MECHANICS 1 SAMPLE EXAM SOLUTIONS m α

MECHANICS 1 SAMPLE EXAM SOLUTIONS
1. (a) A stone of mass m is thrown with initial speed v0 , so that its initial velocity forms
an angle α with the horizontal. In the same direction, at the point where the stone
was thrown there starts a long hill with constant incline β < α. Assume that the only
force acting on the stone before it hits the ground is gravity.
i. Write down the equation of motion. Solve it to find the position of the stone at
time t.
Equation of motion: m¨
r = −mgj, where j is a unit vector pointing upwards. Or,
simply
x¨ = 0, y¨ = −g.
Solving, with initial conditions:
x(t) = v0 cos α t, y(t) = v0 sin α t − gt2 /2.
ii. Write down a Cartesian equation for the path of the stone.
x
Eliminate t: t = v0 cos
, so
α
y(x) = x tan α −
gx2
.
2v02 cos2 α
iii. Find how far from its starting point will the stone hit the ground on the hill.
On the hill y = x tan β. So, the ball will hit the hill when
x tan α −
I.e.
x=
gx2
= x tan β.
2v02 cos2 α
2v02 cos2 α(tan α − tan β)
= xf all .
g
iv. Suppose, after the stone hits the ground it stops immediately, whereupon it may
start sliding down the hill towards its starting point. The friction coefficient
between the hill and the stone µ. Find the total time it will take the stone to
return to the starting point if µ is small enough.
The time it took to fall on the hill, considering that x-motion has been uniform,
is
xf all
2v0 cos α(tan α − tan β)
tf all =
=
.
v0 cos α
g
Now, when the ball slides back, force along the hill acting on it is the projection
of gravity minus friction, i.e. mg sin β − µmg cos β. Thus, the acceleration on the
way backwards is
a = g(sin β − µ cos β).
The time it takes to travel the distance l =
r
tslide =
xf all
cos β
equals
s
2l
=
a
4v02 cos2 α(tan α − tan β)
.
g cos β(sin β − µ cos β)
The return trip time is now t = tf all + tslide above.
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v. Identify for which value of µ will the stone not slide back.
It won’t if there is too much friction:
sin β − µ cos β ≤ 0, mu ≥ tan β,
see the above formula for a.
a
2b
(b) A particle of mass m moves along the positive x-axis in the potential V (x) = 2 −
x
x
for a, b > 0.
i. Sketch the potential. Here is the picture for the problem.
The critical point is where V 0 (x) = 0, so x = a/b. There the value of V (a/b) =
−b2 /a. As x goes to zero V (x) goes to +∞, as x → +∞, V (x) approaches zero
from below.
ii. Find the equilibrium point and show that it is stable.
It is stable because it is a local minimum. For the future, the second derivative
at the critical point: V 00 (x) = 6a/x4 − 4b/x3 , so
V 00 (a/b) =
2b4
> 0.
a3
For the future denote V 00 (a/b) as k.
iii. Find the period of small oscillations near the equilibrium point. Taylor-expanding
V (x) near the critical point
k(x − ab )2
V (x) = −b /a +
+ ....
2
2
In other words, for small x − ab the potential is like the potential for the spring
with constant k, therefore the equation of motion is the spring-like equation
m
d2
a
a
(x
−
)
=
−k(x
−
) + ...,
dt2
b
b
whose solution is
a
= A cos ωt + B sin ωt,
b
p
p
with ω =
k/m, and the period of small oscillations is T = 2π m
, where
k
2b4
00
k = V (a/b) = a3 .
x−
2
iv. Suppose, the total energy of the particle equals E = − 3b
. Show that this value
4a
of energy corresponds to a finite motion.
We have, by energy conservation,
E=
mv 2
+ V (x).
2
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The kinetic energy is non-negative. Hence, one cannot have E < V (x), i.e x must
be smaller than xM on the figure, in particular it is finite.
v. Find the turning points of the above finite motion and give an integral expression
for the period of the trajectory with the above energy, with the turning points as
limits of integration. Do not attempt to calculate the integral.
The turning points are where the kinetic energy is zero, i.e.
E=−
3b2
a
2b
= V (x) = 2 − .
4a
x
x
This is a quadratic equation for x, solving it gives two roots: xm =
xM = 2a
. These are the turning points.
b
From the equation
r
mx˙ 2
dx
2
+ V (x),
=
(E − V (x)),
E=
2
dt
m
2a
3b
and
the time to get from xm to xM is
Z
xM
q
T /2 =
xm
Z
dx
2
(E
m
=
− V (x))
2a
b
q
2a
3b
dx
2
2
(− 3b
m
4a
−
a
x2
+
2b
)
x
.
It takes the same time to return from xM to xm , so the period is twice the integral.
