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Circular Motion and Other Applications
of Newton’s Laws
CHAPTER OUTLINE 6.1 Newton’s Second Law Applied to Uniform Circular Motion 6.2 Nonuniform Circular Motion 6.3 Motion in Accelerated Frames 6.4 Motion in the Presence of Resistive Forces
ANSWERS TO QUESTIONS
*Q6.1 (i) nonzero. Its direction of motion is changing. (ii) zero. Its speed is not changing. (iii) zero: when v = 0, v^2 r = 0. (iv) nonzero: its velocity is changing from, say 0.1 ms north to 0.1 ms south.
Q6.2 (a) The object will move in a circle at a constant speed. (b) The object will move in a straight line at a changing speed.
Q6.3 The speed changes. The tangential force component causes tangential acceleration.
*Q6.4 (a) A > C = D > B = E. At constant speed, centripetal acceleration is largest when radius is smallest. A straight path has infinite radius of curvature. (b) Velocity is north at A, west at B, and south at C. (c) Acceleration is west at A, nonexistent at B, and east at C, to be radially inward.
*Q6.5 (a) yes, point C. Total acceleration here is centripetal acceleration, straight up. (b) yes, point A. Total acceleration here is tangential acceleration, to the right and downward perpendicular to the cord. (c) No. (d) yes, point B. Total acceleration here is to the right and upward.
Q6.6 I would not accept that statement for two reasons. First, to be “beyond the pull of gravity,” one would have to be infinitely far away from all other matter. Second, astronauts in orbit are moving in a circular path. It is the gravitational pull of Earth on the astronauts that keeps them in orbit. In the space shuttle, just above the atmosphere, gravity is only slightly weaker than at the Earth’s surface. Gravity does its job most clearly on an orbiting spacecraft, because the craft feels no other forces and is in free fall.
Q6.7 This is the same principle as the centrifuge. All the material inside the cylinder tends to move along a straight-line path, but the walls of the cylinder exert an inward force to keep everything moving around in a circular path.
Q6.8 The water has inertia. The water tends to move along a straight line, but the bucket pulls it in and around in a circle.
Q6.9 Blood pressure cannot supply the force necessary both to balance the gravitational force and to provide the centripetal acceleration, to keep blood flowing up to the pilot’s brain.
*Q6.10 (a) The keys shift backward relative to the student’s hand. The cord then pulls the keys upward and forward, to make them gain speed horizontally forward along with the airplane. (b) The angle stays constant while the plane has constant acceleration. This experiment is described in the book Science from Your Airplane Window by Elizabeth Wood.
127
128 Chapter 6
Q6.11 The person in the elevator is in an accelerating reference frame. The apparent acceleration due to gravity, “ g ,” is changed inside the elevator. “ g ” = g ± a
Q6.12 From the proportionality of the drag force to the speed squared and from Newton’s second law, we derive the equation that describes the motion of the skydiver:
m
d dt mg y D^ A y
v = − v ρ 2
2
where D is the coefficient of drag of the parachutist, and A is the projected area of the parachu- tist’s body. At terminal speed,
a
d y dt =^ vy^ = 0 and v T
mg D A
1 2 ρ When the parachute opens, the coefficient of drag D and the effective area A both increase, thus reducing the speed of the skydiver.
Modern parachutes also add a third term, lift, to change the equation to
m
d dt mg y D^ A^ L^ A y x
v = − v − v ρ ρ 2 2
2 2
where vy is the vertical velocity, and vx is the horizontal velocity. The effect of lift is clearly seen in the “paraplane,” an ultralight airplane made from a fan, a chair, and a parachute.
Q6.13 (a) Static friction exerted by the roadway where it meets the rubber tires accelerates the car forward and then maintains its speed by counterbalancing resistance forces. (b) The air around the propeller pushes forward on its blades. Evidence is that the propeller blade pushes the air toward the back of the plane. (c) The water pushes the blade of the oar toward the bow. Evi- dence is that the blade of the oar pushes the water toward the stern.
