Download Physics 1. Mechanics Problems

Physics 1. Mechanics
Problems
radius R in the plane x − y. The light frequency in the source frame is f . A distant observer (in the
same x − y plane) measures the frequency of received light as a function of time. Find this function.
10
Problems given at various tests, including midterms and
finals
Problem 10.1.
A particle of the mass m is moving on the elliptic orbit r = p/(1 − cos ϕ). The
particle is attracted to the focus O1 by the force |F1 | = k1 /r12 (where |r1 | is the distance between
the focus O1 and the particle). In addition, it is attracted to the ellipse center O2 with the force
|F2 | = k2 r2 . The third force acting on the particle is always perpendicular to the particle velocity.
When the particle is in A, its velocity is v1 = (0, v1 ). What is its angular momentum in B ?
y
B
A
O1
O2
x
Solution. Points:
First we need the coordinates of the points A, B, O1 , and O2 :
1. O1 is the coordinate origin, so that rO1 = xO1 = yO1 = 0.
2. A corresponds to ϕA = π, so that rA = p/(1 + ), xA = −p/(1 + ), yA = 0.
3. It is convenient to use the point C: ϕC = 0, rC = p/(1 − ), xC = p/(1 − ), yC = 0.
4. O2 is in the middle between A and C, so that xO2 = (xA + xC )/2 = p/(1 − 2 ), yO2 =
(yA + yC )/2 = 0.
5. B is just above O2 , so that xB = xO2 = p/(1 − 2 ). The equation r = p(1 − cos ϕ) can be
written as
r − r cos ϕ = p → r = p + x, x = r cos ϕ
so that rB = p/(1 − 2 ). Now
yB =
q
2
rB
− x2B = √
p
1 − 2
Force vectors:
We denote the position of the particle with r = (x, y), respectively, other points rO1 = (0, 0),
rO2 = (xO2 , 0), rA = (xA , 0), rB = (xB , yB ).
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Physics 1. Mechanics
Problems
Now we introduce the vectors r1 = r − rO1 = r and r2 = r − rO2 . With this notation the forces
(both attractive) will be written as follows
F1 = −
k1
rb1 ,
r12
F2 = −k2 r2 rb2
where rb1 and rb2 are unit vectors, r1 = |r1 | and r2 = |r2 |.
Potential energy:
Both forces are central forces: the direction of F1 is to O1 and the magnitude depends only on
|mbr1 |, the direction of F2 is to O2 and the magnitude depends only on |mbr2 |. Central forces are
conservative, and the relation to the potential energy is the following:
F1 = −
∂U1
rb1 ,
∂r1
F2 = −
∂U2
rb2
∂r2
which gives
∂U1
k1
= 2,
∂r1
r1
∂U2
= k 2 r2
∂r2
and finally
U1 = −
k1
,
r1
U2 =
k2 r22
2
Energy conservation:
Since the third (unknown) force is always perpendicular to the velocity, F3 ⊥ v, it does not
produce work, so that the energy conservation gives
mv 2
mv 2 k1 k2 r22
+ U1 + U2 =
−
+
= E = const
2
2
r1
2
When the particle is in the point A, we have v = (0, v1 ), r1 = p/(1 + ), r2 = p/(1 − 2 ).
When the particle is in the point B, we have v = (0, v) (this v is the unknown we are looking
√
for), r1 = p/(1 − 2 ), r2 = p/ 1 − 2 .
Therefore, we have
mv12 k1 (1 + )
k2 p 2
mv 2 k1 (1 − 2 )
k2 p 2
−
+
=
−
+
2
p
2(1 − 2 )2
2
p
2(1 − 2 )
Finally
1/2
2k1 (1 − )
k2 2 p 2
2
v = v1 −
+
mp
(1 − 2 )2
Angular momentum:
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Physics 1. Mechanics
Problems
J = mr × v. In the point B the velocity is v = (v, 0, 0), so that
mvp
Jz = −mvyB = − √
1 − 2
Problem 10.2. A particle, initially resting in the coordinate origin, suddenly breaks up into three
particles with the masses m1 , m2 , and m3 . The particle m1 has the charge q > 0. It starts moving
into negative x-direction in the homogeneous magnetic field B = (0, 0, B). After having completed
half a circle the particle finds itself at the distance l1 from the starting point. The particle m2 has
the charge −q < 0. It starts moving in the positive x-direction, and after having completed half a
circle is at the distance l2 from the starting point. What is the velocity of the third particle ? The
magnetic force is F = qv × B.
