Answers for Homework III
Wu Yongcheng
Department of Physics
1
KK’s 2.4
Ans:
Assume that the rotational radiuses for particles of mass m and M are rm and rM respectively.
Then we have:
F = mω 2 rm
F = M ω 2 rM
So we get the rotational radiuses for each particle:
rm =
F
ω2 m
rM =
F
ω2 M
So the separation of two particles is:
R = rm + rM =
2
F 1
1 +
ω2 m M
KK’s 2.11
Ans:
Tup
45◦
m
45
◦
Tlow
mg
1
1
1
mg + √ Tlow = √ Tup
2
2
1
1
l
√ Tup + √ Tlow = mω 2 √
2
2
2
So
2
l
√ Tlow = mω 2 √ − mg
2
2
2
l
√ Tup = mω 2 √ + mg
2
2
Finally we have:
Tlow = mω 2
Tup
3
l
−
2
l
= mω +
2
√
√
2
2mg
2
2mg
2
KK’s 2.12
Ans:
When pulling out the cloth, the grass does not move appreciably.
µmg∆t = ∆p = mv
v = µg∆t
To get the longest time, the grass should move the longest distance.
1
mv 2 = µmgl
2
where l = 6in(1in = 2.54cm = 0.0254m)
1
µg(∆t)2 = l
2
s
2l
∆t =
≈ 0.1s
µg
4
KK’s 2.14
Ans:
The force in the same rope is uniform. Assume that the force in the rope connect A and B is
F1 , so the force in the rope connect C is F2 = 2F1 . So we have:
mC g − 2F1 = mC ac
2
F1 = mA aA
F1 = mB aB
And there is a connected condition:
aC =
aA + aB
2
So we get:
2F1
F1 1
1
=
(
+
)
mC
2 mA
mB
1
1 2
g = F1
+
+
mC
2mA
2mB
2mA mB mC g
F1 =
4mA mB + mC mB + mC mA
g−
Finally:
aC = g 1 −
4mA mB
mC mB + mC mA
=
g
4mA mB + mC mB + mC mA
4mA mB + mC mB + mC mA
2mB mC g
4mA mB + mC mB + mC mA
2mA mC g
aB =
4mA mB + mC mB + mC mA
aA =
5
KK’s 2.16
Ans:
N
mA
mg
In the frame of the wedge, the force diagram is as showed. Then in this frame we have:
mA mg
√ + √ =N
2
2
mA mg
ma0k = √ − √
2
2
3
√
a0k
Back to the lab frame:
6
=
2
(A − g)
2
1
1
ax = A − (A − g) = (A + g)
2
2
1
ay = (A − g)
2
KK’s 2.19
Ans:
When M3 is rest, then the whole acceleration is:
a=
F
M1 + M2 + M + 3
In the frame of M3 , for M2 :
M2 a = M3 g
So:
F =
7
M3
(M1 + M2 + M3 )g
M2
KK’s 2.22
Ans:
For the whole rope, it is under 3 forces: mg, Tend1 , Tend2 . Under these forces the rope is at rest:
Tend1 sin θ = Tend2 sin θ
Tend1 cos θ + Tend2 cos θ = mg
So we get:
mg
cos θ
In the middle of the rope, let’s consider about the left(or right) half side of the rope:
Tend1 = Tend2 =
Tmid = T end1 sin θ
Tmid = tan θmg
4
N
f = µN
Tθ+δθ
Tθ
δθ
8
KK’s 2.24
Ans:
From the figure, we have:
δθ
δθ
+ Tθ+δθ sin
≈ Tθ δθ
2
2
Tθ = Tθ+δθ + µN = Tθ+δθ + µTθ δθ
N = Tθ sin
So:
−dT = µT dθ
dT
= −µdθ
T
T
ln
= −µθ
TA
T = TA e−µθ
9
KK’s 2.26
Ans:
f
x
F
r
When we are at radius r, we are under force F :
3
F =G
4πr
Mr m
4πGρmr
3 ρm
=
G
=
2
r
r2
3
5
x
4πGρm
F =
x
r
3
The force is proportional to the distance x, so it is a simple harmonic motion.
r
3
T = 2π
4πGρ
f=
The time needed for satellite:
F =G
4πR3
3 ρm
R2
= m(
2π 2
) R
T
3
T 2
) =
2π
4πGρ
r
3
T = 2π
4πGρ
(
10
KK’s 2.30
Ans:
mB
FB
T
mA
T
FA
In the frame of the disk, the masses are under another forces:
FA = mA ω 2 rA
mA aA = T − mA ω 2 rA
FB = mB ω 2 (l − rA )
mB aB = mB ω 2 (l − rA ) − T
As the two masses are in the same rope, we have the connectional condition:
aA = aB
aA =
mB (l − rA ) − mA rA 2
ω
mA + mB
6
F
m
r
ω
11
KK’s 2.33
Ans:
F = mω 2 r
r̈ = ω 2 r
The generate solution is:
r = Ae−γt + Beγt
where γ = ω and A B are two constants decided by initial condition.
r0 = A + B
v0
= −A + B
ω
v0
1
B = (r0 + )
2
ω
1
v0
A = (r0 − )
2
ω
when B = 0 the r will decrease continually.
v0 = −Aω + Bω
v0 = −r0 ω
The minus sign means that the direction of the velocity points to the origin.
12
KK’s 2.34
Ans:
As the force point to the origin, so the acceleration along the θ direction is zero:
0 = 2ṙω + rω̇
2V ω
r
V dt
dω
=2
ω
r0 − V t
ω̇ =
7
ω
r0
= ln( )2
ω0
r
r0
ω = ω0 (
)2
r0 − V t
ln
The force along the radius is:
F = mω 2 r = mω02
13
r04
r3
KK’s 2.37
Ans:
y
m
2
2
r
m vr = m 4π
T2
x
mg
In the frame of the hovercraft, the force diagram is as show. In this frame, we have the condition:
dy
= tan θ =
dx
4π 2 r
T2
g
=
4π 2 x
gT 2
2
dy =
4π
xdx
gT 2
2π 2 2
x
gT 2
This is the equation of cross section of the bowl shaped track.
y=
14
Additional
Ans:
We consider the potential energy:
1
E = −mgR cos θ − mω 2 R2 sin2 θ
2
The equilibrium position has the extreme potential energy
dE
= 0 = mgR sin θ − mω 2 R2 sin θ cos θ
dθ
8
R
θ
mω 2 ρ
ρ
mg
cos θ =
g
ω2 R
whether the equilibrium is stable or unstable is decided by the 2nd derivative.
d2 E
= mgR cos θ − mω 2 R2 cos 2θ
dθ2
When cos θ =
g
ω2 R
m
d2 E
= 2 (ω 4 R2 − g 2 )
dθ2
ω
If the ring has a equilibrium position, cos θ = ω2gR must have solution. That is:
g ≤ ω2 R
Thus
d2 E
m
= 2 (ω 4 R2 − g 2 ) ≥ 0
dθ2
ω
So it is a stable position.
9