This preview shows pages 1–3. Sign up to view the full content.
Hw #4 solution
Chapter 6
No.5
(a)
In the horizontal plane, the only force that can keep the coin in circular
motion is friction. And because the coin is stationary, so the friction is
static friction.
(b)
22
2
30.0
50.0
/
(50.0
/ )
83.3
/
30.0
r
rr
s
s
R
cm
vc
m
s
m
s
ac
Rc
m
Fm
a
f
m
g
μ
=
=
==
=
=
m
s
m
Cancel
on both sides
2
2
83.3
/
1
/
0.0850
9.8
/
100
rs
s
r
cm s
m
ag
a
g
ms
cm
μμ
=⇒
=
=
=
No.9
(a)
In vertical direction, the net force on
the bob should be zero, because the
bob won’t move in the vertical
direction.
So we have
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document2
cos
0
cos
/ cos
80.0
9.8
/
/ cos5.00
787
y
FT
m
g
Tm
g
T
m
g
k
g
m
s
θ
θθ
=−
=
=⇒
=
=
×
=
∑
N
N
j
So we can write the tension in vector way:
787sin 5.00
787cos5.00
(68.6
)
(784
)
Ti
j
N
i
=+
=
+
JG
G
G
G
G
the vertical component :(784 )
Nj
G
the horizontal component: (68.6 )
Ni
G
(b)
Just consider the horizontal plane. The bob is in uniform circular motion,
so the horizontal component works as the radial force:
sin5.00
sin5.00/
rr
a
a
T
=
m
s
2
68.6
/80.0
0.857
/
r
aN
k
g
m
==
No.19
(a)
At point A, gravity and normal force cause the radial force. And the radial
force should be upward. Since the gravity is downward, so the normal
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '06
 Winnokur
 Physics, Circular Motion, Force, Friction

Click to edit the document details