Physics 8
Spring 2012
NAME:
TA:
Physics 8 Spring 2012
Midterm 2 Solutions
For the midterm, you may use one sheet of notes with whatever you want to put on it, front
and back. Please sit every other seat, and please don’t cheat! If something isn’t clear, please
ask. You may use calculators. All problems are weighted equally. PLEASE BOX YOUR
FINAL ANSWERS! You have the full length of the class. If you attach any additional
scratch work, then make sure that your name is on every sheet of your work. Good luck!
1. While our Sun doesn’t have enough mass to ever become a black hole, suppose that it
could.
(a) What would be the Schwarzschild radius of the Sun?
(b) Using Kepler’s law, determine the orbital period in years of the Earth around this
new black hole.
Hint: for this problem you might find the following information useful:
GN = 6.672 × 10−11 N m2 /kg2 , MSun = 2 × 1030 kg, MEarth = 5.97 × 1024 kg, c =
2.998 × 108 m/s, and the orbital radius of the Earth is 1.5 × 1011 meters.
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Solution
(a) The Schwarzschild radius of the Sun is given by
RS =
2GN MSun
.
c2
Plugging in the values gives
RS =
2GN MSun
2 × 6.672 × 10−11 × 2 × 1030
=
= 2970 meters,
c2
(2.998 × 108 )2
or about 3 kilometers.
(b) Kepler’s law relates the period of the orbit to its radius, r, by
4π 2
r3 .
GN MSun
Since the mass of the Sun didn’t change upon becoming a black hole, the orbital
period is exactly the same as it always was, namely one year. We can check this
just by plugging in the numbers
s
s
2
4π
4π 2 × (1.5 × 1011 )3
T =
r3 =
= 3.16 × 107 seconds,
GN MSun
6.672 × 10−11 × 2 × 1030
T2 =
which is a year!
1
2. Bored, a boy shoots his pellet gun at a piece of cheese that sits on a massive block of
ice. On one particular shot, his 1.2 g pellet gets stuck in the cheese, causing it to slide
25 cm before coming to a stop. If the muzzle velocity of the gun is known to be 65
m/s and the cheese has a mass of 120 g, what is the coefficient of friction between the
cheese and the ice?
————————————————————————————————————
Solution
Let’s start by looking at the momentum. Since the cheese isn’t moving, initially, the
initial momentum is just that of the pellet, pi = ppellet . If the mass of the pellet is m,
and it has an initial velocity v, then pi = mv. After the collision, the pellet is stuck in
the cheese. If the cheese has a mass M , and if the system has a final velocity V , then
the final momentum is pf = (m + M ) V . Solving for the final velocity gives
m
v.
V =
m+M
So, the combined system has an initial kinetic energy of KE = 12 (m + M ) V 2 =
m2
v 2 . After the cheese/pellet system has come to rest, its kinetic energy is zero.
2(m+M )
The energy lost has to go into the work done by friction, which is W = µk FN d =
µk (m + M ) gd, where d is the distance the system slides. So, setting this equal to the
kinetic energy and solving for the coefficient we find
µk =
m2 v 2
.
2 (m + M )2 gd
Plugging in numbers gives
µk =
m2 v 2
0.00122 × 652
=
= 0.085
2 (m + M )2 gd
2 (0.1212)2 × 9.8 × 0.25
2
3. According to the Standard Model of Particle Physics, electrons are pointlike particles
having no spatial extent. (This assumption has been confirmed experimentally, and
the radius of the electron has been shown to be less than 10−18 meters.) The intrinsic
spin of an electron could in principle be due to its rotation. Let us check to see if this
conclusion is feasible.
(a) Assuming that the electron is a uniform sphere whose radius is 1.00 × 10−18 m,
what angular speed would be necessary to produce the observed intrinsic angular
momentum of ~/2? (Hint: Recall that the moment of inertia of a solid sphere is
I = 52 M R2 , ~ = 1.05 × 10−34 J s, and that the mass of an electron is 9.11 × 10−31
kg.)
(b) Using this value of the angular speed, show that the speed of a point on the
“equator” of a “spinning” electron would be moving faster than the speed of light
(which Einstein tells us is impossible). What is your conclusion about the spin
angular momentum being analogous to a spinning sphere with spatial extent?
————————————————————————————————————
Solution
(a) The angular momentum can be expressed in terms of the moment of inertia, I,
and the angular velocity, ω, as L = Iω. Thus, the angular speed is ω = L/I.
Now, for a uniform sphere, I = 25 M R2 , and for L = ~/2, we find
ω=
5~
L
=
.