y
m1
m2
x
Solution. For each particle in the magnetic field ma = qv × B. Since v ⊥ B, we can write
m|a| = |q||v||B|, or
mv 2
= |q|B
r
which gives
|q|Br
v=
m
From this expression we immediately find (l = 2r for a semicircle !)
v1 =
qBl1
,
2m1
v2 =
qBl2
2m2
Momentum conservation gives
m1 v1 + m2 v2 + m3 v3 = 0
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Physics 1. Mechanics
Problems
or
v3 = −
m1 v1 + m2 v2
m3
Substituting v1 = v1 eˆx , v2 = −v2 eˆx , we find
v3 =
qB(l2 − l1 )
m2 v2 − m1 v1
eˆx =
eˆx
m3
2m3
Problem 10.3. A particle of the mass m is connected with the point O with the use of a massless
rod of the length L. The rod can freely rotate around O. When the angle between the rod and the
vertical is θ0 the particle velocity is v0 = v1 θ̂ + v2 ϕ̂. a) What is conserved ? b) Find vθ and vϕ (or
θ̇ and ϕ̇) as functions of θ; c) Find θ̈ and ϕ̈ as functions of θ; d) Find the maximum speed |v|max of
the particle; e) Find the tension in the rod as a function of θ.
g↓
θ
O
Solution. Preparations:
F = −mg ẑ + T r̂
v = vθ θ̂ + vϕ ϕ̂
= rθ̇θ̂ + r sin θϕ̇ϕ̂
ṙ = 0
a):
mv 2
+ U,
U = mgz = mgr cos θ
2
m
E = (vθ2 + vϕ2 ) + mgr cos θ
2
m 2 2
= (r θ̇ + r2 sin2 θϕ̇2 ) + mgr cos θ
2
E=
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Physics 1. Mechanics
Problems
Ė = T · v = T r̂ · (vθ θ̂ + vϕ ϕ̂) = 0
m
E = const = (v12 + v22 ) + mgL cos θ0
2
ṗ = F 6= 0
J = r × mv = mrr̂ × (vθ θ̂ + vϕ ϕ̂)
= mrvθ ϕ̂ − mrvϕ θ̂
J˙ = r × F = −rmg r̂ × ẑ
J˙z = 0
Jz = J · ẑ = mr sin θvϕ = mr2 sin2 θϕ̇
Jz = const = mv2 L sin θ0
b):
Jz
mr sin θ
Jz
vϕ
=
ϕ̇ =
2
r sin θ
mr sin2 θ
m 2
(v + v 2 ) + mgr cos θ = E
2 θr ϕ
2E
vθ = ±
− vϕ2 − 2gr cos θ
m
r
2E
J2
=±
− 2 2 z 2 − 2gr cos θ
m rm r sin θ
vθ
1 2E
J2
θ̇ =
=±
− 2 2 z 2 − 2gr cos θ
r
r m
m r sin θ
vϕ =
c):
dθ̇
· θ̇
dθ
!
r
d
1 2E
Jz2
=
±
− 2 2 2 − 2gr cos θ
dθ
r m
m r sin θ
!
