Download 1. [10 Marks] A train moving with speed V crosses a platform of

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1.
[10 Marks] A train moving with speed V crosses a platform of length L.
In train's
frame, the back end of the train coincides with the rear end of the platform at noon
(t=0). The engine of the train (front) coincides with the front end of the platform at time
τ. Specify units used clearly (time in meters, or time in seconds)."
1.1. What is the time interval between the two events in the platform frame?"
1.2. Compute the length of the train."
1.3. Compute the corresponding quantities in Newtonian framework. Is it consistent
with the relativistic formulas."
2. [10 Marks] An experimenter inside a train shines a laser torch directly upward. The
speed of the train is c/2. Compute the components of velocity of the laser in the
laboratory frame. What is the speed of light in the laboratory frame? Specify units used
clearly (time in meters, or time in seconds)."
3. (a) [5 Marks] Two identical particles 1 and 2 that are moving with velocity v and −v
respectively collide and stick together to make particle 3. What is the mass of particle 3
in the reference frame in which the centre of mass is at rest? Treat v close to speed of
light. Specify units used clearly (energy in kg, or energy in joule)."
Z
3. (b) [5 marks] A cylinder of mass M, radius R, and height
y
h (see figure) is precessing with an angular velocity Ω along
z
the Z (vertical) axis. The centre of mass of the cylinder is at
the origin. Assume constant θ. Compute the torque acting
on the cylinder. !
4. [10 Marks] A ladder is leaning against a frictionless wall
and the ground, which is also frictionless. The ladder starts to slip downward."
4.1. Obtain an expression for the angular velocity of the ladder as a function of time."
4.2. Show that the top of the plank loses contact with the wall when it is at two-thirds
of its initial height."
5. [10 Marks] A charged ball of mass m, radius R, and charge q is resting on a rough
horizontal slab. The coefficient of friction between the ball and the slab is μ. We apply
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an horizontal electric field E on the ball. The direction of the electric field is parallel to
the slab. Describe the motion of the ball (a) when the ball rolls without slipping; (b)
when the ball rolls and slip."
6. [10 Marks] Two infinitely long line charges (charge per unit length = λ (positive)) are
located at (x=-a,y=0) and (x=a,y=0). Treat the problem as two-dimensional."
6.1. Locate the equilibrium point or points (where force is zero)."
6.2. Work out the stability of the equilibrium point(s) for positive test charge q."
7. A pendulum is hanging from the roof of a car, and the car is accelerating with a uniform
acceleration a along the road."
7.1.
What are the forces acting on the pendulum for an observer who is inside the car? [2]"
7.2.
What is the angle of tilt of the pendulum observed by the experimenter inside the car? [3] "
7.3.
What are the forces acting on the pendulum for an observer who is outside the car? [2] "
7.4.
What is the angle of tilt of the pendulum observed by the outside experimenter? [3]"
8. The Earth is rotating about its vertical axis with an angular velocity Ω."
8.1.
What is the horizontal component of the Coriolis force, FH , felt by a particle of mass m moving with a velocity v at latitude λ (treat positive)? [2] "
8.2.
From this expression, compute the direction of wind circulation around a low pressure
centre in the northern hemisphere of the Earth? Explain it with one or two rough
sketches. [2] "
8.3.
What are the forces acting on a parcel of air (small volume element of air) circulating
around the low pressure region with a radius r in the northern hemisphere ? [3] "
8.4.
From the above answers, set up an equation of motion of a parcel of air (small volume
element of air) circulating around the low pressure region in the northern hemisphere.
From your equation determine the wind velocity (in terms of the pressure gradient) far
from the centre of a cyclone. [2+1]
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Solution: Part B
1. Solution:
The two events in this example are
A: The rear of the train coincides with the left end of the platform.
B: The engine of the train coincides with the front end of the platform.
The coordinates of event A in the lab frame are (x1 = 0, t1 = 0), and in the train
frame are (x10 = 0, t10 = 0). The event B has coordinates (x2 = L, t2 = ?) in the lab
frame, and (x20 = l, t20 = ⌧) in the train frame, where l = rest length of the train.
