Download Ch 9 HW Day 1

Ch 9 HW Day 4: p 296 – 308, #’s 58, 59, 60, 62, 65, 68, + 69, 70, 71, 73, 76
69 ••
Picture the Problem The diagrams
show the forces acting on each of the
masses and the pulley. We can apply
Newton’s 2nd law to the two blocks and
the pulley to obtain three equations in
the unknowns T1, T2, and a.
Apply Newton’s 2nd law to the two
blocks and the pulley:
F  T  m a ,
  T  T r  I  ,
x
p
1
(1)
(2)
4
2
1
p
and
F
x
 m2 g  T2  m2 a
(3)
Eliminate  in equation (2) to
obtain:
Eliminate T1 and T2 between
equations (1), (3) and (4) and solve
for a:
Substitute numerical values and
evaluate a:
T2  T1  12 M p a
Using equation (1), evaluate T1:
T1  4 kg  3.11m/s 2  12.5 N
Solve equation (3) for T2:
T2  m2 g  a
Substitute numerical values and
evaluate T2:
(4)
a
m2 g
m2  m4  12 M p
a
2 kg 9.81m/s 2  
2 kg  4 kg  12 0.6 kg 

3.11m/s 2


T2  2 kg  9.81m/s 2  3.11m/s 2

 13.4 N
70 ••
Picture the Problem We’ll solve this problem for the general case in which the mass of the
block on the ledge is M, the mass of the hanging block is m, the mass of the pulley is Mp,
and R is the radius of the pulley. Let the zero of gravitational potential energy be 2.5 m
below the initial position of the 2-kg block. The initial potential energy of the 2-kg block
will be transformed into the translational kinetic energy of both blocks plus rotational
kinetic energy of the pulley plus work done against friction.
(a) Use energy conservation to
relate the speed of the 2 kg block
when it has fallen a distance h to
its initial potential energy, the
K  U  Wf  0
or, because Ki = Uf = 0,
kinetic energy of the system and
the work done against friction:
1
2
m  M v 2  12 I pulley 2
 mgh  k Mgh  0
Substitute for Ipulley and  to
obtain:
1
2
Solve for v:
v2
m  M v   M p  2
R
 mgh  k Mgh  0
v
2
1 1
2 2
2 ghm   k M 
M  m  12 M p
Substitute numerical values and evaluate v:
v


2 9.81m/s 2 2.5 m 2 kg  0.254 kg 
 2.79 m/s
4kg  2 kg  12 0.6 kg 
(b) Find the angular velocity of the
pulley from its tangential speed:

v 2.79 m/s

 34.9 rad/s
R
0.08 m
71 ••
Picture the Problem Let the zero of gravitational potential energy be at the water’s
surface and let the system include the winch, the car, and the earth. We’ll apply energy
conservation to relate the car’s speed as it hits the water to its initial potential energy.
Note that some of the car’s initial potential energy will be transformed into rotational
kinetic energy of the winch and pulley.
Use energy conservation to relate the car’s speed as it
hits the water to its initial potential energy:
K  U  0
or, because Ki = Uf = 0,
1
2
Express w and p in terms of the speed v of the rope,
which is the same throughout the system:
Substitute to obtain:
Solve for v:
mv 2  12 I w w2  12 I p p2  mgh  0
v2
v2
w  2 and p  2
rw
rp
1
2
mv2  12 I w
v
v2 1 v2
 I p  mgh  0
rw2 2 rp2
2mgh
I
I
m  w2  p2
rw rp
Substitute numerical values and evaluate v:
v


21200 kg  9.81 m/s 2 5 m 
320 kg  m 2 4 kg  m 2
1200 kg 

0.8 m2 0.3 m2
 8.21 m/s
73 ••
Picture the Problem The force diagram
shows the forces acting on the sphere
and the hanging object. The tension in
the string is responsible for the angular
acceleration of the sphere and the
difference between the weight of the
object and the tension is the net force
acting on the hanging object. We can use
Newton’s 2nd law to obtain two equations
in a and T that we can solve
simultaneously.

(a)Apply Newton’s 2nd law to the sphere and the
hanging object:
0
x
TR 

Eliminate T between equations (2)
and (3) and solve for a to obtain:
a
g
2M
1
5m
(b) Substitute for a in equation (2)
and solve for T to obtain:
T
2mMg
5m  2M
76 ••
Picture the Problem Choose the
coordinate system shown in the diagram.
By applying Newton’s 2nd law of motion,
we can obtain a system of two equations
in the unknowns T and a. In (b) we can
use the torque equation from (a) and our
value for T to find. In (c) we use the
condition that the acceleration of a point
on the rim of the cylinder is the same as
(1)
 mg  T  ma
(2)
and
F
Substitute for Isphere and  in equation (1)
to obtain:
 TR  I sphere
2
5
MR 2
 Ra
(3)
the acceleration of the hand, together
with the angular acceleration of the
cylinder, to find the acceleration of the
hand.
(a) Apply Newton’s 2nd law to the cylinder about an axis
through its center of mass:

0
 TR  I 0
a
R
and
F
x
Solve for T to obtain:
T  Mg
(b) Rewrite equation (1) in terms
of :
TR  I 0
Solve for :
Substitute for T and I0 to obtain:


TR
I0
MgR
2g

2
R
MR
1
2
(c) Relate the acceleration a of the
hand to the angular acceleration
of the cylinder:
a  R
Substitute for  to obtain:
 2g 
a  R
  2g
 R 
 Mg  T  0