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PHYS-207 Honor’s Section LL, HW8 Solutions
11.7. (a) We find its angular speed as it leaves the roof using conservation of energy. Its initial kinetic
energy is Ki = 0 and its initial potential energy is Ui = Mgh where h  6.0sin30  3.0 m (we are using
the edge of the roof as our reference level for computing U). Its final kinetic energy is (Eq. 11-5)
K f  21 Mv 2  21 I 2 .
Here we use v to denote the speed of its center of mass and  is its angular speed — at the moment it
leaves the roof. Since (up to that moment) the ball rolls without sliding we can set v = R  = v where R =
0.10 m. Using I 
1
2
MR 2 (Table 10-2(c)), conservation of energy leads to
1
1
1
1
3
Mgh  Mv 2  I  2  MR 2 2  MR 2 2  MR 2 2 .
2
2
2
4
4
The mass M cancels from the equation, and we obtain

hb g
c
1 4
1
4
gh 
9.8 m s2 3.0 m  63 rad s .
R 3
010
. m 3
(b) Now this becomes a projectile motion of the type examined in Chapter 4. We put the origin at the
position of the center of mass when the ball leaves the track (the “initial” position for this part of the
problem) and take +x leftward and +y downward. The result of part (a) implies v0 = R  = 6.3 m/s, and
we see from the figure that (with these positive direction choices) its components are
v0 x  v0 cos30  5.4 m s
v0 y  v0 sin 30  3.1 m s.
The projectile motion equations become
x  v0 x t and y  v0 y t 
1 2
gt .
2
We first find the time when y = H = 5.0 m from the second equation:
t
v0 y  v02y  2 gH
g
 0.74s.
b
gb g
Then we substitute this into the x equation and obtain x  5.4 m s 0.74 s  4.0 m.
 
r  F is equal to

yFz  zFy i  zFx  xFz j  xFy  yFx k.

 then (using Eq. 3-30) we find
11.25. If we write r  x i  yj  zk,
d
i b
gd
i
(a) Plugging in, we find
(b) We use Eq. 3-27,
ˆ
   3.0m  6.0N    4.0m  8.0N   kˆ  (50 N  m) k.


 
| r  F |  rF sin  , where  is the angle between r and F . Now
r  x 2  y 2  5.0 m and F  Fx2  Fy2  10 N. Thus,
b gb g
rF  5.0 m 10 N  50 N  m,
the same as the magnitude of the vector product calculated in part (a). This implies sin  = 1 and = 90°.
11.39. (a) Since  = dL/dt, the average torque acting during any interval  t is given by
d
i
 avg  L f  Li t ,
where Li is the initial angular momentum and Lf is the final angular momentum.
 avg 
0.800 kg  m2 s  3.00 kg  m2 s
 1.47 N  m ,
1.50s
or | avg |  1.47 N  m . In this case the negative sign indicates that the direction of the torque is opposite
the direction of the initial angular momentum, implicitly taken to be positive.
(b) The angle turned is
  0t   t 2 / 2.
If the angular acceleration  is uniform, then so is the torque
and  = /I. Furthermore, 0 = Li/I, and we obtain
2
Li t   t 2 / 2  3.00 kg  m s  1.50s    1.467 N  m 1.50s  / 2


I
0.140 kg  m 2
2
 20.4 rad.
(c) The work done on the wheel is
W     1.47 N  m 20.4rad   29.9 J
where more precise values are used in the calculation than what is shown here. An equally good method

for finding W is Eq. 10-52, which, if desired, can be rewritten as W  L f  Li
2
2

2I .
(d) The average power is the work done by the flywheel (the negative of the work done on the flywheel)
divided by the time interval:
Pavg  
W
29.8 J