Q6.14 The larger drop has higher terminal speed. In the case of spheres, the text demonstrates that terminal speed is proportional to the square root of radius. When moving with terminal speed, an object is in equilibrium and has zero acceleration.
*Q6.15 (a) Speed increases, before she reaches terminal speed. (b) The magnitude of acceleration decreases, as the air resistance force increases to counterbalance more and more of the gravita- tional force.
Q6.16 The thesis is false. The moment of decay of a radioactive atomic nucleus (for example) cannot be predicted. Quantum mechanics implies that the future is indeterminate. On the other hand, our sense of free will, of being able to make choices for ourselves that can appear to be random, may be an illusion. It may have nothing to do with the subatomic randomness described by quantum mechanics.
130 Chapter 6
FIG. P6.
P6.5 (a) static friction
(b) (^) ma ˆ i^ = f ˆ i^ + n ˆ j^ + mg (− ˆ j )
∑^ Fy^ =^0 =^ n^ − mg thus n = mg and F m r ∑ r =^^ v =^ f^ =^ n^ = mg
2 μ μ.
Then μ = = ( (^) ) ( )( )
v^2 50 0^2 30 0 980
rg
cm s cm cm s 2
P6.6 Neglecting relativistic effects. F ma m c r = = v
2
F = (^) ( × × ) ( × )
2 1 661 10 − 2 998^10
27 7 2 .
kg
m s m))^
= 6 22. × 10 − 12 N
P6.7 Standing on the inner surface of the rim, and moving with it, each person will feel a normal force exerted by the rim. This inward force causes the 3 00. m s 2 centripetal acceleration:
a (^) c = v^2 / r so^ v^ =^ a rc =^ (^ 3 00.^ m s^2 )(^ 60 0.^ m) =^ 13 4. m s
The period of rotation comes from v = 2 π r T
: T = r = (^
) 2 2 60 0 = 13 4
π π 28 1 v
m. m s
s
so the frequency of rotation is f T
s s
s 1 min
rev miin.
P6.8 T cos 5 00. ° = mg = (80 0. kg)( 9 80. m s (^2) )
(a) T = 787 N:
T = ( 68 6. N ) i ˆ^ + ( 784 N) j ˆ
(b) T sin 5 00°. = mac : ac = 0 857. m s 2 toward the center of the circle.
The length of the wire is unnecessary information. We could, on the other hand, use it to find the radius of the circle, the speed of the bob, and the period of the motion.
P6.9 n = mg since a (^) y = 0
The force causing the centripetal acceleration is the frictional force f.
From Newton’s second law f ma m c r = = v
2 .
But the friction condition is f ≤μ sn
i.e., m r smg
v^2 ≤μ
v ≤ μ s rg = 0 600 35 0. (. m)( 9 80. m s (^2) ) v ≤ 14 3. m s
FIG. P6.
Circular Motion and Other Applications of Newton’s Laws 131
P6.10 (a) (^) v = 235 m= 6 53 36.0 s
. m s
The radius is given by
2 π r = 235 m so r = 150 m
(b) a r r
⎝⎜^
( )
v^2
6 53^2
toward center
m s m
at 35.0 north of west
m s 2
= ( 0 285. )( cos 35 0. °(−− ) + )
= − +
ˆ (^) sin. ˆ
. ˆ^. ˆ
i j
i
m s 2 m s^2 jj
(c) ^
a
v v
j i
a g
f i v (^) t
( 6 53^ −6 53 ) 36 0
. ˆ^. ˆ
m s m s ss = − 0 181. m s 2 ˆ i^ +0 181. m s^2 ˆ j
P6.11 Fg = mg = ( 4 kg )( 9 8. m s (^2) ) =39 2. N
sin. .
θ θ
m 2 m ° r = ( 2 m ) cos 48 6. °=1 32. m
F ma m r
T T
x x
a b
∑ =^ =
v^2
cos 48 6. ° cos 48 6. ° 4 kg^6 m ss m N N
( )
2 1 32 109 48 6
cos. T (^) a Tb °
F ma T T
T T
y y a b a b
∑ =
−
sin 48 6. ° sin 48 6. ° 39 2. N 0
== 39 2 = 48 6
sin.
