Download PY1052 Problem Set 9 – Autumn 2004 Solution

PY1052 Problem Set 9 – Autumn 2004
Solution
(1) A uniform rod with length l = 60 cm and mass M = 1.0 kg rotates about
an axis through one end. As the rod swings through its lowest position, the end
of the rod hits a small wad of putty with a mass of m = 0.20 kg. (a) What is
the moment of inertia of the rod about its rotation axis before it hits the putty?
(b) What is the moment of inertia of the rod+putty? (c) If the angular speed
of the rod just before the collision is 2.4 rad/s, what is the angular speed of the
rod+putty immediately after the collision?
(a) We can use the parallel-axis theorem, which relates the moment of inertia we are looking
for, I, with the moment of inertia about an axis parallel to ours and going through the centre
of mass, Icom . The moment of inertia of a rod about an axis passing through its centre of
1
mass is Icom = 12
M l2 , where M and l are the mass and length of the rod. According to the
parallel-axis theorem (h is the distance between the two axes),
I = Icom + M h2
1
1
=
M l2 + M ( l)2
12
2
1
3
1
2
=
M l + M l2 = M l2
12
12
3
1
2
=
(1.0 kg)(0.60 m) = 0.12 kg m2
3
(b) The moment of inertia of the rod+putty will just be the sum of the two moments of
inertia. Treating the putty like a point located a distance l from the axis of rotation, we thus
have
Itot =
=
=
=
Irod + Iputty
Irod + ml2
0.12 kg m2 + (0.20 kg)(0.60 m)2
0.19 kg m2
(c) The key here is that the angular momentum of the rod+putty is conserved during their
collision. Therefore,
Lbef ore = Laf ter
Irod ωi = Itot ωf
Irod
ωi
ωf =
Itot
0.12 kg m2
(2.4 rad/s)
=
0.19 kg m2
= 1.5 rad/s
(2) (a) What is the magnitude of the gravitational force exerted on the Earth
by the Sun? (b) What is the acceleration of the Earth due to this force? (c) What
is the magnitude of the gravitatonal force exerted on the Sun by the Earth? (d)
What is the acceleration of the Sun due to this force? (e) The Earth orbits the
Sun once per year. What is the linear velocity of the Earth in its orbital motion?
(f ) What is the acceleration of the Earth due to this circular motion?
(a) According to Newton’s shell theorem, a uniform spherical shell of matter attracts an object
that is outside the shell as if all the shell’s mass were concentrated at its centre. Since we can
treat the Sun as a collection of such uniform shells, the Sun will attract the Earth as if all the
Sun’s mass were concentrated at its centre:
GMSun MEarth
r2
(6.67 × 10−11 N m2 /kg2 )(2 × 1030 kg)(5.98 × 1024 kg)
=
(1.5 × 1011 m)2
= 3.54 × 1022 N
F =
(b) The acceleration of the Earth due to this force is
3.54×1022 N
F
= 5.98×10
= 5.92 × 10−3 m/s 2
a = MEarth
24 kg
(c) The gravitational force exerted on the Sun by the Earth is equal and opposite to the
gravitational force exerted on the Earth by the Sun from part (a): 3.54 × 10 22 N.
(d) The acceleration of the Sun due to this force is
22 N
a = MFSun = 3.54×10
= 1.77 × 10−8 m/s 2
2.0×1030 kg
(e) The linear velocity of the Earth as it orbits the Sun is
2π(1.5 × 1011 m)
2πr
=
T
1 year
11
2π(1.5 × 10 m)
=
= 2.99 × 104 m/s
3.156 × 107 s
(f) The centripetal acceleration associated with the Earth’s motion around the Sun is
2
(2.99×104 m/s)2
ac = vr = 1.5 ×1011 m = 5.96 × 10−3 m/s2
Note the similarity with the answer to part (b)! - this is the case because it is the gravitational
force that provides the centripetal force for the Earth’s orbital motion.
v =
(3) A solid sphere of uniform density has a mass of 1.4 × 10 4 kg and a radius
of 1.0 m. What is the magnitude of the gravitational force due to the sphere on
a particle of mass m located at a distance of (a) 1.5 m and (b) 0.5 m from the
center of the sphere?
The key here is Newton’s shell law, which states that “a uniform shell of matter attracts a
particle that is outside the shell as if all the shell’s mass were concentrated at its centre; the
shell exerts no net gravitational force on a particle that is located inside it”. Looking at the
solid sphere as being made up of a whole lot of shells with various radii, from zero out to
1.0 m, this means that only the mass inside the radius where the particle is located, M in exerts
a net gravitational force on the particle.
