HRW Hints for exercise. Chapter 12.pdf

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C
HAPTER
12
H
INT FOR
P
ROBLEM
2
(a) through (f) As viewed by a passenger, the tire is simply rotating. Every point on the rim
has speed
v
=
v
c
, where
v
c
is the speed of the car. The point on top is moving forward, the
point on the bottom is moving backward. Their accelerations have magnitude
v
2
/R,
toward
the center. The velocity and acceleration of the center of the wheel are zero.
(g) and (l) Simply add vectorially
v
c
, in the forward direction, to each of the velocities found
above. Since the observer is not accelerating with respect to the car, the accelerations are
the same.
C
HAPTER
12
H
INT FOR
E
XERCISE
7
(a) Here an acceleration is given, not a speed, so we use a dynamical method, not an energy
method. Write Newton’s second law equations for the motion of the center of mass and
for rotation about the center of mass. Apply the condition
a
com
=
αR,
where
a
com
is the
acceleration of the center of mass,
α
is the angular acceleration about the center of mass,
and
R
is the radius. Solve for the incline angle.
Draw a force diagram for the sphere. Take the
x
axis to be parallel to the incline and down
the incline. Take the
y
axis to be perpendicular to the incline. Draw the sphere and show
the forces: the force of friction
f
, up the incline and acting at the point of contact of the
sphere with the plane; the normal force
N
, in the
y
direction and acting at the point of
contact, and the force of gravity
mg,
down and acting at the center of the sphere. Let
θ
be
the angle of incline and write the
x
component of the second law:
mg
sin
θ
−
f
=
ma
com
,
where
m
is the mass of the sphere and
a
com
is its acceleration.
Choose the direction for positive rotation, consistent with rolling down the plane, and write
the second law for rotation:
f R
=
Iα ,
where
I
is the rotational inertia of the sphere. The normal force and the force of gravity
do not exert torques about the center of the sphere; only the force of friction does. Finally
write the condition for no sliding:
a
com
=
αR ,
where
R
is the radius of the sphere. Systematically eliminate
α
and
f
, then solve for sin
θ.
You should get
a
com
sin
θ
= (1 +
I/mR
2
)
.
g
Substitute
I
=
2
mR
2
(see Table 11–2) and calculate
θ.
5
(b) Use Newton’s second law and solve for
a
com
or put
I
= 0 in the result for the sphere.
You should get
a
com
=
g
sin
θ .
Your result should be greater than 0.1g. To understand why, think about the forces on the
rolling sphere that have components along the incline.
ans:
(a) 8.1
◦
; (b) more
C
HAPTER
12
H
INT FOR
P
ROBLEM
8
(a) Draw a force diagram for the wheel. Four forces act on it : the applied force
F
, at the
center and directed to the right, the frictional force
f
, at the point of contact with the surface
and directed to the left, the force of gravity
mg,
at the center and directed downward, and
the normal force of the surface, at the point of contact with the surface and directed upward.
Take the positive direction to be to the right. Then the horizontal component of Newton’s
second law is
F
−
f
=
ma
com
,
where
m
is the mass of the wheel and
a
com
is the acceleration of its center of mass. Solve
for
f
. If you get a positive result then the frictional force is indeed to the left. If you get a
negative result it is to the right.
(b) The angular acceleration of the wheel about its center is
α
=
−
a
com
,
R
where counterclockwise was chosen to the positive. Here
R
is the radius of the wheel.
According to Newton’s second law for rotation
τ
net
=
Iα ,
where
τ
net
is the net torque on the wheel and
I
is the rotational inertia of the wheel. The
only torque about the center of the wheel is the torque of friction and this is
−f
R,
so
−f
R
=
−I
Solve for
I.
a
com
.
R
C
HAPTER
12
H
INT FOR
P
ROBLEM
12
(a) First find the speed that the marble must have at the top in order to stay on the track.
At the top the marble experiences the smallest possible inward force when the normal force
of the track is zero and the only force acting is the force of gravity
mg.
The inward force
is greater than this if the track exerts a normal force but it cannot be less than this. Thus,
the minimum speed
v
of the center of mass at the top is determined by
mv
2
.
mg
=
R
−
r
Note that the radius of the circle traversed by the center of mass,
R
−
r,
is used.
Now, use conservation of mechanical energy to find the initial height
h
of the center of mass
that will result in speed
v
at the top. If the zero of gravitational potential energy is at the
bottom of the loop, then the initial potential energy is
mgh.
The initial kinetic energy is 0
since the marble starts from rest. The potential energy at the top of the loop is
mg(2R
−
r)
since the center of mass is then 2R
−
r
above the bottom. The kinetic energy is given by
1
mv
2
+
1
Iω
2
, where
I
is the rotational inertia of the marble. When the condition
R
r
2
2
2
2
is used and when
I
=
5
mr
(see Table 11–2) and
ω
=
v/r
(no sliding) are substituted, the
conservation of mechanical energy equation becomes
mgh
= (7/10)mv
2
+ 2mgR
.
Use
mv
2
=
mgR
to eliminate
v,
then solve for
h.
(b) When the marble is at Q the horizontal component of Newton’s second law is
N
=
mv
2
/R,
where
v
is the speed of the marble and
N
is the normal force of the track on the
marble.
R
r
was assumed. Use conservation of mechanical energy, with
h
= 6R, to find
v.
C
HAPTER
12
H
INT FOR
P
ROBLEM
14
Draw a force diagram for the ball. Three forces act on it: the force of gravity
mg,
down and
at the center of the ball, the normal force of the lane
N
, up and at the point of contact with
the lane, and the frictional force
f
, backward and at the point of contact with the lane.
Take the forward direction to be positive. Then the horizontal component of Newton’s
second law for the center of mass yields
−f
=
ma
com
and the vertical component yields
N
−
mg
= 0
.
The only torque about the center of the ball is the frictional torque. Its magnitude is
f R,
where
R
is the radius of the ball, and it is clockwise (the negative direction). Newton’s
second law for rotation about the center of the ball is
−f
R
=
Iα ,
where
I
is the rotational inertia of the ball about its center and
α
is the angular acceleration
of the ball.
(a) Notice that the acceleration of the center of mass of the ball is directed opposite the
velocity. The ball slows as it slides. At the same time the frictional torque causes the
angular speed of the ball to increase. The ball stops sliding and starts to roll smoothly when
ωR
+
v
com
= 0.
(b) and (c) While the ball is sliding the frictional force is kinetic in nature, so
f
=
µ
k
N
,
where
µ
k
is the coefficient of kinetic friction. Substitute this expression into the horizontal
component of Newton’s second law for the center of mass and into Newton’s second law for
rotation, then solve for
a
com
and
α.
Also use
I
= (2/5)mR
2
, which you can find in Table
11–2 of the text.
(d) Use
v
com
=
v
0
+
a
com
t
to find the velocity of the center of mass as a function of time and
ω
=
αt
to find the angular velocity as a function of time. Solve
ωR
+
v
com
= 0
for the time
t
when the rolling becomes smooth.

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