1.
Apply conservation of energy to the sphere, as done in Example 8-13 (textbook).
(a) The work of Example 8-13 is exactly applicable here. The symbol d is to represent the
distance the sphere rolls along the plane. The sphere is rolling without slipping, so
v CM = ω R.
vCM =
10
7
gh =
10
7
gd sin θ =
10
7
(9.80 m s )(10.0 m )(sin 30.0º ) = 8.367
2
= 8.37 m s
ω = vCM R = 8.367 m s (2.00 × 10 −1 m ) = 41.8 rad s
(b)
K E trans
=
K E rot
1
2
1
2
M v C2 M
=
I CMω 2
M v C2 M
=
2
2 vC M
MR
R2
1
2
1
2
(
2
5
)
2 .5
(c) Only the angular speed depends on the radius. None of the results depend on the mass.
2. The beam is in equilibrium. Use the conditions of equilibrium to calculate the tension in the
wire and the forces at the hinge. Calculate torques about the hinge, and take
counterclockwise torques to be positive.
∑ τ = (F
T
1
2
FT =
sin θ )l2 − m1gl1 2 − m2 gl1 = 0
m1gl1 + m2 gl1
=
l2 sin θ
1
2
→
(155 N )(1.70 m ) + (245 N )(1.70 m )
(1.35 m )(sin 35.0º )
= 708.0 N ≈ 7.08 × 10 2 N
FH x = FT cos θ = (708 N ) cos 35.0º = 579.99 N ≈ 5.80 × 10 2 N
x
= FH x − FT cos θ = 0
y
= FH y + FT sin θ − m1g − m2 g = 0
∑F
∑F
→
→
FH y = m1g + m2 g − FT sin θ = 155 N + 245 N − (708 N ) sin 35.0º = −6.092 N ≈ −6 N (down )
3. Draw a force diagram for the cable that is
supporting the right-hand section. The forces
r
will be the tension at the left end, FT2 , the
r
tension at the right end, FT1 , and the weight of
r
the section, mg.
The weight acts at the midpoint of the horizontal span of the cable. The
system is in equilibrium. Write Newton’s 2nd law in both the x and y directions to find the
tensions.
∑F
x
= FT1 cos19º − FT 2 sin 60º = 0 →
FT2 = FT1
∑F
y
cos 19º
sin 60º
= FT 2 cos 60º − FT1 sin 19º − mg = 0 →
FT1
FT 2 cos 60º − mg
=
sin 19º
sin 60º
= mg
= 4.539 mg ≈ 4.5 mg
(cos 19º cos 60º − sin 19º sin 60º )
cos 19º
cos 19º
= FT1
= 4.539
mg = 4.956mg ≈ 5.0mg
sin 60º
sin 60º
FT1 =
FT1
FT2
cos 19º
cos 60º − mg
sin 19º
→
sin 19º
To find the height of the tower, take torques about the
point where the roadway meets the ground, at the right
r
side of the roadway. Note that then FT1 will exert no
torque. Take counterclockwise torques as positive. For
r
purposes of calculating the torque due to FT2 , split it into x
and y components.
∑ τ = mg ( d )+ F
(F
h=
1
2
T2 y
1
− 12 mg
FT2 x
T2 x
h − FT2 y d1 = 0
)d = (F
T2
1
→
cos 60º − 12 mg )
FT2 sin 60º
d1 =
(4.956 mg cos 60º
− 0.50 mg )
4.956mg sin 60º
(343 m )
= 158 m
4.
This problem is solved observing the following facts:
1. The total initial angular and linear momentum of the system merry-goround/person/rock is zero;
2. There is an instantaneous (infinitesimally small interval of time) torque applied to the
system when the person throws the rock tangentially to the outer edge of the merrygo-round;
3. During this infinitesimally small interval of time, both the merry-go-round/person and
the rock final system will experience such torque;
4. At this point, there is still conservation of total angular momentum, considering the
torque as applied for only an infinitesimally interval of time. In this situation, the
rock, and mentioned in the precious item, also feels this torque and experience an
instantaneous angular motion; We can still consider it as an isolated system where the
objects that compound the system can interact with each other without losing
momentum or energy to the “outside world”;
5. Note, however, that the rock won’t have any angular momentum after this small
interval of time, it won’t be subjected to any torque and it is not related to any axis of
rotation (contrary to the merry-go-round/person system which has an axis of rotation).
a) Considering the above items, we can state the system will conserve total angular
momentum at the exact infinitesimally small amount of time the torque is applied.
0
total initial angular momentum
inal total angular momentum
Where,
and momentum of inertia and angular velocity of the merry-go-round;
and momentum of inertia and angular velocity of the person;
and momentum of inertia and angular velocity of the rock;
Using conservation of total angular momentum:
0
;
! ;
! Since the person and merry-go-round make a single system, both will have the same
angular velocity: " Then,
# $ ! %
0
' ! 2
( % 2 '
is related to the tangential linear velocity, ) 7.82 /. , of the rock by:
)
!
We then have:
) ⁄!
( (
% 2 '
1.13 23
5
7.82 /.
43.72 827 63
50.6 239
2
(0.005118 !:;/.
(5.12 < 10=> !:;/.
b) The linear speed of the girl, ) , is:
) ! 5(0.005118
!:;
9 3.72 (1.90 < 10= /.
.