Download Ch 9 HW Day : p 296 – 308, #`s 1, 3, 5, 7, 8, 9, 11, 13, 14, 15, 16, 17

Ch 9 HW Day 6: p 296 – 308, #’s 79, 81, 83, 85, 88, 90*, 93, 94, 96*, 99
Notes for #81
Hi, kids! Just to let you know for yo-yo problems, the torque (T∙r) depends on “little” r, while the
moment of inertia, I, assumes the disk (yo-yo) has radius “big” R (because it does!). That just means you
can’t cancel all of the r’s out in the beginning.
81 ••
Picture the Problem The forces acting
on the yo-yo are shown in the figure.
Apply Newton’s 2nd law in both
translational and rotational form to
obtain simultaneous equations in T, a,
and  from which we can eliminate 
and solve for T and a.
Apply Newton’s 2nd law to the
yo-yo:
F
y
 mg  T  m(a y ) so, T = mg – ma
(1)
and

: Tr  I 0
(2)
Use a  r to eliminate  in
equation (2)
Tr  I 0
Eliminate T between
equations (1) and (3) to
obtain:
(mg – ma)r = (½ mR2)(a/r)
(4)
Cancel out the m’s, multiply both sides by r, distribute on
the left.
a
r
gr2 – ar2 = ½ R2a
move – ar2 to the right side
Substitute 12 mR 2 for I0 in
equation (4):
gr2 = ½ R2a + ar2
Factor out the a on the right side, and divide
Solve for a:
gr2 = (½ R2+ r2)a
gr2
= a
2
2
(½ R + r )
(3)
Substitute numerical values
and evaluate a:
Use equation (1) to solve for
and evaluate T:
a
(9.81 m/s 2 )(0.01) 2
 0.192 m/s 2
1
2
2
 (0.1)  (0.01) 
2

T  mg  a 

 0.1kg  9.81m/s 2  0.192 m/s 2

 0.962 N
88 ••
Picture the Problem The three forces
acting on the basketball are the weight of
the ball, the normal force, and the force of
friction. Because the weight can be
assumed to be acting at the center of
mass, and the normal force acts through
the center of mass, the only force which
exerts a torque about the center of mass is
the frictional force. We can use Newton’s
2nd law to find a system of simultaneous
equations that we can solve for the
quantities called for in the problem
statement.
(a) Apply Newton’s 2nd law in both
translational and rotational form to the
ball:
F
F
x
 mg sin   f s  ma ,
y
 Fn  mg cos  0
and

 f s r  I 0
0
Because the basketball is rolling
without slipping we know that:

Substitute in equation (3) to obtain:
fsr  I 0
From Table 9-1 we have:
I 0  23 mr 2
Substitute for I0 and  in equation
(4) and solve for fs:
Substitute for fs in equation (1) and
solve for a:
(b) Find fs using equation (5):
(1)
(2)
a
r
fsr 
a
(3)
a
r
(4)

2
3
3
5
g sin 
mr 2
 ar  f
s
f s  23 m 53 g sin   
 23 ma
2
5
mg sin 
(5)
(c) Solve equation (2) for Fn:
Fn  mg cos
Use the definition of fs,max to obtain:
f s,max  s Fn  s mg cos  max
Use the result of part (b) to obtain:
2
5
Solve for max:
mg sin  max  s mg cos  max
max  tan 1  52 s 
94 •••
Picture the Problem Let M represent the
combined mass of the two disks and their
connecting rod and I their moment of
inertia. The object’s initial potential
energy is transformed into translational
and rotational kinetic energy as it rolls
down the incline. The force diagram
shows the forces acting on this composite
object as it rolls down the incline.
Application of Newton’s 2nd law will
allow us to derive an expression for the
acceleration of the object.
(a) Apply Newton’s 2nd law to the
disks and rod:
F
F
x
 Mg sin   f s  Ma ,
y
 Fn  Mg cos  0 ,
(1)
(2)
and