r
1 2E
Jz2
· ±
− 2 2 2 − 2gr cos θ
r m
m r sin θ
−1/2
1 2E
Jz2
− 2 2 2 − 2gr cos θ
=±
r m
m r sin θ
2
Jz cos θ
·
+ gr sin θ θ̇
m2 r2 sin3 θ
2
1
Jz cos θ
= 2
+ gr sin θ
r
m2 r2 sin3 θ
θ̈ =
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Physics 1. Mechanics
Problems
dϕ̇
· θ̇
dθ r
Jz2
2Jz cos θ
1 2E
− 2 2 2 − 2gr cos θ
=∓
r m
m r sin θ
mr2 sin3 θ
ϕ̈ =
d):
v2 =
2
(E − mgr cos θ)
m
d 2
v = 2gr sin θ > 0
dθ
v = vmax → θ = θmax > π/2
θ = θmax ⇒ θ̇ = 0 ⇒
2E
J2
− 2 2 z2
− 2gr cos θmax = 0
m
m r sin θmax
e):
mv 2
(T + mg) · r̂ = −
r
mv 2
2E − 2mgr cos θ
T − mg cos θ = −
=−
r
r
2E
T = 3mg cos θ −
r
Problem 10.4.
A particle moves according to r = r0 exp(ϕ) in the central force. Find U (r).
Solution. In the central force energy and angular momentum are conserved. Energy conservation
gives
E = 12 mṙ2 + 21 mr2 ϕ̇2 + U = const.
Angular momentum conservation gives:
mr2 ϕ̇ = J = const.
and therefore
ϕ̇ =
J
.
mr2
From the given relation r = r0 exp(ϕ) we have
ṙ = (d/dt)(r0 exp(ϕ)) = r0 exp(ϕ)ϕ̇ = rϕ̇ =
J
.
mr
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Physics 1. Mechanics
Problems
Substituting (10.4) and (10.4) into the energy conservation (10.4) we finally obtain
E=
J2
+U
mr2
and
U =E−
Problem 10.5.
J2
mr2
A particle moves according to
x = A exp(−γt) cos(ωt)
y = A exp(−γt) sin(ωt)
Find tangential acceleration.
Solution. The problem is solved by direct differentiating: vx = ẋ, vy = ẏ, ax = v̇x , ay = v˙y , so
that the velocity and acceleration are:
vx = ẋ = −γA exp(−γt) cos(ωt) − ωA exp(−γt) sin(ωt) = −γx − ωy,
vy = −γy + ωx,
ax = −γvx − ωvy = (γ 2 − ω 2 )x + 2γωy,
ay = −γvy + ωvx = (γ 2 − ω 2 )y − 2γωx.
Tangential acceleration is parallel to the velocity vector, so that ak = a · v/|v|:
ak = (ax vx + ay vy )/(vx2 + vy2 )1/2
p
= −γ γ 2 + ω 2 A exp(−γt)
Problem 10.6. A particle of the mass m moves in the potential energy U (x) = −Ax2 /2 + Bx3 /3,
A > 0, B > 0. Find the frequency of small oscillations.
Solution. Equilibrium is where Fx = −(dU/dx) = 0:
−
dU
= −Ax + Bx2 = 0 → x1 = 0, x2 = A/B.
dx
Whether the equilibrium is stable or not is determined by the second derivative (d2 U/dx2 ) = −A +
2Bx. If (d2 U/dx2 ) < 0 the point is a maximum and the equilibrium is unstable, if (d2 U/dx2 ) > 0
the point is a minimum and the equilibrium is stable. In x = x1 = 0 we have (d2 U/dx2 ) = −A < 0
so that this is an unstable equilibrium. In x = x2 = A/B we have (d2 U/dx2 ) = A > 0 and this is a
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Physics 1. Mechanics
Problems
stable equilibrium.
Near the equilibrium point the potential energy can be Tayor expanded:
U (x) ≈ U (x2 ) + 21 A(x − x2 )2 .
Let us denote X = x − x2 then vx = ẋ = Ẋ and the energy conservation is written as
1
mẊ 2
2
+ 21 AX 2 = const
which immediately gives (according to the general rule)
ω 2 = A/m.
Problem 10.7.
A hollow cylinder is sliding without friction (no rolling) with the velocity v. The
cylinder comes to a surface with friction. What is the final velocity of the cylinder ?
Solution. When the cylinder comes to the surface with friction it is decelerated by the friction
force and at the same time its rotation is acceletrated until the cylinder begins to roll without sliding.