N . N
(a) To solve simultaneously, we add the equations in Ta and Tb : T (^) a + T (^) b + T (^) a − Tb = 165 N +52 3. N
Ta = =
N
N
(b) T (^) b = 165 N − Ta = 165 N − 108 N = 56 2. N
FIG. P6.
39.2 N
Ta
Tb forces
motion
v a (^) c
Circular Motion and Other Applications of Newton’s Laws 133
continued on next page
FIG. P6.
P6.15 Let the tension at the lowest point be T.
F ma T mg ma m r
T m g r
T
∑ =^ −^ =^ c =
= +
⎝⎜^
v
v
2
2
85 0 kg)) +
⎡ ( ) ⎣
2 .
m s. m s m
(^2) kN 0 000 N
He doesn’t make it across the river because the vine breaks.
*P6.16 ( a) Consider radial forces on the object, taking inward as positive.
Σ F ma T r
r = r − = =
( .0 5 kg)(. 9 8 m s 2 ) cos 20 ° mv^2 0 5. kkg m N N N
m s^2 T = + = (b) We already found the radial component of acceleration, ( 8 m s) 2 2 m =32 0. m s^2 inward. Consider tangential forces on the object. Σ F ma a a
t t t t
( .0 5 kg)(. 9 8 m s 2 ) sin 20 0 5. kg 3 3 35. m s 2 downward tangent to the circle
(c) a = [ 32 2 +3 35. 2 ]1 2^ m s^2 inward and below the cord at angle tan −^1 ( 3 35 32. ) = 32 2. m s 2 inward and below the cord at 5.98°
(d) No change. If the object is swinging down it is gaining speed. If the object is swinging up it is losing speed but its acceleration is the same size and its direction can be described in the same terms.
P6.17 F^
m r ∑ y =^ =^ mg^ + n
v^2
But n = 0 at this minimum speed condition, so m r mg gr v v
2 = ⇒ = = (^) ( 9 80. m s (^2) )(1 00. m ) = 3 13. m s
P6.18 (a) a^ c =^ r
v^2 r ac
( ) ( )
v^2 13 0^2 2 9 80
m s m s 2 m (b) Let n be the force exerted by the rail. Newton’s second law gives
Mg n
M
r
v^2
n M r = − g M g g Mg
= (^) ( − ) = v^2 2 , downward
FIG. P6.
FIG. P6.
134 Chapter 6
(c) a c (^) r = v
2 ac = (^ )^ =
.^2
m s m
m s 2
If the force exerted by the rail is n 1
then n^ Mg^
M
r 1 Mac
2
n 1 (^) = M ( a (^) c − g ) which is < 0 since ac = 8 45. m s 2
Thus, the normal force would have to point away from the center of the curve. Unless they have belts, the riders will fall from the cars.
To be safe we must require n 1 to be positive. Then a (^) c > g. We need v^2 r > g or v > rg = ( 20 0. m )( 9 80. m s (^2) ), v > 14 0. m s
Section 6.3 Motion in Accelerated Frames
P6.19 (a) (^) ∑ Fx = Ma ,
a
T
M
N
5.00 kg m s 2 to the
right.
(b) If v = const, a = 0 , so T = 0 (This is also an equilibrium situation.)
(c) Someone in the car (noninertial observer) claims that the forces on the mass along x are T and a fictitious force (– Ma ). Someone at rest outside the car (inertial observer) claims that T is the only force on M in the x -direction.
P6.20 The water moves at speed
v = = ( ) =
π r π T
m s m s.
The top layer of water feels a downward force of gravity mg and an outward fictitious force in the turntable frame of reference, m r
m m v^2 0 104^22 0 12
( ) = × −
m s m m s 2
It behaves as if it were stationary in a gravity field pointing downward and outward at
tan
m s m s
2 2 ° Its surface slopes upward toward the outside, making this angle with the horizontal.
FIG. P6.