Eliminate fs and  between
equations (1) and (3) and solve for a
to obtain:
Express the moment of inertia of the
two disks plus connecting rod:
a
0
 f s r  I
(3)
Mg sin 
I
M 2
r
(4)
I  2 I disk  I rod
 212 mdisk R 2   12 mrod r 2
 mdisk R 2  12 mrod r 2
Substitute numerical values and
evaluate I:
I  20 kg 0.3 m   12 1 kg 0.02 m 
2
 1.80 kg  m 2
2
Substitute in equation (4) and
evaluate a:

41 kg  9.81 m/s 2  sin30 
a
1.80 kg  m 2
41 kg 
0.02 m 2
 0.0443 m/s 2
(b) Find  from a:

a 0.0443 m/s 2

 2.21 rad/s 2
r
0.02 m
(c) Express the kinetic energy of
translation of the disks-plus-rod
when it has rolled a distance s
down the incline:
K trans  12 Mv 2
Using a constant-acceleration
equation, relate the speed of the
disks-plus-rod to their acceleration
and the distance moved:
v 2  v02  2as
or, because v0 = 0,
Substitute to obtain:
K trans  Mas
v 2  2 a s


 41 kg  0.0443 m/s 2 2 m 
 3.63 J
(d) Express the rotational kinetic
energy of the disks after rolling 2 m
in terms of their initial potential
energy and their translational kinetic
energy:
Substitute numerical values and
evaluate Krot:
K rot  U i  K trans  Mgh  K trans


K rot  41kg  9.81m/s 2 2 m sin30 
 3.63 J
 399 J
93 •••
Picture the Problem The kinetic energy of the wheel is the sum of its translational and
rotational kinetic energies. Because the wheel is a composite object, we can model its
moment of inertia by treating the rim as a cylindrical shell and the spokes as rods.
K  K trans  K rot
Express the kinetic energy of the wheel:
 12 M tot v 2  12 I cm 2
 12 M tot v 2  12 I cm
v2
R2
where Mtot = Mrim + 4Mspoke
I cm  I rim  I spokes
Express the moment of inertia of the
wheel:

 M rim R 2  4 13 M spoke R 2
 M rim  43 M spoke R 2


K  12 M tot v 2  12 M rim  43 M spoke R 2
Substitute for Icm in the equation for
K:

 M
1
2
tot

 M rim   23 M spoke v 2
 Rv
2
2
K  12 7.8 kg  3 kg   23 1.2 kg 6 m/s 
Substitute numerical values and
evaluate K:
2
 223 J
99 ••
Picture the Problem The forces
responsible for the rotation of the
gears are shown in the diagram to
the right. The forces acting through
the centers of mass of the two gears
have been omitted because they
produce no torque. We can apply
Newton’s 2nd law in rotational form
to obtain the equations of motion of
the gears and the not slipping
condition to relate their angular
accelerations.
(a) Apply   I to the gears
to obtain their equations of
motion:
Because the gears do not slip
relative to each other, the
tangential accelerations of the
points where they are in contact
must be the same:
Divide equation (1) by R1 to obtain:
Divide equation (2) by R2 to obtain:
2 N  m  FR1  I11
(1)
and
FR2  I 2 2
(2)
where F is the force keeping the gears from slipping
with respect to each other.
R11  R2 2
or
2 
R1
1  12 1
R2
2 Nm
I
 F  1 1
R1
R1
F
I2
2
R2
(3)
Add these equations to obtain:
2 N  m I1
I
 1  2  2
R1
R1
R2
Use equation (3) to eliminate 2:
2 N  m I1
I
 1  2 1
R1
R1
2 R2
Solve for 1 to obtain:
Substitute numerical values and
evaluate 1:
1 
1 
2Nm
R
I1  1 I 2
2 R2
2Nm
0.5 m
1 kg  m 2 
16 kg  m 2
21 m 


 0.400 rad/s 2
Use equation (3) to evaluate 2:
 2  12 0.400 rad/s 2   0.200 rad/s 2
(b) To counterbalance the 2N·m torque, a counter torque of
2 N·m must be applied to the
first gear. Use equation (2) with
1 = 0 to find F:
2 N  m  FR1  0
and
F
2Nm 2Nm

 4.00 N
R1
0.5 m