Download Physics I - Rose

Physics I
Homework VII
Chapter 7: 6, 15, 28, 40, 47, 54, 62
CJ
7.6. Solve: Speed, radial velocity, radial acceleration, tangential acceleration, and the
magnitude of the net force are constant. Furthermore, the radial velocity and tangential
acceleration are zero.
7.15. Model: The motion of the moon around the earth will be treated through the particle
model. The circular motion is uniform.
Visualize
:
Solve: The tension in the cable provides the centripetal acceleration. Newton’s second law is
 2
 Fr  T  mr  mr  T
 moon
2



2
 2
1 day
1 hr 
 7.36  1022 kg 3.84  108 m 



27.3
days
24
hrs
3600
s




2
 1020 N
Assess: This is a tremendous tension, but clearly understandable in view of the moon’s large
mass and the large radius of circular motion around the earth.
7.28. Model: We will use the particle model for the test tube which is in nonuniform circular
motion.
Solve: (a) The radial acceleration is
2
rev 1 min 2 rad 

ar  r 2   0.1 m   4000


 1.75  10 4 m/s2
min 60 s
1 rev 

(b) An object falling 1 meter has a speed calculated as follows:


v12  v02  2ay  y1  y0   0 m  2 9.8 m/s2  1.0 m   v1  4.43 m/s
When this object is stopped in 1103 s upon hitting the floor,


v2  v1  ay  t2  t1   0 m s  4.43 m s  ay 1  10 3 s  ay  4.43  10 3 m/s2
This result is one-fourth of the above radial acceleration.
Assess: The radial acceleration of the centrifuge is large, but it is also true that falling objects
are subjected to large accelerations when they are stopped by hard surfaces.
7.40. 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,
mv2
r
 F  n cos 
r
F  n sin  mg  0  n sin  mg
z
Dividing the two force equations gives
tan 
From the geometry of the cone,
tan  r y .
gr
v2
Thus
r gr

 v  gy
y v2
7.47. Model: Model a passenger as a particle rotating in a vertical circle.
Visualize
:
Solve: (a) Netwon’s second law at the top is
F  n
r
T
 w  mar 
mv2
mv2
 nT  mg 
r
r
The speed is
v
2 r 2  8 m 

 11.17 m/s
T
4.5 s
 11.17 m/s 2

 v2

 nT  m   g    55 kg  
 9.8 m/s2 
8m


 r



That is, the ring pushes on the passenger with a force of 319 N at the top of the ride. Newton’s
second law at the bottom:
 v2

mv2
mv2
 nB 
 mg  m   g 
r
r
 r

2
 11.17 m/s

  55 kg  
 9.8 m/s2   1397 N
8m




F  n
r
B
 w  mar 
This is the force with which the ring pushes on the rider when she is at the bottom of the ring.
(b) To just stay on at the top, n
r-equation at the top in part (a). Thus,
mg 
 2
mv2
 mr 2  mr 
r
 Tmax
2

r
8m
 2
 5.68 s
  Tmax  2
g
9.8 m/s2

7.54. Model: Model the steel block as a particle and use the model of kinetic friction.
Visualize
:
Solve: (a) The components of thrust ( F ) along the r-, t-, and z-directions are
Fr  F sin20   3.5 N sin20  1.20 N
Ft  F cos20   3.5 N cos20  3.29 N
Fz
Newton’s second law is
 Fnet r  T  Fr  mr 2
 Fnet t  Ft  fk  mat
 Fnet z  n  mg  0 N
The z-component equation means n
mg. The force of friction is


fk   k n   k mg   0.60  0.5 kg  9.8 m/s2  2.94 N
Substituting into the t-component of Newton’s second law
 3.29 N   2.94 N   0.5 kg at  at  0.70 m/s2
Having found at
follows:
rad as
1 a 
1   t  t12  t1 
2 r 
 at
r
1   0  
(b) Substituting
1
2r1
 18.95 s
at

 t1  6.63 rad/s

into the r-component of Newton’s second law yields:
T1  Fr  mr12  T1  1.20 N    0.50 kg  2.0 m  6.63 rad/s   T1  42.8 N
2
7.62. Model: Use the particle model for a ball in motion in a vertical circle and then as a
projectile.
Visualize
:
Solve: For the circular motion, Newton’s second law along the r-direction is
F  T  w 
r
mvt2
r
Since the string goes slack as the particle makes it over the top, T
w  mg 
mvt2
 vt  gr 
r
 9.8 m/s   0.5 m   2.21 m/s
2
The ball begins projectile motion as the string is released. The time it takes for the ball to hit the
floor can be found as follows:
2
2
y1  y0  v0 y  t1  t0   21 ay  t1  t0   0 m  2.0 m  0 m  21  9.8 m/s2   t1  0 s  t1
The place where the ball hits the ground is
x1  x0  v0 x  t1  t0   0 m   2.21 m/s 0.639 s  0 s  1.41 m
The ball hits the ground 1.41 m to the right of the point beneath the center of the circle.