Download LSU Physics Exam 3: Angular Momentum, Gravity, Buoyancy and more Exams Physics in PDF only on Docsity!
Name: KEY SS#:_________________________
Instructor:
Giaime Lehner Sprunger (9:40) Sprunger(8:40) Svoboda(10:40) Svoboda(1:40)
Louisiana State University, Physics 2101, Exam 3
November 9, 2004
- Be sure to write your name, student ID number, and circle your instructor’s name
- For the 17 point problems: you must show all your work and use correct SI units.
State your assumptions, starting equations, and the meaning of any variables; then
write an understandable sequence of steps.
- You may use scientific or graphing calculators, but you must derive and explain
your answer fully on the paper to receive full credit.
- You may detach and use the formula sheet provided at the back of this test. No
other reference materials are allowed.
- You may not answer or use cell phones during the exam. Turn them off now.
- Good Luck!
The gravitational potential energy is given by:
U
grav
GM
Omega
m
Alpha
R
. In this case, R final
>
R initial
, so that U grav, final
is less negative (greater) than U grav,initial
Since KE =-1/2 U grav
, KE final
< KE initial
Kepler’s 3
rd
law states that for a orbiting planet, T
2
∝ R
3
, so as R ↑ then T ↑
1. Question (8 points)
A man, with his arms at his sides, is spinning on a light frictionless turntable. When he
then extends his arms outward …
A. ….. his angular momentum _________.
(a) must increase
(b) must decrease
(c) must remain the same
(d) may increase or decrease depending on his initial angular velocity
B. ….. his rotational inertia _________.
(a) must increase
(b) must decrease
(c) must remain the same
(d) may increase or decrease depending on his initial angular velocity
C. ….. his angular velocity _________.
(a) must increase
(b) must decrease
(c) must remain the same
(d) may increase or decrease depending on his initial angular velocity
2. Question (8 points)
A small moon, Alpha, is in a circular orbit around a large planet, Omega. Omega's
resident mad scientists wish to move their moon further away, as they find ocean tides
inconvenient. They attach gigantic rockets to Alpha and fire them for several years in
appropriate directions. The distance between Alpha's center and Omega's center is
doubled; Omega is moved between one stable circular orbit and another (hint: remember
the sign of the gravitational potential)
A. The rockets _____________the gravitational potential energy of the Alpha-Omega
system (circle one).
(a) increase (b) decrease (c ) leave unchanged
B. The rockets___________ the kinetic energy of Alpha (circle one).
(a) increase (b) decrease (c ) leave unchanged
C. The rockets___________ the period of Alpha about Omega (circle one).
(a) increase (b) decrease (c ) leave unchanged
Frictionless turntable = no net
external torque Angular
momentum is conserved
Rotational inertia is given by:
I
total
" I
body
arm
r
2
, where the 2 is because he
has 2 arms. As he puts his arms out, r goes up and
I
body
goes up (increases).
For a fixed axis of rotation, angular momentum is
given by:
L = I ". Because we have conservation
of angular momentum if I goes up (part B), the
angular velocity, ω , must decrease.
4. Problem (17 points)
Three uniform spheres, with masses m A
= 300 kg, m B
= 700 kg, and m C
= 550 kg, have
( x , y ) coordinates of (0, 3.5 m), (0, 0), and (8 m, - 2 m), respectively.
(a) What is the net gravitational force
r
F
net
on sphere B due to the other spheres? Express
your answer in terms of
!
ˆ
i and
j.
From the figure, we find:
r
BA
= 3 " m
r
BC
2
2
" m
$ = tan
1
Now computing the net force on B:
r
F
net
r
F
BA
r
F
BC
GM
B
M
A
r
BA
2
r
BA
GM
B
M
C
r
BC
2
r
BC
= GM
B
M
A
2
j +
M
C
2
2
cos "
i # sin "
j
7
+ N
i + 10.5 * 10
7
+ N
j
(b) How much would the gravitational potential of this system change if sphere B is
completely removed?