5.00 kg
136 Chapter 6
P6.24 In an inertial reference frame, the girl is accelerating horizontally inward at
v^2 5 70^2 2 40
r
( )
m s m m s 2
In her own non-inertial frame, her head feels a horizontally outward fictitious force equal to its mass times this acceleration. Together this force and the weight of her head add to have a magni- tude equal to the mass of her head times an acceleration of
g r
2 2
2
⎠⎟^
= ( ) + ( ) = v
.. m s 2. m s^2
This is larger than g by a factor of
Thus, the force required to lift her head is larger by this factor, or the required force is
F = 1 71 55 0. (. N) = 93 8. N.
P6.25 a
R
r T = ⎛ e ⎝⎜^
⎠⎟^
2 2
π cos. °. m s 2
We take the y axis along the local vertical. a a a
y r^ y x
net 2
net
m s m
( (^) ) = − ( ) = ( ) =
. ss 2
θ = arctan^ a =. a
x y
Section 6.4 Motion in the Presence of Resistive Forces
P6.26 m = 80 0. kg, vT = 50 0. m s, mg D A (^) T D A mg T
ρ v ρ v
2 2 2 2 0 314.^ kg m (a) At v = 30 0. m s
a g
D A
m
( )( )
ρ v^2 2 9 80
. m ss downward^2
(b) At v = 50 0. m s, terminal velocity has been reached. F mg R R mg
∑ (^) y =^ =^ − ⇒ = = ( )( ) =
80 0. kg 9 80. m s 2 784 N diirected up (c) At v = 30 0. m s D ρ A v^22 2 = ( 0 314. )(30 0. ) = 283 N upward
FIG. P6.
N
ar
g 0 a net
35.0° Equator
θ
35.0° (exaggerated size)
Circular Motion and Other Applications of Newton’s Laws 137
P6.27 (a) a = g − bv
When v = vT , a = 0 and g = b vT b g T
v The Styrofoam falls 1.50 m at constant speed vT in 5.00 s.
Thus,
vT y t
= = 1 50.^ m=0 300. 5.00 s
m s
Then
b = 9 80 = − 0 300
m s. m s
s
2
(b) At t = 0 , v = 0 and a = g = 9 80. m s 2 down
(c) When v = 0 150. m s, a = g − b v = 9 80. m s 2 − ( 32 7. s− (^1) )( 0 150. m s ) = 4 90. mss 2 down
P6.28 (a) ρ = m V
, A = 0 020 1. m 2 , R =^1 AD (^) T = mg 2
ρ^2 air v
m = V = ⎡ ( ) ⎣⎢^
ρbead 0 830 g cm^3 4 π cm = 3
. 8 00. 3 1 78. kg
Assuming a drag coefficient of D = 0 500. for this spherical object, and taking the density of air at 20°C from the endpapers, we have
vT = ( )( ) ( )
kg m s kg m
2 (^3) ( 0 0 1 m (^2) ) =^ 53 8. m s
(b) v^2 f^ = vi^2^ + 2 gh = 0 + 2 gh : h g
= f = (^ ) ( )
v^2 2
m s m s 2 m
P6.29 Since the upward velocity is constant, the resultant force on the ball is zero. Thus, the upward applied force equals the sum of the gravitational and drag forces (both downward): F = mg + bv.
The mass of the copper ball is
m = 4 r = ⎛⎝ ⎞⎠ ( × ) ( × − 3
πρ (^33 ) π. kg m 3. m)) = 3 0 299. kg
The applied force is then F = mg + b v = ( 0 299. )( 9 80. ) + ( 0 950. ) (^) ( 9 00. × 10 − (^2) ) = 3 .001 N
P6.30 The resistive force is
R = D A = ( )( )( )
ρ v^2 0 250. 1 20. kg m 3 2 20. m^2 27 8. m s N N 1200 kg m s
( )
= − = − = −
2
255 255 0 212
R
a
R
m
Circular Motion and Other Applications of Newton’s Laws 139
(c) dx dt
mg b
bt m
⎣⎢^
1 exp ;^ dx mg b
bt m dt x
x t
0
0 ∫ =^ ∫
⎠ −^
⎣⎢^
exp
x x mgt b
m g b
bt m
mgt b
t m g − = +
⎝⎜^
⎠⎟^
0 ⎠ =^ +
2 2 0
2 exp bb
bt (^2) m
⎝⎜^
⎠⎟^
⎣⎢^
exp
At t = 5 54. s,
x = ( ) +
( ) 9 00
kg 9.80 m s s 1.13 kg s
2 kg^22 2
exp.