(a) If the particle is at r = 1.5 m, then the particle is outside the sphere, and the entire mass
of the sphere acts on the particle, as if it were all concentrated at its centre:
GM m
r2
(6.67 × 10−11 N m2 /kg2 )(1.4 × 104 kg)m
=
(1.5 m)2
= 4.15 × 10−7 m N
F =
(b) If the particle is at r = 0.5 m, it is inside the sphere, and only the mass of the sphere
inside the radius of the particle exerts a net gravitational force on the particle:
Min = ρVin
M
Vin
=
Vtot
M 4
= 4 3 πr3
πR 3
3
0.5 m3
r3
4
= 1.75 × 103 kg
= M 3 = 1.4 × 10 kg
3
R
1.0 m
GMin m
r2
(6.67 × 10−11 N m2 /kg 2 )(1.75 × 103 kg)m
=
(0.5 m)2
= 4.67 × 10−7 m N
F =
(4) The mean diameters of Mars and the Earth are DM = 6.9 × 103 km and
DE = 1.3 × 104 km. The mass of Mars is 0.11 times the mass of the Earth. (a)
What is the gravitational acceleration at the surface of Mars? (b) What is the
escape speed on Mars?
(a) The surface gravitational acceleration of a planet is given by the relation (note that
according to Newton’s shell theorem, the planet attracts a particle that is at the surface with
all the planet’s mass, acting as if it is concentrated at the planet’s centre):
GM m
= mg
R2
GM
g =
R2
where M and m are the masses of the planet and particle and R is the radius of the planet.
Thus, we have for the Earth and Mars:
gE =
gM
gE
=
=
=
gM
=
=
GME
GMM
,
gM =
2
2
RE
RM
MM RE 2
ME RM
M M DE 2
M E DM
!2
1.3 × 104 km
0.11
6.9 × 103 km
0.39
0.39gE = 0.39g = 3.82 m/s2
(b) The escape speed on Mars is given by the relation
vesc =
s
2GMM
RM
v
u
u 2(6.67 × 10−11
t
N m2 /kg 2 )[0.11(5.98 × 1024 kg)]
/2
=
6.9 × 106 m
= 5.0 × 103 m/s = 5.0 km/s
This is lower than the escape speed of the Earth, vesc = 11.2 km/s, due to the smaller
gravitational attraction of Mars.
(5) In a certain double-star system, each star has the mass of the Sun (M =
2 × 1030 kg) and they revolve about their centre of mass. The distance between the
two stars is d = 5 × 1011 m. (a) What is their period of revolution (hint: do NOT
use Kepler’s 3rd law!)? (b) What is the angular momentum of the double-star
system due to their orbital motion about each other?
(a) Here, each of the stars moves in a circle with radius r = d/2. This circular motion is due
to the stars’ mutual gravitational attraction – in other words, the centripetal force is provided
by the gravitational force. For each star,
M v2
GM 2
2M v 2
GM 2
=
−→
=
(d/2)
d2
d
d2
v =
=
s
GM
2d
v
u
u (6.67 × 10−11
t
N m2 /kg2 )(2 × 1030 kg)
2(5 × 1011 m)
= 1.15 × 104 m/s
2πr
T
2πr
T =
v
2π(2.5 × 1011 m)
= 1.36 × 108 s = 4.31 years
= =
1.15 × 104 m/s
(b) The angular momentum of each of the stars (treating them like points moving in a circle
with radius r) will be
v =
L = Iω = M r2 ω = M r2
2π
T
= (2.0 × 1030 kg)(2.5 × 1011 m)2
= 5.77 × 1045 kg m2 /s
2π
1.36 × 108 s
The total angular momentum of the two stars will be twice this, 1.15 × 10 46 kg m2 /s.
F
f
a
A
(6) The circular piston of small radius r = 2.00 cm above is used to exert a force
f on the enclosed liquid. The radius of the larger piston is R = 26.0 cm. (a) What
force f must be exerted on the small piston to balance a 20.0 kN force on the
larger piston? (b) If the smaller piston is pushed down a distance of d = 0.850 m,
how much will the larger piston be raised?
(a) Because the pressure is the same everywhere in the fluid, the pressure of the two pistons
must be the same:
f
F
=
a
A
F
f
=
2
πr
πR2
f = F
r2
R2
(2.00 cm)2
(26.0 cm)2
= 0.118 kN = 118 N
= 20.0 kN
(b) The volume pushed down by the smaller piston is equal to the volume pushing up on the
larger piston:
da = DA
dπr2 = DπR2
r2
D = d 2
R
(2.00 cm)2
(26.0 cm)2
= 5.03 × 10−3 m = 5.03 mm
= (0.850 m)
(7) What is the difference in blood pressure between the brain and foot of a
person of height 1.83 m? Where is the pressure higher – in the brain or in the
foot? The density of blood is 1.06 × 103 kg/m3 .
