
The distance traveled during the 2.5 seconds is obtained by integrating v dx/dt. Thus, v 10/(5 2t) so that
x
0
dx 2.5
0
10
dt
5 2t
x
10
ln (5 2t)
2
R
B=W
2.5
3.47 m
0
Ans.
Sample Problem 3/5
Helpful Hints
Be careful to observe the minus sign
for R.
Suggestion: Express the distance x
after release in terms of the velocity
v and see if you agree with the resulting relation x 5 ln (v0 /v).
θ
The collar of mass m slides up the vertical shaft under the action of a force
F of constant magnitude but variable direction. If kt where k is a constant
and if the collar starts from rest with 0, determine the magnitude F of the
force which will result in the collar coming to rest as reaches /2. The coefficient of kinetic friction between the collar and shaft is k.
F
µk N θ
m
F
N
µk
Solution.
After drawing the free-body diagram, we apply the equation of motion in the y-direction to get
[ΣFy may]
F cos k N mg m
dv
dt
Helpful Hints
where equilibrium in the horizontal direction requires N F sin . Substituting
kt and integrating first between general limits give
(F cos kt F sin kt mg) dt m t
0
k
v
dv
0
For /2 the time becomes t /2k, and v 0 so that
and
F
mg
2(1 k)
the vertical displacement y instead
of the time t, the acceleration would
become a function of the displacement and we would use v dv a dy.
pend on k, the rate at which the
force changes direction.
F
[sin kt k(cos kt 1)] mgt mv
k
mg
F
[1 k(0 1)] 0
k
2k
If were expressed as a function of
We see that the results do not de-
which becomes
mg
Ans.
115
13.4 EQUATIONS OF MOTION: RECTANGULAR COORDINATES
EXAMPLE 13.4
A smooth 2-kg collar C, shown in Fig. 13–9a, is attached to a spring
having a stiffness k = 3 N>m and an unstretched length of 0.75 m. If
the collar is released from rest at A, determine its acceleration and the
normal force of the rod on the collar at the instant y = 1 m.
SOLUTION
Free-Body Diagram. The free-body diagram of the collar when it is
located at the arbitrary position y is shown in Fig. 13–9b. Note that the
weight is W = 219.812 = 19.62 N. Furthermore, the collar is assumed
to be accelerating so that “a” acts downward in the positive y direction.
There are four unknowns, namely, NC , Fs , a, and u.
-NC + Fs cos u = 0
(1)
+ T ©Fy = may ;
19.62 - Fs sin u = 2a
(2)
From Eq. 2 it is seen that the acceleration depends on the magnitude
and direction of the spring force. Solution for NC and a is possible
once Fs and u are known.
The magnitude of the spring force is a function of the stretch s of the
spring; i.e., Fs = ks. Here the unstretched length is AB = 0.75 m,
Fig. 13–9a; therefore,
Since k = 3 N>m, then
s = CB - AB = 4y2 + 10.7522 - 0.75.
Fs = ks = 3 A 4y2 + 10.7522 - 0.75 B
u
y
B
k $ 3 N/m
C
Equations of Motion.
+ ©F = ma ;
:
x
x
0.75 m
A
(a)
a
x
19.62 N
y
(b)
Fs
u
NC
Fig. 13–9
(3)
From Fig. 13–9a, the angle u is related to y by trigonometry.
tan u =
y
0.75
(4)
Substituting y = 1 m into Eqs. 3 and 4 yields Fs = 1.50 N and
u = 53.1°. Substituting these results into Eqs. 1 and 2, we obtain
NC = 0.900 N
Ans.
2
a = 9.21 m>s T
Ans.
NOTE: This is not a case of constant acceleration, since the spring force
changes both its magnitude and direction as the collar moves downward.
Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River,
New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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129
13.5 EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES
EXAMPLE 13.6
Determine the banking angle u for the race track so that the wheels of
the racing cars shown in Fig. 13–12a will not have to depend upon
friction to prevent any car from sliding up or down the track. Assume
the cars have negligible size, a mass m, and travel around the curve of
radius r with a speed v.
u
(a)
b
SOLUTION
Before looking at the following solution, give some thought as to why
it should be solved using t, n, b coordinates.