I
4M R2
Taking M = me = 9.11 × 10−31 kg, we have
ω=
5 × 1.05 × 10−34
5~
=
= 1.44 × 1032 rad/sec.
4M R2
4 × 9.11 × 10−31 × (10−18 )2
(b) The speed is v = Rω, so using our results from part (a) gives
v = Rω = 10−18 × 1.44 × 1032 = 1.44 × 1014 m/s!
This is much faster than light. So, if we require that these “points” on the “equator”
can’t spin faster than light, we have to abandon the model of the electron as a tiny
little spinning sphere and think of spin in a different way.
3
4. A basketball rolls without slipping down an incline of angle θ. The coefficient of static
friction is µs . Model the ball as a thin spherical shell. Find
(a) the acceleration of the center of mass of the ball,
(b) the frictional force acting on the ball from the torque, and
(c) the maximum angle of the incline for which the ball will roll without slipping (this
Picture
the (b)
Problem
The Fthree
happens when the friction
from part
just equals
f = µforces
s FN ). acting on the basketball are
the ball, the normal force, and the force of friction. Because the
assumed to be acting at the center of mass, and the normal force a
————————————————————————————————————
center of mass, the only force which exerts a torque about the ce
the frictional
force. Let the mass of the basketball be m and apply
Solution
law to find a system of simultaneous equations that we can solve fo
(a) We can begin by writing down Newton’s laws for
called for in the problem statement.
Hint: The moment of inertia of a spherical shell is I = 23 mR2 .
the ball, including the torques:
P
P Fx = −mg sin θ + Ff = −ma
0
PFy = −mg cos θ + Fn =
τi =
Ff R
= Iα.
y
Furthermore, since the ball is rolling without slipping, we have that a = αR. So, we have enough
to solve this system of equations. Plugging in
Ff = Iα/R = Ia/R2 to the first equation and
solving for a gives
m
x
r
Fn
r
0
r
mg θ
mg sin θ
.
m + I/R2
(a) Apply Newton’s 2nd law in2both2
The moment of inertia of the spherical shell is just I = 3 mR , and so
a=
translational and rotational form to
mg sin θ
3
thea ball:
=
= g sin θ.
2
m + 2m/3R
(b) Since Ff =
I
a
R2
=
2
mR2
3
R2
5
a = 23 ma, we have
2
mg sin θ. is rolling
BecauseFthe
f =basketball
5
without slipping we know that:
r
f
∑F
∑F
x
= mg sin θ − f s = m
y
= Fn − mg cosθ =
and
∑τ
α=
= f s r = I 0α
0
a
r
(c) Now, we know that the frictional force is Ff = µs FN = µs mg cos θ. Setting this
equal to our result from part (b) we find
a
Substitute in equation (3) to
fs r = I 0
2
5
mg
sin θ = µs mg cos θ ⇒ tan θ = µs .
r
obtain:
5
2
So, the maximum angle of the incline that the ball will roll without slipping is 2
From Table 9-1 we
I 0 = 23 mr
have:
5
θ = tan−1
µs .
2
Substitute for I0 and α in equation
(4) and solve for fs:
The rolling of the ball increases the angle from the sliding case.
Substitute4for fs in equation (1)
and solve for a:
fs r =
a=
(
2
3
3
5
g sin θ
mr 2
) ar ⇒ f
s
=
2
3
Extra Credit Question!!
The following is worth 10 extra credit points!
Some people think that the shuttle astronauts are “weightless” because they are “beyond the pull of Earth’s gravity.” In fact, this is completely untrue.
(a) What is the magnitude of the gravitational field in the vicinity of a shuttle orbit?
A shuttle orbit is about 400 km above the ground.
(b) Given the answer in Part (a), explain why shuttle astronauts suffer from adverse biological effects such as muscle atrophy even though the are not actually
“weightless.”
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Solution
(a) The shuttle orbits at a distance of r = RE + h from the center of the Earth, where
RE is the radius of the Earth, and h is the height above the surface. If the height
is h = 400 km, then r = 6400 + 400 = 6800 km, or 6.8 × 106 meters. At this
point, the gravitational field has a magnitude
6.67 × 10−11 × 5.98 × 1024
GME
=
= 8.6 m/s2 ,
g=
2
6
2
r
(6.8 × 10 )
which is still almost 90% of the acceleration at the surface of the Earth.
(b) Remember that the weight that we feel is due to the normal force. The astronauts
in orbit are in constant free-fall, and don’t feel their weight. So, without the
compensating normal force to fight against the muscles begin to weaken, not
needing to do as much anymore.
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