Download Rotary Homework #1

Rotary Homework #8:
1. A model airplane with mass 0.750 kg is tethered by a wire so that it flies in a circle
30.0 m in radius. The airplane engine provides a net thrust of 0.800 N perpendicular to
the tethering wire. (a) Find the torque the net thrust produces about the center of the
circle. (b) Find the angular acceleration of the airplane when it is in level flight. (c) Find
the linear acceleration of the airplane tangent to its flight path.
2. {optional, not required} A potter’s wheel having a radius of 0.50 m and a moment of
inertia of 12 kg ∙ m2 is rotating freely at 50 rev/min. The potter can stop the wheel in 6.0 s
by pressing a wet rag against the rim and exerting a radially inward force of 70 N. Find
the effective coefficient of kinetic friction between the wheel and the wet rag.
3. A 150-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius
1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the
rope. What constant force must be exerted on the rope to bring the merry-go-round from
rest to an angular speed of 0.500 rev/s in 2.00 s?
4. A 5.00-kg cylindrical reel with a radius of 0.600 m and a frictionless axle starts from
rest and speeds up uniformly as a 3.00-kg bucket falls into a well, making a light rope
unwind from the reel (figure below). The bucket starts from rest and falls for 4.00 s. (a)
What is the linear acceleration of the falling bucket? (b) How far does it drop? (c) What
is the angular acceleration of the reel?
5. A 10.0-kg cylinder rolls without slipping on a rough surface. At an instant when its
center of gravity has a speed of 10.0 m/s, determine (a) the translational kinetic energy of
its center of gravity, (b) the rotational kinetic energy about its center of gravity, and (c) its
total kinetic energy.
6. Use conservation of energy to determine the angular speed of the spool shown in the
figure for problem #4 after the 3.00-kg bucket has fallen 4.00 m, starting from rest. The
light string attached to the bucket is wrapped around the spool and does not slip as it
unwinds.
7. A horizontal 800-N merry-go-round of radius 1.50 m is started from rest by a constant
horizontal force of 50.0 N applied tangentially to the merry-go-round. Find the kinetic
energy of the merry-go-round after 3.00 s. (Assume it is a solid cylinder.)
8. A uniform solid cylinder of mass M and radius R rotates on a frictionless horizontal
axle (figure below). Two objects with equal masses m hang from light cords wrapped
around the cylinder. If the system is released from rest, find (a) the tension in each cord
and (b) the acceleration of each object after the objects have descended a distance h.
Hint: This actually behaves as if there were a single hanging mass of 2m!
Solutions:
1.
(a)   F  r  0.800 N  30.0 m   24.0 N  m
(b)  
(c)

I



2
mr
24.0 N  m
 0.750 kg  30.0 m

  0.0356 rad

at  r   30.0 m
2. The angular acceleration is  
 0.0356 rad s2
s2  1.07 m s2
 f  i
t
2
 
   i since  f  0 .
 t
 I 
Thus, the torque is   I    i . But, the torque is also    fr, so the
 t 
magnitude of the required friction force is
f


12 kg  m 2  50 rev m in  2 rad   1 m in 
I i

  21 N


r t
1 rev   60 s 
 0.50 m  6.0 s
k 
Therefore, the coefficient of friction is
3. I
1
1
2
M R 2  150 kg 1.50 m   169 kg  m
2
2
and

 f  i
t

2
 0.500 rev s 0   2 rad   
2.00 s


1 rev 
Thus,   F  r  I gives
I
F

r
f 21 N

 0.30
n 70 N
169 kg  m   2
2
1.50 m

rad s2 

 177 N
2
rad s2
4.
The moment of inertia of the reel is
I
1
1
M R 2   5.00 kg  0.600 m
2
2
2  0.900 kg  m 2
Applying Newton’s second law to the falling bucket gives
29.4 N  T   3.00 kg at
(1)
Then, Newton’s second law for the reel gives
 at 
 R 
  T R  I  I
or


0.900 kg  m 2
Iat
T 2 
at  2.50 kg at
R
 0.600 m  2
(2)
(a) Solving equations (1) and (2) simultaneously gives
at  5.35 m s2 dow nw ard


1
1
2
(b) y  v0 t at t2  0  5.35 m s2  4.00 s  42.8 m
2
2
5. (a)
(c)  
at 5.35 m s2

 8.91 rad s2
R
0.600 m
KEtrans 
2
1 2 1
m vt  10.0 kg10.0 m s  500 J
2
2
2
1
1 1
 v 
(b) KErot  I 2   m R 2  t2 
2
2 2
 R 

(c)
1 2 1
2
m vt   10.0 kg  10.0 m s  250 J
4
4
KEtotal  KEtrans  KErot  750 J
6.
As the bucket drops, it loses gravitational potential energy.
The spool gains rotational kinetic energy and the bucket gains
translational kinetic energy. Since the string does not slip on
the spool, v  r where r is the radius of the spool. The
moment of inertia of the spool is I 12 M r2 , where M is the
mass of the spool. Conservation of energy gives
 KE  KE  PE    KE  KE  PE 
t
r
g
f
t
r
g
i
1 2 1 2
m v  I  m gyf  0  0  m gyi
2
2
or

1
1 1
2

m  r    M r2   2  m g yi  yf

2
2 2
This gives

2m g yi  yf

m 
1
2
M r

2



2 3.00 kg 9.80 m s2  4.00 m
 3.00 kg+
1
2

 5.00 kg  0.600 m 2
 10.9 rad s
7. The moment of inertia of the cylinder is
I
1
1 w 
1  800 N 
M R2    R2  
1.50 m
2
2 g
2  9.80 m s2 
2
 91.8 kg  m
The angular acceleration is given by


I

F  R  50.0 N 1.50 m 

 0.817 rad s2
2
I
91.8 kg  m
At t = 3.00 s, the angular velocity is
  i   t 0   0.817 rad s2  3.00 s  2.45 rad s
and the kinetic energy is
KErot 

1 2 1
I  91.8 kg  m
2
2
2
  2.45 rad s
2
 276 J
2
8. 8.73 Choose an axis perpendicular to the page and passing through
the center of the cylinder. Then, applying   I to the
cylinder gives
 2T   R  
1
1

1
a
M R 2     M R 2   t  , or T  M at





2
2
R
4
(1)
Now apply Fy  m ay to the falling objects to obtain
 2m  g  2T   2m  a , or a  g  mT
t
t
(a) Substituting equation (2) into (1) yields
T
M mg
M g  M 

T , which reduces to T 

M  4m
4  4m 
(b) From equation (2) above,
at  g 
M g
4m g
1 M mg 
 g



m  M  4m 
M  4m
M  4m
(2)