Download Ch 9 HW Day 1

Ch 9 HW Day 4: p 296 – 308, #’s 58, 59, 60, 62, 65, 68, 70, 73
62. Picture the Problem The earth’s rotational kinetic energy is given by
K rot  12 I 2 where I is its moment of inertia with respect to its axis of rotation. The
center of mass of the earth-sun system is so close to the center of the sun and the earthsun distance so large that we can use the earth-sun distance as the separation of their
centers of mass and assume each to be point mass.
Express the rotational kinetic energy
of the earth:
K rot  12 I 2
Find the angular speed of the earth’s
rotation using the definition of :

From Table 9-1, for the moment of
inertia of a homogeneous sphere, we
find:
I  52 MR 2
Substitute numerical values in
equation (1) to obtain:
K rot 
(1)

2 rad

t 24 h  3600 s
h
5
 7.27 10 rad/s

2
5
6.0 10
24

kg 6.4  106 m

2
 9.83  1037 kg  m 2
9.83 10 kg  m 
 7.27 10 rad/s 
1
2
37
2
2
5
 2.60 1029 J
Express the earth’s orbital kinetic
energy:
2
K orb  12 I orb
Find the angular speed of the center
of mass of the earth-sun system:

(2)

t
2 rad

365.25 days  24
h 3600 s

day
h
 1.99  10 7 rad/s
Express and evaluate the orbital
moment of inertia of the earth:
2
I  M E Rorb


 6.0 1024 kg 1.50 1011 m
 1.35 1047 kg  m 2
Substitute in equation (2) to obtain:
K orb 
1.35 10 kg  m 
 1.99 10 rad/s 
1
2
47
7
 2.67 1033 J
2
2

2
Evaluate the ratio
K orb
:
K rot
K orb 2.67  1033 J

 10 4
29
K rot 2.60  10 J
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
kinetic energy of the system and the
work done against friction:
Substitute for Ipulley and  to obtain:
Solve for v:
K  U  Wf  0
or, because Ki = Uf = 0,
1
2
m  M v 2  12 I pulley 2
 mgh  k Mgh  0
1
2
m  M v
v
2


1 1
2 2
Mp
v
2
R2
 mgh  k Mgh  0
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