2. (a) A downward external force of magnitude F acts on a solid ball of mass m suspended
from the lower end of the weightless spring with Hooke’s constant k. There is a
constant gravitational force acting, with acceleration g due to gravity, and no air
resistance.
i. Identify all the forces, acting as the ball and describe their values and direction.
Gravity mg acts downwards; if x is the position of the mass increasing downwards
with x = 0 for the unstretched spring, then the spring force is −kx. In addition
there is F . So
m¨
x = −kx + F + mg
ii. Show that the equation of motion can be written in the form
x¨ + ω 2 x = f,
where x is the deformation of the spring. Express
ω, f via k, m, g, and F .
p
Clearly, from above (divide it by m), ω = k/m, f = g + F/m.
iii. Let the external force be periodic, namely F = F0 cos Ωt with the constant amplitude F0 > 0 and frequency Ω : Ω − ω = ² > 0, where ² is small compared to
Ω.
Denote for the future f0 = Fm0 .
A. Find the the oscillatory particular solution of the above ODE.
The right-hand side f has two terms. The equation
x¨ + ω 2 x = g
, which physically corresponds to the
has a particular solution x0 = ωg2 = mg
k
mass on the spring equilibrium under gravity.
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The equation
x¨ + ω 2 x = f0 cos Ωt
has a particular solution x1 (t) = C cos Ωt, where C is determined after sub0
stitution x(t) = C cos Ωt into the above equation: we have C = ω2f−Ω
2.
So, the particular solution of
x¨ + ω 2 x = g + f0 cos Ωt
is
f0
cos Ωt
ω 2 − Ω2
,
B. Find the trajectory x(t) which satisfies the initial conditions x(0) = mg
k
x(0)
˙
= 0. What is the meaning of the condition x(0) = mg
?.
k
The general solution, or trajectory, is xp (t) above plus the solution of the
homogeneous equation
xp (t) = x0 +
xh (t) = A cos ωt + B sin ωt,
the constants A, B being determined from the initial conditions. So
x(t) = A cos ωt + B sin ωt + x0 +
ω2
f0
cos Ωt.
− Ω2
0
Since x(0) = 0, we have A = − ω2f−Ω
˙
= 0, we have B = 0. So the
2 . Since x(0)
trajectory in question is
x(t) = x0 +
f0
(cos Ωt − cos ωt).
ω 2 − Ω2
The meaning of x0 is once again the equilibrium position, the variable x(t)−x0
satisfies the differential equation without g in the right-hand side.
C. Given that the end of the un-stretched spring was at the height H (much
greater than mg
) above ground, suppose that as the rest of the parameters
k
are the same, ² is driven to zero. Find the value of ² when the oscillating ball
is guaranteed to touch the ground. If this happens for ² = ²0 and does not
happen for ² > ²0 , give an approximate expression for the time t when the
ground be touched for the first time.
HINT: You may use the formula
cos a − cos b = 2 sin
a+b
b−a
sin
.
2
2
Using the above formula, the latter x(t) can be rewritten as
x(t) = x0 +
²
2ω + ²
2f0
sin(
t)
sin
t.
2ω² + ²2
2
2
Approximately, if ² is small,
µ
x(t) ≈ x0 +
¶
f0
²
sin t sin ωt.
2ω²
2
The expression in brackets can be viewed as a slowly changing amplitude
multiplying the oscillating function sin ωt.
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From now on we disregard ”fast” oscillations of sin ωt. relative to the ”slow”
f0
amplitude 2ω²
sin( 2² t). Namely we assume that over the period 2π/ω of sin ωt
the amplitude is constant.
The ball is guaranteed to touch the ground when x(t) ≥ H. Given that H is
much greater than x0 , the latter can also be disregarded, so the condition to
surely touch the ground is
f0
≥ H,
2ω²
²<
2ωH
.
f0
the first touchdown will occur at the time t where sin 2² t reaches its maximum,
i.e when 2² t = π2 , so t = 1/².
(b) A bee sits on the bottom of a small sealed test tube of length l. The test tube is hung
vertically over a table by a thread, so that its bottom is at height l above the table.
The thread is cut, and as the test tube falls (there is no air resistance), the bee flies
up to the tube’s sealed top. Assuming that the mass of the bee equals the mass of
the test tube, find the time it takes for the lower end of the tube to hit the table.
Let x be the position of the mass centre of the empty tube relative to its bottom. Then the
mass centre of the system bee in the tube, relative to the table was at the height m(l+x)+ml
2m
when the tube was hanging; when the tube landed the mass centre is at the height mx+ml
.
2m
So the mass centre p
has traveled the distance l/2 down, with constant acceleration g, and
the time it takes is l/g.