Free-Body Diagram. As shown in Fig. 13–12b, and as stated in the
problem, no frictional force acts on the car. Here NC represents the
resultant of the ground on all four wheels. Since an can be calculated,
the unknowns are NC and u.
Equations of Motion. Using the n, b axes shown,
+ ©F = ma ;
:
n
n
+ c ©Fb = 0;
an
NC
n
u
W $ mg
v2
NC sin u = m
r
(1)
NC cos u - mg = 0
(2)
(b)
Fig. 13–12
Eliminating NC and m from these equations by dividing Eq. 1 by
Eq. 2, we obtain
v2
tan u =
gr
v2
u = tan-1 a b
Ans.
gr
NOTE: The result is independent of the mass of the car. Also, a force
summation in the tangential direction is of no consequence to the
solution. If it were considered, then at = dv>dt = 0, since the car
moves with constant speed. A further analysis of this problem is
discussed in Prob. 21–48.
Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River,
New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.
131
13.5 EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES
EXAMPLE 13.8
Design of the ski jump shown in the photo requires knowing the type of
forces that will be exerted on the skier and his approximate trajectory. If
in this case the jump can be approximated by the parabola shown in
Fig. 13–14a, determine the normal force on the 150-lb skier the instant he
arrives at the end of the jump, point A, where his velocity is 65 ft>s. Also,
what is his acceleration at this point?
SOLUTION
Why consider using n, t coordinates to solve this problem?
y
y$
Free-Body Diagram. The free-body diagram for the skier when he
is at A is shown in Fig. 13–14b. Since the path is curved, there are two
components of acceleration, a n and a t . Since an can be calculated, the
unknowns are at and NA .
x
200 ft
A
Equations of Motion.
+ c ©Fn = man ;
(a)
2
NA - 150 =
+ ©F = ma ;
;
t
t
0 =
150 1652
a
b
r
32.2
(1)
150
a
32.2 t
(2)
[1 + 1dy>dx22]3>2
ƒ d2y>dx2 ƒ
`
x=0
=
n
an
The radius of curvature r for the path must be determined at point
1
1
1
A(0, -200 ft). Here y = 200
x2 - 200, dy>dx = 100
x, d2y>dx2 = 100
,
so that at x = 0,
r =
1 x2 # 200
200
[1 + 1022]3>2
1
ƒ 100
ƒ
t
at
= 100 ft
Substituting into Eq. 1 and solving for NA , we have
NA = 347 lb
150 lb
(b)
NA
Ans.
Fig. 13–14
Kinematics. From Eq. 2,
at = 0
Thus,
an =
16522
v2
=
= 42.2 ft>s2
r
100
aA = an = 42.2 ft>s2 c
Ans.
NOTE: Apply the equation of motion in the y direction and show that
when the skier is in mid air the acceleration is 32.2 ft>s2.
Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River,
New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.
132
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
A C C E L E R AT I O N
AND
EXAMPLE 13.9
v0 $ 1 m/s
ds $ 0.5 du
SOLUTION
Free-Body Diagram. The free-body diagram for a package, when it
is located at the general position u, is shown in Fig. 13–15b.The package
must have a tangential acceleration a t , since its speed is always
increasing as it slides downward.The weight is W = 219.812 = 19.62 N.
Specify the three unknowns.
Equations of Motion.
v2
+b©Fn = man;
-NB + 19.62 cos u = 2
(1)
0.5
+R©Ft = mat;
19.62 sin u = 2at
(2)
u
du r $ 0.5 m
(a)
NB
an
at
u
19.62 N
t
n
(b)
Fig. 13–15
Packages, each having a mass of 2 kg, are delivered from a conveyor to
a smooth circular ramp with a velocity of v0 = 1 m>s as shown in
Fig. 13–15a. If the effective radius of the ramp is 0.5 m, determine the
angle u = umax at which each package begins to leave the surface.
At the instant u = umax , the package leaves the surface of the ramp so
that NB = 0. Therefore, there are three unknowns, v, at , and u.