Hence,
x 0 = l; and
x = L;
t 0 = ⌧.
Using the Lorentz transformation
x = ( x 0 + V t 0 ),
we obtain
L = (l + V⌧)
where
=p
l=
L
1
1 - (V/c)2
. Therefore,
- V⌧.
For the time di↵erence, we employ the transformation
✓
◆
V
t0 =
t- x 2
c
✓
◆
V
which implies that ⌧ =
t - 2 L . Consequently,
c
t=
⌧
+
V
L
c2
In Lab frame, train would cross the platform at time
t.
In Newtonian framework, l + v⌧ = L (see Fig. 20.4) Therefore,
l = L - v⌧.
v
l
l
L
Figure 1: A train of length l travelling wrt a platform of length L with velocity V.
When V ⌧ c,
expected.
! 1, and we recover this limit from the relativistic expression, as
Alternate soluton: In the train reference frame, the platform is shrunk to L/ .
Hence the total distance the the train has to cover in time ⌧ is L/ - l, Hence,
L/ - l = V⌧
which yields
l=
L
- V⌧.
In the platform reference frame, the train is shrunk to l/ , and the train moves a
distance L - l/ in time t. Therefore,
V t=L-
l
=L-
L
2
+
V⌧
which yields
⌧
V
t = + 2 L,
c
which are same as the earlier results.
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2. Vy0 = c
0
y
Vr = 0.5c
r
Vx0 = 0
)
y
x
=
=
0
x
0
x
+
0
y
(1 +
= 0.5
=0
0.5
= 0.5
1 + 0.5.0
p
1 ⇥ 1 - 0.52
=
0
1
x)
p
Vy = 1 - 0.52 c
r
0
r x
1+
Vx = 0.5c
=1
r
=
The net speed m the lab frame =
p
(0.5)2 + (1 - 0.52 )c = c
which is expected since the speed of light is same in all reference frames.
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3. SOLUTION 3(a): In the CM frame, the net momentum is zero. The total energy
is
E= p
2m
1-
2
,
where = v/c. After the collision, the new particle has zero velocity, so the mass of
the particle is (using E2 - p2 = m2 )
2m
m3 = E = p
1-
2
.
SOLUTION 3(b):
The moment of inertia Izz = I3 = (1/2)MR2 and Ixx = I1 = (1/4)MR2 + (1/12)Mh2 .
The angular velocity of the cylinder is
⌦ = ⌦ sin ✓x̂ + ⌦ cos ✓ẑ.
Therefore, the angular momentum of the rigid body about the CM is
L = I1 ⌦ sin ✓x̂ + I3 ⌦ cos ✓ẑ.
The torque acting on the rigid body is
N=
dL
= L⇢ ⇢ˆ˙ .
dt
where
L⇢ = Lz sin ✓ - Ly cos ✓ = (I3 cos ✓ sin ✓ - I1 sin ✓ cos ✓)⌦.
Using ⇢ˆ˙ = ⌦ ˆ , we obtain
N=
dL
= (I3 - I1 )⌦2 sin ✓ cos ✓ ˆ
dt
Alternate solution:
Euler equations:
I1
d!x
+ (I3 - I2 )!y !z = Nx ,
dt
I2
d!y
+ (I1 - I3 )!z !x = Ny ,
dt
I3
d!z
+ (I2 - I1 )!x !y = Nz .
dt
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The angular velocity in the body frame, ! = ! sin ✓ŷ + ! cos ✓ẑ.
Using the Euler equation we deduce
Nx = (I3 - I2 )!2 sin ✓ cos ✓,
Ny = Nz = 0.
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4. Solution:
(a) Let us denote the coordinate of the CM of the ladder as (X, Y). Then
X=
l
2
sin ✓,
Y=
l
2
cos ✓.