 19.9 W .
t
1.50s
11.58. The initial rotational inertia of the system is Ii = Idisk + Istudent, where Idisk = 300 kg  m2 (which,
incidentally, does agree with Table 10-2(c)) and Istudent = mR2 where m  60 kg and R = 2.0 m.
The rotational inertia when the student reaches r = 0.5 m is If = Idisk + mr2. Angular momentum
conservation leads to
I disk  mR 2
I i i  I f  f   f   i
I disk  mr 2
which yields, for i = 1.5 rad/s, a final angular velocity of f = 2.6 rad/s.
12.5. The object exerts a downward force of magnitude F = 3160 N at the midpoint of the rope, causing a
“kink” similar to that shown for problem 10 (see the figure that accompanies that problem in the text). By

analyzing the forces at the “kink” where F is exerted, we find (since the acceleration is zero) 2T sin = F,
where  is the angle (taken positive) between each segment of the string and its “relaxed” position (when
the two segments are collinear). In this problem, we have
 0.35m 
  11.5.
 1.72 m 
  tan 1 
Therefore, T = F/(2sin ) = 7.92 × 103 N.
12.24. As shown in the free-body diagram, the forces on the climber consist of T from the rope, normal
force FN on her feet, upward static frictional force
f s , and downward gravitational force mg .
Since the climber is in static equilibrium, the net force acting on her is zero. Applying Newton’s second
law to the vertical and horizontal directions, we have
0   Fnet, x  FN  T sin 
0   Fnet, y  T cos   f s  mg .
In addition, the net torque about O (contact point between her feet and the wall) must also vanish:
0   net  mgL sin   TL sin(180     )
O
From the torque equation, we obtain
T  mg sin  / sin(180     ).
Substituting the expression into the force equations, and noting that f s   s FN , we find the coefficient of
static friction to be
s 

With
f s mg  T cos  mg  mg sin  cos  / sin(180     )


FN
T sin 
mg sin  sin  / sin(180     )
1  sin  cos  / sin(180     )
.
sin  sin  / sin(180     )
  40 and   30 , the result is
1  sin  cos  / sin(180     ) 1  sin 40 cos30 / sin(180  40  30)

sin  sin  / sin(180     )
sin 40 sin 30 / sin(180  40  30)
 1.19.
s 
12. 35. We examine the box when it is about to tip. Since it will rotate about the lower right edge, that is
where the normal force of the floor is exerted. This force is labeled FN on the diagram that follows. The
force of friction is denoted by f, the applied force by F, and the force of gravity by W. Note that the force
of gravity is applied at the center of the box. When the minimum force is applied the box does not
accelerate, so the sum of the horizontal force components vanishes: F – f = 0, the sum of the vertical force
components vanishes: FN  W  0 , and the sum of the torques vanishes:
FL – WL/2 = 0.
Here L is the length of a side of the box and the origin was chosen to be at the lower right edge.
(a) From the torque equation, we find
F
W 890 N

 445 N.
2
2
(b) The coefficient of static friction must be large enough that the box does not slip. The box is on the
verge of slipping if s = f/FN. According to the equations of equilibrium
FN = W = 890 N, f = F = 445 N,
s 
f
445 N

 0.50.
FN 890 N
(c) The box can be rolled with a smaller applied force if the force points upward as well as to the right. Let
 be the angle the force makes with the horizontal. The torque equation then becomes
FL cos  + FL sin  – WL/2 = 0,
F
W
.
2(cos   sin  )
We want cos + sin to have the largest possible value. This occurs if  = 45º, a result we can prove by
setting the derivative of cos + sin equal to zero and solving for . The minimum force needed is
F
W
890 N

 315 N.
2(cos 45  sin 45) 2(cos 45  sin 45)
Note: The applied force as a function of  is plotted below. From the figure, we readily see that
corresponds to a maximum and   45  to a minimum.
  0
12.60. (a) Equation 12-8 leads to T1 sin40º + T2 sin = mg . Also, Eq. 12-7 leads to
T1 cos40º  T2 cos = 0.
Combining these gives the expression
T2 
mg
.
cos  tan 40  sin 
To minimize this, we can plot it or set its derivative equal to zero. In either case, we find that it is at its
minimum at  = 50.
(b) At  = 50, we find T2 = 0.77mg.