Kinematics. The third equation for the solution is obtained by noting
that the magnitude of tangential acceleration at may be related to the
speed of the package v and the angle u. Since at ds = v dv and
ds = r du = 0.5 du, Fig. 13–15a, we have
v dv
at =
(3)
0.5 du
To solve, substitute Eq. 3 into Eq. 2 and separate the variables. This gives
v dv = 4.905 sin u du
Integrate both sides, realizing that when u = 0°, v0 = 1 m>s.
v
L1
v dv = 4.905
u
L0°
sin u du
u
v2 v
` = -4.905 cos u ` ; v2 = 9.8111 - cos u2 + 1
2 1
0°
Substituting into Eq. 1 with NB = 0 and solving for cos umax yields
2
19.62 cos umax =
[9.8111 - cos umax2 + 1]
0.5
43.24
cos umax =
58.86
umax = 42.7°
Ans.
NOTE: The speed of the package is increasing because its tangential
acceleration is increasing with u, Eq. 2.
Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River,
New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.
141
13.6 EQUATIONS OF MOTION: CYLINDRICAL COORDINATES
EXAMPLE 13.10
The 2-lb block in Fig. 13–19a moves on a smooth horizontal track,
such that its path is specified in polar coordinates by the parametric
equations r = 110t22 ft and u = 10.5t2 rad, where t is in seconds.
Determine the magnitude of the tangential force F causing the
motion at the instant t = 1 s.
SOLUTION
F
Free-Body Diagram. As shown on the block’s free-body diagram,
Fig. 13–19b,the normal force of the track on the block,N,and the tangential
force F are located at an angle c from the r and u axes. This angle can be
obtained from Eq. 13–10. To do so, we must first express the path as
r = f1u2 by eliminating the parameter t between r and u. This yields
r = 40u2. Also, when t = 1 s, u = 0.511 s2 = 0.5 rad. Thus,
r
40u2
=
= 0.25
`
dr>du
4012u2 u = 0.5 rad
c = 14.04°
r
u
(a)
tan c =
Because c is a positive quantity, it is measured counterclockwise from
the r axis to the tangent (the same direction as u) as shown in
Fig. 13–19b. There are presently four unknowns: F, N, ar and au .
Equations of Motion.
2
+ T ©Fr = mar ;
F cos 14.04° - N sin 14.04° =
a
(1)
32.2 r
2
c + ©Fu = mau ;
F sin 14.04° + N cos 14.04° =
a
(2)
32.2 u
Kinematics. Since the motion is specified, the coordinates and the
required time derivatives can be calculated and evaluated at t = 1 s.
r = 10t2 `
#
r = 20t `
t=1 s
t=1 s
= 10 ft
u = 0.5t `
t=1 s
u
14.04'
N
Tangent
au
F
r
c = 14.04'
ar
u
(b)
Fig. 13–19
= 0.5 rad
#
= 20 ft>s u = 0.5 rad>s
$
$
r = 20 ft>s2
u = 0
#
$
ar = r - ru2 = 20 - 1010.522 = 17.5 ft>s2
$
##
au = ru + 2ru = 10102 + 2120210.52 = 20 ft>s2
Substituting into Eqs. 1 and 2 and solving, we get
F = 1.36 lb
Ans.
N = 0.942 lb
NOTE: The tangential axis is in the direction of F, and the normal axis
is in the direction of N.
Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River,
New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.
142
C H A P T E R 13
KINETICS
OF A
PA R T I C L E : F O R C E
AND
A C C E L E R AT I O N
EXAMPLE 13.11
The smooth 2-kg cylinder C in Fig. 13–20a has a peg P through its
center which passes through the slot# in arm OA. If the arm rotates in
the vertical plane at a constant rate u = 0.5 rad>s, determine the force
that the arm exerts on the peg at the instant u = 60°.
u
0.4 m
O
·
u $ 0.5 rad/s
r
C
SOLUTION
Why is it a good idea to use polar coordinates to solve this problem?
Free-Body Diagram. The free-body diagram for the cylinder is
shown in Fig. 13–20b. The force on the peg, FP , acts perpendicular to
the slot in the arm. As usual, a r and a u are assumed to act in the
directions of positive r and u, respectively. Identify the four unknowns.