In the rolling state the friction force is zero and the velocity does not change. Let the friction force
magnitude be Fs . Then the deceleration is
mV̇ = −Fs → V (t) = v −
Fs
t,
m
whereV (t) is the velocity of the center-of-mass in the moment t. The moment of the friction force
(torque) accelerates the rotation around the axis passing through the center-of-mass, as follows:
Fs r
t,
I
I ω̇ = Fs r → ω(t) =
where ω(t) is the angular velocity of rotation and I is the moment of inertia around the axis passing
P
through the center-of-mass. According to the definition I =
mi ri2 , where ri is the distance from
the rotating mass (small part of the body) from the rotation axis. All points of the cylinder are at
the same distance r from the rotation axis, so that we have
I=
X
mi ri2 =
X
mi r 2 = (
i
X
mi )r2 = mr2 .
i
From (10.7) and (10.7) we find
Fs
t.
mr
The velocity and angular velocity stop changing when sliding stops. The condition of the rolling
ω(t) =
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Physics 1. Mechanics
Problems
without sliding is V = ωr which gives
v−
Fs
Fs
Fs
t=
tr = t,
m
mr
m
and therefore, in the moment when the velocity stops changing
Fs
t = v/2.
mr
Substituting into (10.7) we have
Vf inal = v/2.
Problem 10.8.
A rod of the mass M and length l can rotate in the vertical plane around the
axis which passes through the point at a < l/2 from its upper end. A bullet of the mass m M
v
with the horizontally directed velocity v strikes the rod at the upper end and remains in it. What is
m, v
the rotation angle of the rod ?
Solution. Angular momentum is conserved during the collision, which means
Iω = mva → ω = mva/I,
E ! mc2
where ω is the angular velocity of the rotation just after the collision, and I is the momentλof inertia.
After the collision the energy is conserved, that is, the initial kinetic energy goes into the potential
energy as the bar rotates to the angle θ and its center-of-mass rises. Therefore, for the maximum
angle we have
Iω 2 /2 = M gL(1 − cos θ) → cos θµ= 1 − Iω 2 /2M gL.
"m"M
M and center-of-mass of the mbar (we neglect the
where L = l/2 − a is the distanceµ between
the axis
mass of the bullet).
R positive direction downwards.
Calculation of I: Let us choose coordinate origin in the axis, and the
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Physics 1. Mechanics
Problems
Then
Z
l−a
I=
x2 (M/l)dx =
−a
M
[(l − a)3 + a3 ]
3l
Combine all calculations.
Problem 10.9. An electron moving with the energy E mc2 toward the coordinate origin emits
a photon with the wavelength λ forward. What wavelength measures a non-moving observer in the
coordinate origin ?
Solution. We shall use directly the expression for the Doppler effect
1
λ0
=
.
λ
γ(1 − v cos θ/c)
In our case the source is moving towards the receptor, so that θ = 180◦ and
0
λ
=
λ
s
1 − v/c
.
1 + v/c
We need velocity which can be found as follows: γ = E/mc2 and v = c
p
1 − 1/γ 2 .
Problem 10.10.
Potential energy (2D) is given by U = A cos ϕ/ρ2 . A particle of the mass m is
in the point r = (a, a) and its velocity is v ⊥ r. Find the normal acceleration.
Solution. By definition an ⊥ v. Since r ⊥ v we have an k r. Thus, the magnitude an = |a · r̂|.
√
√
a) The long way. r̂ = (ax̂ + aŷ)/ 2a2 = (x̂ + ŷ)/ 2. Express the potential energy in Cartesian
coordinates x = r cos ϕ, y = sin ϕ: U = Ax(x2 + y 2 )−3/2 . Then
∂U
∂U
x̂ −
ŷ
∂x
∂y
A 3Ax2
3Axy
=− 3 − 5
x̂ +
ŷ
r
r
r5
A(3 cos2 ϕ − 1)
3A sin ϕ cos ϕ
=
x̂ +
ŷ.
3
r
r3
F =−
√
Now a = F /m, r = a 2, and ϕ = 45◦ , so that an = A/2ma3 .
b) The short way. F = −(∂U/∂r)r̂−(1/r)(∂U/∂ϕ)ϕ̂, so that an = |(∂U/∂r)|/m = |2A cos ϕ/r3 |/m =
A/2ma3 .
