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Physics 141 Problem Set 2 Corrected Solutions
Nataliya Yufa
(2.6) Gravel mixer
We want to ﬁnd
ω
such that at least for a moment the particles are not stuck to the
cylinder. This is most likely to happen at the top, where the normal force is in the direction
of gravity. For the particle to leave the cylinder, N=0. Hence the ycomponent of the total
force on the particle is
F
g
+
N
=
F
g
=
mv
2
r
.
(1)
Using the relation
v
=
ωR
we obtain
mg
=
m
v
2
r
=
mω
2
R,
(2)
⇒
ω
c
=
r
g
R
.
(3)
Thus for all values of
ω
less than
ω
c
the particles will not be stuck to the walls all the time.
(2.9) Particle in a cone
The particle happens to be moving in a horizontal circle inside a cone of halfangle
θ
.
We want to ﬁnd the radius
r
of the circle in terms of
v
0
,θ
and
g
. Consider the x and
ycomponents of the net force acting on the mass.
Since there’s no movement in the ydirection,
F
y
is zero, i.e.
F
y
=
N
y

mg
= 0
⇒
N
y
=
mg.
(4)
From the geometry we ﬁnd that tan
θ
=
N
y
N
x
,
hence
N
x
=
N
y
tan
θ
=
mg
tan
θ
.
(5)
Circular motion in the horizontal direction implies that
F
x
=
N
x
=
mg
tan
θ
=
mv
2
0
r
,
(6)
and solving for
r
we have
r
=
v
2
o
tan
θ
g
.
(7)
1
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Initially, the glass is accelerated by the force of friction due to the tablecloth for some
time
t
max
. In this time it will reach velocity
v
o
v
o
=
F
f
t
max
m
=
μgt
max
.
(8)
On the other hand, the glass will be slowed down by the friction due to the table (which,
because the coeﬃcients of friction happen to be the same for the table and the tablecloth,
has the same magnitude as the frictional force of the tablecloth). Therefore, during both the
acceleration and the deceleration the glass will travel exactly half of the available distance.
We want the glass to slow down in a given distance
d/
2, thus
v
0
=
q
μgd.
(9)
Combine the two expressions for
v
0
to get
v
0
=
μgt
max
=
q
μgd
(10)
⇒
t
max
=
s
μgd
μg
=
s
d
μg
=
s
0
.
5
ft
0
.
5
·
32
ft/s
2
=
1
4
√
2
s.
(11)
(2.25) Shortest possible period
Intuitively, the shortest period of rotation will occur when the two solid spheres are as
close as possible to each other. Let’s try to prove this. Because gravity is responsible for the
circular motion, we have
Gm
2
4
r
2
=
mv
2
r
(12)
⇒
v
=
s
Gm
4
r
.
(13)
The period is given by
T
=
2
πr
v
=
4
π
√
Gm
r
3
/
2
.
(14)
Note that the smaller the radius, the shorter the period. However, we’re limited by the
radius of the spheres, as they cannot come closer than their radius
R
. Hence the shortest
possible period would be
T
=
4
π
√
Gm
R
3
/
2
.
(2.28) Car going around a bend
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This note was uploaded on 02/03/2012 for the course PHY F1118 taught by Professor Landora during the Spring '11 term at Andhra University.
 Spring '11
 Landora

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