The components of the velocity of the CM are
y
N2
l/2
mg
l/2
N1
/2x
Figure 2:
˙
Ẋ = l2✓ cos ✓,
˙
Ẏ = - l2✓ sin ✓
and the components of the acceleration of the CM are
¨
˙2
Ẍ = l2✓ cos ✓ - l✓2 sin ✓,
¨
˙2
Ÿ = - l2✓ sin ✓ - l✓2 cos ✓,
The equations of motion for the CM are
mẌ = N2 ,
mŸ = N1 - mg.
For rotation about the CM,
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I✓¨ = N1 2l sin ✓ - N2 2l cos ✓.
l2 ¨
12 ✓
l¨
6✓
= (g + Ÿ) 2l sin ✓ - Ẍ 2l cos ✓.
¨
= g sin ✓ - 2l ✓.
l✓¨ = 32 g sin ✓.
Thus the ladder accelerates faster and faster as it falls. The solution of the above
equation is similar to an inverted pendulum involving pendulum. The system unstable
like inverted pendulum.
(b) The plank loses contact with the vertical wall when N2 = 0 or Ẍ = 0, which yields
✓¨ cos ✓ = ✓˙ 2 sin ✓.
¨ we obtain
Using the derived result for ✓,
✓˙ 2 =
3g
2l
cos ✓.
Note that the conservation of energy implies that
l
l
mg cos ✓0 = mg cos ✓ + KE
2
2
l⇤
)⇢
m
⇢ g (cos ✓0 - cos ✓) =
2
m
⇢ l2 ˙2
⇢
✓ ,
63
(1)
whose substitution in the above equation yields
cos ✓ = (2/3) cos ✓0 ,
or y(t) = (2/3)y0 .
Note that we could also derive l✓¨ = (3g/2l) sin ✓ by taking a time derivative of the
energy equation (Eq. (1)).
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5. Solution:
The ball experiences a force (qE - f)x̂, where f is the frictional force (see figure xxx).
The frictional force opposes the motion induced by the electric field, so it is in the
opposite direction that that of E. Hence, The equations of motion of the charged ball
wrt the CM of the ball are
ma = qE - f,
(1)
I↵ = fR,
(2)
where a, ↵ are the linear and angular accelerations respectively. Since the forces are
constant, both a and ↵ are constants. For small E, f < µN = µmg, and the ball
rolls without slipping. The ball starts to slip and roll after f = fc = µmg. For rolling
without slipping a = ↵R, which yields
f
qE/m
=
< µg
m
1 + 1/k
(3)
and
a=
f
qE/m
=
,
mk
1+k
(4)
where k = I/(mR2 ) = 2/5. The Eq. (3) gives the condition for transition from rolling
motion to roll-slip, which is
qEc
= µg(1 + k).
m
The ball rolls without slipping for E < Ec , and starts to slip after E crosses Ec . For
the later motion,
(5)
a = qE/m - µg,
and
(6)
R↵ = µg/k.
Note that a > R↵ which makes the ball slip and roll.
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6. Solution:
The two infinitely long line charges are located at (x = -a, y = 0) and (x = a, y = 0).
Naturally they are parallel to the z axis. The motion of the test charge would be in
xy plane. The problem has been solved in the PPTs of energy (slide no. 30).
The equilibrium point, where the force is zero, is (0,0). We compute the potential
energy at a displaced point (x, y).
h p
i
p
q
2 + y2 + log
2 + y2
U(x, y) = - 2⇡✏
log
(x
+
a)
(x
a)
0
= -↵ log (a4 + 2a2 (y2 - x2 ) + x4 + y4 + 2x2 y2 )
where ↵ =
q
4⇡✏0
For small x, y, the potential to second order order is
2
U(x, y) = const + 2↵ x a-y
2
2
Hence, the equations of motion for the test charge are
mẍ = - 4↵
x,
a2
mÿ =
4↵
y.
a2
Since > 0 and q > 0, ↵ > 0. Hence, the test particle would execute oscillations
along x and unstable motion along y.
Some students have solved the problem partially using force method, basically along
x and y directions. This is not a complete solution. One needs to work out stability
in general direction. Partial credit has been given for this.
Also note that the potential method is superior to the force method since potential is
a single function. Potential provides an elegant solution to the problem.
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