Problem 10.11. A satellite of the mass m is moving on an elliptical orbit round the Earth (mass
M m), so that rmax = 2rmin . The satellite energy E < 0 is known. Find the angular momentum.
Solution. When r = rmin or r = rmax the radial velocity vanishes ṙ = 0, so that in these points
42
Physics 1. Mechanics
Problems
the energy and momentum conservation give
E=
J2
GM m
.
−
2
2mr
r
Thus,
r1,2 =
GM m ±
p
(GM m)2 − 2|E|J 2 /m
2|E|
(here we used E = −|E| < 0) and
p
(GM m)2 − 2|E|J 2 /m
p
= 2.
GM m − (GM m)2 − 2|E|J 2 /m
GM m +
Solving this equation we get
√
2GM m m
p
J=
.
3 |E|
Another way of solving: from the energy expression we have
J 2 /2m = r2 E + GM mr = (2r)2 E + GM m(2r),
so that r = −GM m/3E and
J=
p
2m(r2 E + GM mr)
same as above.
Problem 10.12.
Two disks of the masses m1 and m2 and radii R1 and R2 are connected to a
massless rod which can rotate in the vertical plane. The disk centers are at the distances l1 and l2
from the rotation axis. Find the frequency of small oscillations.
m2 , R2
l2
l1
m1 , R1
R
h
M
r
43
m2 , R2
Physics 1. Mechanics
Problems
l2 moved to the angle θ 1 and the angular velocity is θ̇, the energy is
Solution. If the system
I θ̇2 mgLθ2
I θ̇2
+ mgL(1 − cos θ) =
+
,
E=
2
2
2
where
l1
m1 R12
m2 R22
+ m1 l12 ) + (
+ m2 l22 )
2
2
m1 ,relative
R1
is the moment of inertia
to the axis, and
I=(
L=
m1 l1 − m2 l2
m1 + m2
is the distance of the center of mass from the axis. The frequency is ω 2 = mgL/I.
m
R
M
Problem 10.13. To a disk of the mass M and radius R a smaller disk of the mass mv and radius
h
r
r is connected. At what height a horisontal force should be applied in order that the body start (in
the very first moment) to roll without sliding. There is no friction.
Solution. Let the force be F . The acceleration of the center of mass is a = F/(M + m). The
angular acceleration around the contact point is α = N/I = F h/I, where the moment of inertia is
I = (M R2 /2 + M R2 ) + (mr2 /2 + mr2 ) = (3/2)(M R2 + mr2 ). If there is no sliding, the linear and
angular acceleration are related by a = αL, where L = (M R + mr)/(M + m) is the distance of the
B = (0, B, 0)
x−y
center of mass from the rotation axis (contact point). Thus, we have F/(M + m) = (F h/I)L, so
q>0
m
that h = I/L(M + m).
Problem 10.14.
There is a homogeneous magnetic field B = (0, B, 0) between two 2plates,
parallel to the x − y plane. A particle of the mass m and electric charge q > 0 enters the space
between the planes through the lower plate, when its velocity is v = (v cos 45◦ , 0, v sin 45◦ ). What
is the minimal distance between the plates for which the particle will come back to the lower plate
44
Physics 1. Mechanics
Problems
v = (v cos 45◦ , 0, v
z
v
x
(without touching the upper one) ? Gravity is negligible.
Physics 1. Mechanics
Solutions 2
Solution. The particle moves on a circle of the radius r which is obtained from F = qvB =
mv 2 /r, that is, r = mv/qB, as shown in the figure.
v1 = (0.5c, 0.5c, 0)
z
|vrel |
v2 = (−0.5c, 0.5c,
R
v
M
x
√
From the figure it is clear that the distance between the two planes should be L > r + r/ 2.
√
From the figure it is clear that the distance between the two planes should be L > r + r/ 2.