The difference in blood pressure is due to the difference in height, and so is given by the
relation between the pressure P2 at depth h2 and the pressure P1 at depth h1 :
P2 = P1 + ρg(h1 − h2 )
Inserting the density of blood and h1 − h2 = 1.83 m, we find
P2 − P1 = ρg(h1 − h2 ) = (1.06 × 103 kg/m3 )(9.8 m/s2 )(1.83 m) = 1.90 × 104 N/m2
The pressure at the lower depth increases due to the extra fluid that is sitting above that
depth. Thus, the pressure in the feet is higher.
(8) A block of wood floats in fresh water (ρ = 103 kg/m3 ) with two-thirds of
its volume submerged. In oil, the block floats with 0.80 of its volume submerged.
Find the density of the (a) wood and (b) oil.
When the block floats, it feels two forces which are equal and opposite: the downward force of
gravity and the upward bouyant force due to the fluid in which it floats – this buoyant force
is equal to the volume of the fluid displaced by the block. Therefore, when the block is in the
water:
Fg − F b = 0
mwood g − mwater disp g = 0
2
ρwood V g − ρwat ( V )g = 0
3
2
ρwood = ρwat ( ) = 667 kg/m3
3
Now considering the wood floating in the oil:
Fg − F b = 0
mwood g − moil disp g = 0
ρwood V g − ρoil (0.80V )g = 0
1
)
0.80
667 kg/m3
= 834 kg/m3
= =
0.80
ρoil = ρwood (
(1b) Suppose that a straight tunnel is drilled through the Earth as indicated
for a space-age railway system. (a) Assuming that the rails are frictionless and
that air resistance can be neglected, determine the net force on a car of mass
m sitting on the rails when it is a distance x from the mid-point of the tunnel.
(b) Show that the car will execute simple harmonic motion back and forth in the
tunnel, and that the period of this motion is independent of where the tunnel is
drilled. (c) Determine the period of the oscillations.
m
x
r
φ
Re
(a) The net force felt by the car will be the component of the gravitational force it feels toward
the centre of the tunnel, along the rails. By Newton’s shell theorem, only the mass inside
the radius where the car is located will exert a net gravitational force on the car. This total
gravitational force Fg will point toward the centre of the Earth, and the component of this
force toward the centre of the tunnel will be Fg cos φ = Fg (x/r). Therefore,
GMin m
GρVin m
=
2
r
r2
Gm ME 4 3
=
πr
3
r2 43 πRE
3
GME m
r
=
3
RE
x
GME m
= Fg =
x
3
r
RE
Fg =
Fnet
(b) Since G, ME , m and RE are all constants, the force Fnet is proportional to x, the distance
from the centre of the tunnel. Since F ∝ x, we can see that the car will undergo simple
harmonic motion. We haven’t done anything to specify where the tunnel is drilled, so this
will be true for all such tunnels.
(c) We can find the period of the oscillations as follows, using the values R E = 6.37 × 106 m,
ME = 5.98 × 1024 kg:
GME m
x = ma
3
RE
GME
x ≡ ω2x
a =
3
RE
Fnet =
GME
R3
s E
GME
2π
=
ω =
3
RE
T
ω2 =
T = 2π
s
3
RE
= 805 s = 13.4 min
GME
(2b) Estimate the difference in pressure at the bottom of the River Lee at high
and low tide. What assumptions do you have to make to do this? Try to make
them as reasonable as possible.
We can apply the same analysis as in problem (7) above for the River Lee. The difference in
the depth of the river at high and low tide is about 2 m. The difference in the pressure at the
bottom at high and low tide will accordingly be
P2 − P1 = ρg(h1 − h2 )
If we assume that h1 − h2 = 2.0 m and that the density of the river is equal to the density of
water, we obtain P2 − P1 = (1.00 × 103 kg/m3 )(9.8 m/s2 )(2.00 m) = 1.96 × 104 N/m2
(3b) Three children, each with mass m = 36 kg, make a log raft by lashing
together logs of diametre 0.30 m and length 1.80 m. How many logs are needed
for the raft and children to float in fresh water? Take the density of the logs to
be 830 kg/m3 .
Suppose the number of logs needed to float the children and their raft is N . In this case,
the total weight of the children + raft will be equal to the total upward bouyant force. Let’s
suppose that the limiting case is when the water just comes up to the top of the raft (i.e., the
logs are just completely submerged, but still floating). Then,
Fg
3mg + N mlog g
3m + N ρlog Vlog
N (ρwat − ρlog )Vlog
=
=
=
=
Fb
mwat,disp g = N Vlog ρwat g
N Vlog ρwat
3m
3m
N =
(ρwat − ρlog )Vlog
3m
=
(ρwat − ρlog )(πr2 l)
Inserting m = 36 kg, ρwat = 1000 kg/m3 , ρlog = 830 kg/m3 , r = 0.15 m and l = 1.8 m yields
N = 4.99, so at least 5 logs are needed to float the children.