From above, the change in the potential energy is given by:
" U
grav
= U
final
# U
init
= U
AC
[ ]
final
# U
BA
+ U
BC
+ U
AC
[ ]
init
= # U
BA
+ U
BC
[ ]
GM
B
M
A
r
BA
GM
B
M
C
r
BC
6
J
Buoyant force depends ONLY on the mass of the fluid displaced or (ρ fluid
)(V displaced
)
k
64.1$ rad/m ( )
v
wave
"
k
(2.3 # rad/s)
64.1# rad/m ( )
5. Question (8 points)
A lead sphere of mass M and volume V is completely submerged in water
(ρ water
= 1000 kg/m
3
) and a force F 1
is needed to keep it from sinking. Here, both lead
and water can be considered incompressible. (circle one answer for each part)
A. If we moved the sphere to a deeper point in the water, the force F
2
needed to keep it
from sinking is __________ F 1
LARGER THAN SMALLER THAN THE SAME AS
B If, without changing its mass, we are able to increase the volume of the sphere. The
force required to hold it in place would now have to be _____ F 1
LARGER THAN SMALLER THAN THE SAME AS
C. The buoyant force on the sphere is independent of ______ of the sphere.
THE MASS THE VOLUME THE DENSITY OF THE
SURROUNDING LIQUID
D. If we place our original sphere in a tank filled with ethyl alcohol (whose density is
ρ = 800 kg/m
3
). The force required to keep the sphere from sinking must be
_____ F 1
.
LARGER THAN SMALLER THAN THE SAME AS
6. Question (8 points)
A wave traveling along a string is described by:
y ( x , t ) = [0.0042 " m]sin [64.1" rad/m] x # [2.3 " rad/s] t
A. The amplitude of the wave is _____________ m. (circle one):
(a) 0.0021 (b) 0.0042 (c) 0.0980 (d) 28.
B. The wavelength is ____________ m. (circle one):
(a) 0.0042 (b) 27.9 (c) 0.0980 (d) 64.
C. The wave is traveling in the ____________ direction. (circle one):
(a) +x (b) – x (c) cannot be determined
D. The wave speed is _____________ (m/s) (circle one):
(a) 2.3 (b) 27.9 (c) 64.1 (d) 0.
8. Problem (17 points)
(a) Assuming the board is near-horizontal at all times, what is the torque about the hinge
on the board as a function of x due to the spring?
The spring has a restoring force of F
spring
= – kx. Thus, the torque about the hinge
due to the spring:
@ hinge : spring
= r # F
spring
= $ Lkx
(b) If θ is the angle to the horizontal of the board, what is the torque on the board about
the point of the hinge due to the spring in terms of θ?
If θ is the angle to the horizontal of the board, then assuming θ is small we can
use simple arc length, namely
arc length = L " # x. Thus, the torque about the hinge
due to the spring in terms of θ is:
@ hinge : spring
= r # F
spring
$ % Lk L & ( )
= % L
2
k &
(c) After the penguin jumps off the board, what is the angular acceleration of the board in
terms of θ? Assume the board can be treated as a thin rod rotating about the hinge.
From the angular form of Newton’s 2
nd
Law, we know that the net torque is
proportional to the angular acceleration about that point, or :
I " =
net
. Thus, once the
penguin jumps off the only torque is due to the restoring force of the spring. Moreover,
the moment of inertia of the board about the end point (using the Parallel Axis Theorem)
is
I
end
1
3
M
board
L
2
. Thus we can write the angular acceleration of the board about the
hinge as:
@ hinge : spring
I
end
% L
2
k &
1
3
M
board
L
2
3 k
M
board
= % 250 - s
% 2
( )
(d) What is the period of oscillation of the board after the penguin jumps off?
In order that we have SHM, we must have the equation of motion of the form:
2
%. As seen in part (c ), this is indeed the case here. Hence we have:
T =
M
board
3 k
= 0.397 $ s
A penguin stands at the end of a uniform board
that is hinged to the wall at the left and
attached to a spring at the right. The diving
board has a length L = 2.50 m and mass m =
12.0 kg; the spring constant is 1100 N/m. The
board is depressed from the horizontal a small
distance x by the weight of the penguin.