Problem 1.6. The easiest way is to rotate the coordinates so that the new x axis will be along v1 ,
√
√
then the new y axis will be along v2 . In the new coordinates v1 = (0.5 2c, 0, 0), v2 = (0, 0.5 2c, 0).
Problem
An
observer
on (S
Earth
sees
one galaxy moving with the velocity v1 = (0.5c, 0.5c, 0)
!
Taking v1 ≡ 10.15.
v0 as the velocity
of the
moving frame
), and v
2 as the velocity of a body we are
looking
at,
we
have
and another with v2 = (−0.5c, 0.5c, 0). Find their relative velocity |vrel |.
√
vx − v0
vx! =
−v0 rotate
= −0.5 2c,the coordinates so(11)
Solution. The easiest
way is= to
that the new x axis will be along v1 ,
1 − vx v0 /c2
√
√
√
v
v
0.5
2c
y
y
then the new y axis will
1 = (0.5 2c, 0, 0), v2 = (0, 0.5 2c, 0).
vy! = be along2 v=2 . In
= the .new coordinates v(12)
γ(1 − vx v0 /c )
γ
γ
Taking v1 ≡ v0 as the velocity of the moving frame (S 0 ), and v2 as the velocity of a body we are
where we used vx = (v2 )x = 0, and
looking at, we have
√
γ = (1 − v12 /c2 )−1/2 =
2.
(13)
√
v −v
= −v0 = −0.5 2c,
2
For advanced only !
1 − vx v0 /c
√
0.5
y
y
Problem 1.7.
Let us consider a small part of 0the mass of the v
length
dr and crossvsection
dA at 2c
vy = the density ρ = 3M/4πR
=
=
.
the radius r. This part has a mass dm = ρdAdr, where
γ(1 − vx v0 /c2 ) 3. Is γis attracted toγ
!
3
!
√
√
Thus, we have vx! = −0.5 2c, vy! = 0.5c, and v ! =0 vx! 2 + vy! 2x= 0.5 0
3c.
vx =
the center by the mass M = ρ(4πr /3) inside the radius r. This force is balanced by the pressure
forces. From the inside the pressure pushed it outwards with the force p[r]dA. From the outside the
where
used
vx =
and Thus, we have
2 )force
x =p[r0,
pressurewe
pushes
it inward
with(v
the
+ dr]dA.
p[r]dA − p[r + dr]dA =
GM ! dm
GMrρ
=
dAdr.2
r2
R3
γ = (1 − v1 /c2 )−1/2 =
√ (14)
2.
q
√
√
3
Thus, we have vx0 = −0.5 2c, vy0 = 0.5c, and v 0 = vx0 2 + vy0 2 = 0.5 3c.
Problem 10.16.
Potential energy is given as U = −A/r4 , A > 0. When the particle (mass m)
is in the point (a, 0, 0) its velocity is v = (0, v, 0) and its energy is negative. What is the maximum
45
Physics 1. Mechanics
Problems
distance between the particle and the coordinate origin ?
Solution. Central force means that J = mr ×v = const and E = mvr2 /2+J 2 /2mr2 +U = const.
From the initial conditions we have J = J ẑ, J = mav, E = mv 2 /2 − A/a4 . Since E < 0 we have
v 2 < 2A/ma4 . In the closest and farthest points vr = 0 so that we have
J2
A
− 4 =E
2
2mr
r
where J, m, E < 0, and A are known. The equation is easily solved to give
J2
±
r2 = −
4m|E|
s
(
A
J2 2
) +
4m|E|
|E|
2
dve‘zd
z‘ ‘vn
/2 physical
, y = r0 sin(ωt)
, xwhich
= r0 cos(ωt)
wiwlg
The solution
with. z−=isatnot
(r2 < 0)
means itl
thatrp
there
is no minimum radius: the
particle reaches the maximum radius, bounces back and falls into r .zilnxepd
= 0.
Solution .
Problem 10.17.
A particle moves according to x = r0 cos(ωt), y = r0 sin(ωt), z = kt2 /2. Find
the normal acceleration.
v = ωr0 (− sin(ωt)x̂ + cos(ωt)ŷ) + atẑ
Solution.