Equations of Motion. Using the data in Fig. 13–20b, we have
+R©Fr = mar ;
+b©Fu = mau ;
P
(1)
(2)
Kinematics. From Fig. 13–20a, r can be related to u by the equation
A
r =
(a)
FP
u
u
NC
r
(b)
Fig. 13–20
0.4
= 0.4 csc u
sin u
Since d1csc u2 = -1csc u cot u2 du and d1cot u2 = -1csc2 u2 du, then
r and the necessary time derivatives become
#
u = 0.5
r = 0.4 csc u
$
#
#
u = 0
r = -0.41csc u cot u2u
19.62 N
ar
19.62 sin u - NC sin u = 2ar
19.62 cos u + FP - NC cos u = 2au
= -0.2 csc u cot u
#
#
$
r = -0.21-csc u cot u21u2 cot u - 0.2 csc u1-csc2 u2u
au
= 0.1 csc u1cot2 u + csc2 u2
u
Evaluating these formulas at u = 60°, we get
#
u = 0.5
r = 0.462
$
#
u = 0
r = -0.133
$
r = 0.192
#
$
ar = r - ru2 = 0.192 - 0.46210.522 = 0.0770
$
##
au = ru + 2ru = 0 + 21-0.133210.52 = -0.133
Substituting these results into Eqs. 1 and 2 with u = 60° and
solving yields
NC = 19.4 N
FP = -0.356 N
Ans.
The negative sign indicates that FP acts opposite to the direction
shown in Fig. 13–20b.
Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River,
New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.
143
13.6 EQUATIONS OF MOTION: CYLINDRICAL COORDINATES
EXAMPLE 13.12
A can C, having a mass of 0.5 kg, moves along a grooved horizontal slot
shown in Fig. 13–21a.The slot is in the form of a spiral, which is defined by
the equation r = 10.1u2 m, where u is in radians. If the arm OA is
#
rotating at a constant rate u = 4 rad>s in the horizontal plane, determine
the force it exerts on the can at the instant u = p rad. Neglect friction
and the size of the can.
SOLUTION
Free-Body Diagram. The driving force FC acts perpendicular to the
arm OA, whereas the normal force of the wall of the slot on the can, NC ,
acts perpendicular to the tangent to the curve at u = p rad, Fig. 13–21b.
As usual, ar and a u are assumed to act in the positive directions of r and
u, respectively. Since the path is specified, the angle c which the
extended radial line r makes with the tangent, Fig. 13–21c, can be
determined from Eq. 13–10. We have r = 0.1u, so that dr>du = 0.1,
and therefore
r
0.1u
tan c =
=
= u
dr>du
0.1
u
r $ 0.1 u
r
A
O
·
u $ 4 rad/s
C
(a)
FC
ar
r
f
NC
f
au
Tangent
-1
When u = p, c = tan p = 72.3°, so that f = 90° - c = 17.7°, as
shown in Fig. 13–21c. Identify the four unknowns in Fig. 13–21b.
Equations of Motion.
Fig. 13–21b, we have
+ ©F = ma ;
;
r
r
+ T ©Fu = mau ;
u
Using f = 17.7° and the data shown in
NC cos 17.7° = 0.5ar
FC - NC sin 17.7° = 0.5au
Kinematics. The time derivatives of r and u are
#
u = 4 rad>s
r = 0.1u
$
#
#
u = 0
r = 0.1u = 0.1142 = 0.4 m>s
$
$
r = 0.1u = 0
At the instant u = p rad,
#
$
ar = r - ru2 = 0 - 0.11p21422 = -5.03 m>s2
$
##
au = ru + 2ru = 0 + 210.42142 = 3.20 m>s2
(b)
(1)
(2)
r $ 0.1 u
u$p
r
c
f
Tangent
u
(c)
Fig. 13–21
Substituting these results into Eqs. 1 and 2 and solving yields
NC = -2.64 N
FC = 0.800 N
Ans.
What does the negative sign for NC indicate?
Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River,
New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the
publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.
© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
For the exclusive use of adopters of the Hibbeler series of books.
Problem 13-12
The particle of weight W is subjected to the action of its weight and forces F1 = (ai+bj+ct k),
F2 = (dt2i+et j+fk) and F 3 = hti. Determine the distance the ball is from the origin a time t
after being released from rest.