m s kg s
( 2 ) ( )
⎟ [^ (−^ ) − ]]
x = 434 m + 626 m (− 0 500. ) = 121 m
P6.34 (^) ∑ F = ma
− =
− =
∫ ∫
−
km m d dt kdt d
k dt d
k t
t
v v
v v
v v v
v
2
2
0
2 0
0 )) = −
v − v v
v v
v v
v v
v
1 v 0
0
0 0 0
0 kt kt
vv 0 kt
P6.35 (a) From Problem 34,
v v v
v v
v
∫ ∫ =
dx dt kt
dx dt kt k
x t kdt
0 0
0
0 0 0
0
= ( + )
− = ( + )
∫ (^) v
v
v
0 0
(^0 0 )
0
kt
x k kt
x k kt
t
x (^) ln t
⎡⎣ln −− ⎤⎦
= ( + )
ln
ln
x (^1 ) k vkt
(b) We have ln 1( + v (^) 0 kt ) = kx
1 + v 0 kt = e kx^ so v v v
v = v v
0
(^00) 1 kt e kx ekx
140 Chapter 6
P6.36 We write − km v^2 = −^1 D Av^2 2
ρ so
k D^ A m
= = (^ )^ ρ ( × − ) 2
...^3
kg m 3 m^2 kg
m
m s
( )
= ×
= = ( )
−
− −^ ×^ −
3
0
5 3 10
v v e kx^. e.
(^33) 18 3 ( m^ )(^ .m)^ = 36 5. m s
P6.37 (a) v ( ) = t vie − ct^ v ( 20 0. s) = 5 00. = v ie − 20 0. c , vi = 10 0. m s
So 5 00. = 10 0. e −20 0. c^ and − 20 0 = ⎛⎝^1 ⎞⎠ 2
. c ln c = − ln( ) = × −^ − .
(^12 2 ) 20 0
3 47 10 s
(b) At t = 40 0. s v = ( 10 0. m s) e − 40 0. c = ( 10 0. m s (^) )( 0 250. ) = 2 50. m s
(c) v = v ie − ct a d dt
= v^ = − c vi e − ct = − cv
P6.38 In R = 1 D A 2
ρ v^2 , we estimate that D = 1 00. , ρ = 1 20. kg m 3 ,
A = ( 0 100. m )( 0 160. m ) = 1 60. × 10 −^2 m 2 and v = 27 0. m s. The resistance force is then
R = 1 ( )( ) ( × − )( 2
1 00. 1 20. kg m 3 1 60. 10 2 m 2 27 0. m s)) 2 =7 00. N or R ~ 101 N
Additional Problems
*P6.39 Let v 0 represent the speed of the object at time 0. We have
d b m
dt b m
t v t v v v
v v
v ∫ 0 ∫ 0 0 0 = − ln = − tt b m
t bt m e
ln ln /
/
v v v v
v v
− = − ( − ) (^) ( ) = −
=
0 0
0
0 ln −− bt m / (^) v = ve − bt m / 0
From its original value, the speed decreases rapidly at first and then more and more slowly, asymptotically approaching zero.
In this model the object keeps losing speed forever. It travels a finite distance in stopping.
The distance it travels is given by dr e dt
r m b e b m dt
r t bt m
t bt m
0 0 0
(^0 )
∫ ∫
∫
−
−
v
v
/
/ ⎠⎠ = −^ = −^ (^ − ) =^ −
m − − − b e m b e m b
bt m e t (^) bt m bt v v v 0 0 0 / / 1 01 // m
t
( )
As goes to infinity, the distance appproaches m b m b v (^0) ( 1 − 0 ) = v 0 /
FIG. P6.