2
a = −ω r0 (cos(ωt)x̂ + sin(ωt)ŷ) + aẑ
an = |a × v̂|
! v = ωr0 (− sin(ωt)x̂ + cos(ωt)ŷ) + ktẑ
ω2 a2 t2 + a2 + ω4 r02
= ωr0 a!= −ω 2 r0 (cos(ωt)
x̂ + sin(ωt)ŷ) + k ẑ
ω2 r02 + a2t2
an = |a × v̂|
p
ω 2 k 2 t2 + k 2 + ω 4 r02
p
= ωr0
ω 2 r02 + k 2 t2
3. dl‘y
ly zeiexicz ‘vn . U (x) = A(x − a)2(x − b)2 i’’r dpezp zil‘ivphet dibxp‘
Problem 10.18.
Potential energy
is U xzei)
(x) =lwyn
A(x −ieeiy
a)2 (x
b)2 , Azecepz
> 0. Find the frequency of
.(cg‘n
cil−zephw
small oscillations.
Solution
.
Equilibrium:
dU/dx
= 0=→
x =x a=or
to find
thatthat U = A(b − a)2 X 2
Solution.
Equilibrium:
dU/dx
0→
ax
or−xb.− It
b. isIteasy
is easy
to find
2 2
2
2
U = A(b − a) X where X = x − a or X = x2 − b. Thus, ω2 = 2A(b − a) /m.
where X = x − a or X = x − b. Thus, ω = 2A(b − a) /m.
Problem 10.19.
Three identical cylinders are in equilibrium 4.
(see
figure). Find the minimum
dl‘y
friction
between
cylinders.
mcwn
z‘coefficient
‘va .xeiva
d‘xpythe
itk
lwyn ieeiya mi‘vnd midf mililb dyely
.mililbd oia ilnipind jekigd
Solution. Let the normal force acting on the lower cylinder from the upper is N , and the friction
Solution .
Let the normal force acting on the lower cylinder from the upper is
N , and the friction force between the upper and lower cylinders is fs , fs /N ≤ µ.
From the torque balance on the lower cylinder the friction force between the lower
2
46
Physics 1. Mechanics
Problems
force between the upper and lower cylinders is fs , fs /N ≤ µ. From the torque balance on the lower
cylinder the friction force between the lower cylinder and the table is fs . The force balance on the
lower cylinder in the horisontal direction is
fs + fs sin 30◦ = N sin 60◦ → µ ≥
fs √
= 3/3
N
Problem 10.20.
A body is built of two small identical balls connected with a massless rod of
the length l. Initially the body is at rest. Another identical ball, moving with the velocity v, collides
elastically with one of these two (see figure). Find the angular velocity of the body rotation after
the collision.
v
Solution. Elastic collision means that the momentum, angular momentum, and energy are
N yig‘ ote‘a
(dgepn
zkxrna)
a, b, after
c zecin
daiza6.
conserved.
Let mibltzn
the velocity
of the
third ball
thezlra
collision
be v 0dl‘y
(it will be in the same direction
v zexidn
rpd velocity
dtevd cecni
xtqn)
miwiwlgd
zetitv
as
the initial
or in(gtp
the zcigil
opposite
direction
- in the
last efi‘
case.miwiwlg
v 0 < 0). Let the velocity of the
a rlv jxe‘l
center of the rod be V and the angular velocity of the rotation?around
this center (counterclockwise)
be
ω. Momentum conservation gives
Solution . a! = a/γ, V ! = V /γ → n! = nγ
mv = mv 0 + 2mV
Angular momentum conservation (relative to the collision point):
0 = 2mV (l/2) − 2m(l/2)2 ω
Energy conservation gives
mv 0 2 2mV 2 2m(l/2)2 ω 2
mv 2
=
+
+
2
2
2
2
Solving these equations we obtain ω = v/l.
Problem 10.21.
In a box of the size a × b × c (in its rest frame) N particles are homogeneously
distributed. Find the particle density as viewed by an observer moving with the velocity v along a.
Solution. a0 = a/γ, V 0 = V /γ → n0 = nγ
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