3. (a) Consider two attracting to each other particles with masses m1 and m2 , such that
the force of their interaction depends only on the distance r between the particles, is
directed along the line connecting the particles, and equals in magnitude F (r).
i. Use the Newton laws to show that
A. the relative motion of the particles, in terms of r = r 2 − r 1 can be described
by the equation
r
µ¨
r = −F (r) ,
r
find what µ is;
Let F 12 be the force exerted by particle 2 on particle 1. We have by the
Second and Third laws:
m1 r¨1 = F 12 ,
m2 r¨2 = −F 12 .
Dividing the first equation by m1 , the second by m2 and then subtracting the
first from the second yields
µ
¶
1
1
r¨2 − r¨1 = −
+
F 12 .
m1 m2
m2
This yields the desired result, given that F 12 = F (r) rr , with µ = mm11+m
.
2
B. the mass centre of the system moves with constant velocity, find the trajectory
of the mass centre.
Adding the two above equations rather than subtracting yields
m1 r¨1 + m2 r¨2 = 0,
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or
m1 r 1 + m2 r 2
.
M
So R(t) = R(0) + V t, for some constant velocity V , determined by initial
conditions.
ii. Let (r, θ) be polar coordinates associated with the vector r above. Use the equations of motion in the central force field
¨ = 0, with M = m1 + m2 ,
MR
m(¨
r − rθ˙2 )
R=
= f (r)
˙ =
m(rθ¨ + 2r˙ θ)
0
to show that L = µr2 θ˙ is an integral of motion for the two-particle system in
question. Show that for r 6= 0, the first equation can then be rewritten as
µ¨
r = −F (r) +
L2
.
µr3
The fact that L = µr2 θ˙ is an integral of motion (alias constant of motion) follows
after multiplying the second of the given equations by r, whereupon it becomes
˙ = 0. So L = const. determined by the initial conditions r(0) and
L˙ = dtd (mr2 θ)
˙
θ(0).
L
Expressing then θ˙ = mr
2 and plugging it into the first of the quoted equations
yields the desired
L2
µ¨
r = −F (r) + 3 .
µr
iii. Show that the constant of motion L equals the absolute value of the angular
momentum of a particle with mass µ, and radius-vector r.
Let r = (x, y, z). We know that a particle of mass µ in central field moves in the
(x, y) plane, so the angular momentum is directed collinear with the z axis. For
the z-component of the angular momentum we have Lz = µ(xy˙ − y x).
˙ Besides,
x = r cos θ, y = r sin θ by definition of polar coordinates.
Then
x˙ = r˙ cos θ − rθ˙ sin θ, y˙ = r˙ sin θ + rθ˙ cos θ.
Plugging these polar coordinates expressions into Lz = µ(xy˙ − y x)
˙ yields
2
˙
˙
θ + cos2 θ) = µr2 θ.
Lz = µr2 θ(sin
α
iv. For F (r) = n , with α > 0, show that for any nonzero value of L, there exists
r
a solution where the distance between the two interacting particles is constant.
Find this constant distance and the period of revolution of one particle around
the other.
The equation for r(t) is now
µ¨
r=−
α
L2
+
.
rn µr3
If n 6= 3 the right-hand side is zero, provided that
µ
r = rC =
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L2
µα
1
¶ 3−n
,
well defined for each L 6= 0. So, the initial condition r(0) = rC , r(0)
˙
= 0, together
with the equation µ¨
r = 0 will result in r = rC = const.
The angular frequency of revolution is θ˙ = mrL2 = const, the period is then
C
2
˙ = 2πmrC .
T = 2π/|θ|
|L|
v. Consider now the case n = 3 and show that for any, but one special values of |L|
there are no circular orbits.
If n = 3, the effective force is
µ
¶
1
L2
−α +
.
r3
µ
√
This expression is zero if |L| = αµ and nonzero otherwise. In the latter case it
always retains its sign, positive or negative. So, in the latter case always either
r¨ > 0 or r¨ < 0 which is incompatible with r(t) = const. Yet in the former
√
case |L| = αµ any initial condition with r(0)
˙
= 0 will result in the solution
r = r(0) = const of the equation m¨
r = 0.
(b) A person of mass M is sitting on a chair of mass m, and the coefficient of friction
between their trousers’ material and the chair is µ. The chair can be moved without
friction along the floor (say, the floor is ice). What horizontal force F would suffice
to pull the chair from under the person?
Suppose the man and the chair sustain in motion together. Then they move, driven by
F
the force F , with acceleration a = m+M
. The only horizontal force acting on the man
himself is the static friction force Fsf impelled by the chair, and its value is always
such that
Fsf ≤ µM g.
So for the two to move together we must have µM g ≥ M a, which is sustainable only
if
F
µg ≥
,
M +m
otherwise the man will find himself on the floor.
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