0^ t
0
142 Chapter 6
P6.42 When the cloth is at a lower angle θ, the radial component of (^) ∑ F = ma reads
n mg m r
At θ = 68 0. °, the normal force drops to zero and g r sin 68
2 ° = v .
v = rg sin 68 ° = ( 0 33. m )( 9 8. m s (^2) ) sin 68 °=1 73. m s
The rate of revolution is
angular speed = ( )⎛⎝ ⎞⎠ ( )
⎝⎜^
m s rev π m
π r π
r .. 835 rev s =50 1. rev min
P6.43 (a) v = ( )
⎝⎜^
km h 8 3 h 3 600 s
m 1 km
. 33 m s
∑^ Fy^ =^ may :^ +^ n^ −^ mg^ = −
m r
v^2
n m g r
⎝⎜^
v = − ( ) 2 2 1 800 9 8 8 33 20
kg. m s 2. m s .4 4 1 15 10 4
m N up
=. ×
(b) Take n = 0. Then mg m r
v^2 .
v = gr = (^) ( 9 8. m s (^2) )( 20 4. m ) = 14 1. m s =50 9. km h
P6.44 (a) F ma m ∑ y y R
v^2
mg n m R
v^2 n mg m R
v^2
(b) When n = 0 mg m R
v^2
Then,
v = gR
A more gently curved bump, with larger radius, allows the car to have a higher speed with- out leaving the road. This speed is proportional to the square root of the radius.
P6.45 (a) slope =
N
m s 2 2 kg m
(b) slope = = =
R D A
D A
v
v (^2) v
12 2 2
ρ ρ
FIG. P6.
n
mg
FIG. P6.
R
p mg
p mg cos 68°
mg sin 68°
continued on next page
Circular Motion and Other Applications of Newton’s Laws 143
(c) 1 2
D ρ A = 0 016 2. kg m
D =
( (^) ) ( ) (^ )^
kg m kg m 3 π m
(d) From the table, the eighth point is at force mg = 8 1 64(. × 10 −^3 kg (^) )( 9 8. m s (^2) ) =0 129. N and horizontal coordinate ( 2 80. m s)^2. The vertical coordinate of the line is here ( 0 016 2. kg m )( 2 8. m s )^2 = 0 127. N. The scatter percentage is 0 129.^ N^ 0 127.^ N 1 5. % 0.127 N
(e) The interpretation of the graph can be stated thus: For stacked coffee filters falling at terminal speed, a graph of air resistance force as a function of squared speed demonstrates that the force is proportional to the speed squared within the experimental uncertainty estimated as 2%. This proportionality agrees with that described by the theoretical equation R = 1 D A 2
ρ v^2. The value of the constant slope of the graph implies that the drag coefficient for coffee filters is (^) D = 0 78. ± 2 %.
*P6.46 (a) The forces acting on the ice cube are the Earth’s gravitational force, straight down, and the basin’s normal force, upward and inward at 35° with the vertical. We choose the x and y axes to be horizontal and vertical, so that the acceleration is purely in the x direction. Then ∑ F (^) x = max : n sin 35° = mv^2 R ∑ F (^) y = may : n cos 35° − mg = 0 Dividing eliminates the normal force: n sin 35° n cos 35° = mv^2 Rmg tan 35° = v^2 Rg v = Rg tan 35 0. ° = (^) ( 6 86. m s (^2) ) R
(b) The mass is unnecessary.
(c) The answer to (a) indicates that the speed is proportional to the square root of the radius, so doubling the radius will make the required speed increase by 2 times.
(d) The period of revolution is given by T
R R
Rg
= = = ( ) R
π π v tan.
s m
When the radius doubles, the period increases by 2 times.
(e) On the larger circle the ice cube moves 2 times faster but also takes longer to get around, because the distance it must travel is 2 times larger. Its period is also proportional to the square root of the radius.