Given:
lb
a := 2lb
e := −4
b := 6lb
f := −1lb
lb
c := −2
d := 1
h := −2
s
lb
s
lb
s
t := 2s
2
s
W := 6lb
ft
g := 32.2
2
s
Solution:
x - direction
W
2
a + d⋅ t + h⋅ t =
vx =
⎛
W ⎝
g
⋅ ⎜ a⋅ t +
g
⋅ ax
ax =
h 2 d 3⎞
⋅t + ⋅t
2
3 ⎠
sx :=
g
W
(
⋅ a + h⋅ t + d⋅ t
)
2
⎛ a 2 h 3 d ⋅ t4⎞
⋅⎜ ⋅t + ⋅t +
6
12 ⎠
W ⎝2
g
sx = 14.31 ft
y - direction
b + e⋅ t =
vy =
W
g
⋅ ay
ay =
⎛
W ⎝
g
e 2⎞
⋅t
2 ⎠
⋅ ⎜ b⋅ t +
sy :=
g
W
⋅ ( b + e⋅ t)
⎛ b 2 e 3⎞
⋅⎜ ⋅t + ⋅t
6 ⎠
W ⎝2
g
sy = 35.78 ft
z - direction
c⋅ t + f − W =
vz =
W
g
⋅ az
⎛
W ⎝
g
⋅ ⎜ f⋅ t − W⋅ t +
Total distance
s :=
az =
c 2⎞
⋅t
2 ⎠
2
sz :=
2
2
sx + sy + sz
g
W
⋅ ( f − W + c⋅ t)
⎛ f 2 W ⋅ t2 + c ⋅ t3⎞
⋅⎜ ⋅t −
2
6 ⎠
W ⎝2
g
s = 97.39 ft
sz = −89.44 ft
c03.qxd
6/15/06
12:32 PM
Page 143
Article 3/5
143
Curvilinear Motion
Sample Problem 3/9
t
S
Compute the magnitude v of the velocity required for the spacecraft S to
maintain a circular orbit of altitude 200 mi above the surface of the earth.
h
Solution.
mme
F = G ——–––
(R + h)2
n
The only external force acting on the spacecraft is the force of gravi-
S
tational attraction to the earth (i.e., its weight), as shown in the free-body dia-
R
gram. Summing forces in the normal direction yields
[ΣFn man]
G
mme
(R h)
2
m
v2 ,
(R h)
v
(R h) R (R h)
Gme
g
where the substitution gR2 Gme has been made. Substitution of numbers gives
v (3959)(5280)
25,326 ft/sec
(3959 32.234
200)(5280)
Ans.
Helpful Hint
Note that, for observations made within an inertial frame of reference, there is no such quantity as “centrifugal force” acting in the minus n-direction. Note also that neither the spacecraft nor its occupants are “weightless,” because the weight
in each case is given by Newton’s law of gravitation. For this altitude, the weights are only about 10 percent less than the
earth-surface values. Finally, the term “zero-g” is also misleading. It is only when we make our observations with respect
to a coordinate system which has an acceleration equal to the gravitational acceleration (such as in an orbiting spacecraft)
that we appear to be in a “zero-g” environment. The quantity which does go to zero aboard orbiting spacecraft is the familiar normal force associated with, for example, an object in contact with a horizontal surface within the spacecraft.
Sample Problem 3/10
O
·
θ =ω
Tube A rotates about the vertical O-axis with a constant angular rate ˙ and contains a small cylindrical plug B of mass m whose radial position is controlled by the cord which passes freely through the tube and shaft and is wound
around the drum of radius b. Determine the tension T in the cord and the horizontal component F of force exerted by the tube on the plug if the constant angular rate of rotation of the drum is 0 first in the direction for case (a) and
second in the direction for case (b). Neglect friction.
Solution.
With r a variable, we use the polar-coordinate form of the equations
of motion, Eqs. 3/8. The free-body diagram of B is shown in the horizontal plane
and discloses only T and F. The equations of motion are
[ΣFr mar]
[ΣF ma]
T m(r
¨ r ˙2)
r
B
ω0
case (a)
·
θ =ω
O
F 2mb0
+θ
B
+r
r
Fθ
Case (a). With ˙r b0, ¨r 0, and ¨ 0, the forces become
T mr2
r
ω0
case (b)
b
T
F m(r ¨ 2r
˙ ˙)
A
Ans.
Helpful Hint
The minus sign shows that F is in
Case (b). With ˙r b0, ¨r 0, and ¨ 0, the forces become
T mr2
F 2mb0
Ans.
the direction opposite to that shown
on the free-body diagram.