P6.47 Take x -axis up the hill F ma T mg ma
a
T
m g F m
x x
y
∑
∑
: sin sin
sin sin
θ φ
θ φ aa T mg
T mg
a g
y :^ cos^ cos cos cos cos sin c
θ φ φ θ φ θ
oos
sin
cos tan sin
θ
φ
φ θ φ
= ( − )
g
a g
Circular Motion and Other Applications of Newton’s Laws 145
*P6.
(a) The only horizontal force on the car is the force of friction, with a maximum value determined by the surface roughness (described by the coefficient of static friction) and the normal force (here equal to the gravitational force on the car).
(b) ∑ Fx = max − f = ma a = − f m = ( v^2 − v 02 )2( x − x 0 )
x − x 0 = ( v^2 − v 02 ) m 2 f = (0 2 − [20 ms] 2 )1200 kg2(−7000 N) = 34 3. m
(c) Now f = mv^2 r r = mv^2 f = 1200 kg [20 ms] 2 7000 N = 68 6. m
A top view shows that you can avoid running into the wall by turning through a quarter-circle, if you start at least this far away from the wall.
(d) Braking is better. You should not turn the wheel. If you used any of the available fric- tion force to change the direction of the car, it would be unavailable to slow the car, and the stopping distance would be longer.
(e) The conclusion is true in general. The radius of the curve you can barely make is twice your minimum stopping distance.
P6.52 v = = (^ ) ( )
π r π 3 77 T
m. s
m s
(a) a r (^) r =^ v =
2 1 58. m s 2
(b) F low (^) = m g ( + a (^) r ) = 455 N
(c) F high (^) = m g ( − ar ) = 328 N
(d) F mid (^) = m g^2 + ar^2 = 397 N upward and at θ = tan −^ = tan−. = .
a 9 15 g
r (^) ° inward.
P6.53 (a) The mass at the end of the chain is in vertical equilibrium. Thus T cos θ = mg.
Horizontally T ma m r r sin θ = = v
2
r r
= ( + ) = ( + ) =
. sin. . sin..
θ m ° m 5 5 17. m
Then ar = v
2 5 17. m
By division tan .
θ =^ a = g g
r v 2 5 17 v v
= = ( )( )( )
. g tan θ.. tan .° m 2 s^2 ..19 m s (b) T cos θ = mg
T = mg = ( )( ) °
cos
θ cos.
kg m s N
2
FIG. P6.
T
R = 4.00 m θ
l = 2.50 m
r
mg
146 Chapter 6
P6.54 (a) The putty, when dislodged, rises and returns to the original level in time t. To fi nd t , we use
v (^) f = vi + at : i.e., − v = + v − gt or t g
2 v where v is the speed of a point on the rim of the wheel.
If R is the radius of the wheel, v = 2 π R t , so t g
R
2 v 2 v
π .
Thus, v^2 =π Rg and v = π Rg.
(b) The putty is dislodged when F , the force holding it to the wheel, is
F m R = = m g v^2 π
*P6.55 (a) n m R
v^2 f − mg = 0
f =μ s n v = 2 π R T
T
R
g = 4 s π 2 μ
(b) T = 2 54. s
#. rev min
rev 2.54 s
s min
rev min
(c) The gravitational and frictional forces remain constant. The normal force increases. The person remains in motion with the wall.
(d) The gravitational force remains constant. The normal and frictional forces decrease. The person slides relative to the wall and downward into the pit.
P6.56 Let the x -axis point eastward, the y -axis upward, and the z -axis point southward.
(a) The range is Z g = vi^ i
(^2) sin 2 θ
The initial speed of the ball is therefore
vi i
gZ = = ( )( ) °
sin
sin.
θ m s
The time the ball is in the air is found from ∆ y = viy t + a ty
(^2) as
0 = (53 0. m s)( sin 48 0. ° ) − ( t (^) 4 90. m s (^2) ) t^2
giving t = 8 04. s.
continued on next page
FIG. P6.
f
m g
nn
