Q11.1.a:
What is the direction of
< 0, 0, 3> x < 0, 4, 0>?
1)
2)
3)
4)
5)
6)
7)
+x
–x
+y
–y
+z
–z
zero magnitude
Q11.1.b:
What is the direction of
< 0, 4, 0> x < 0, 0, 3>?
1)
2)
3)
4)
5)
6)
7)
+x
–x
+y
–y
+z
–z
zero magnitude
Q11.1.c:
What is the direction of
< 0, 0, 6> x < 0, 0, -3>?
1)
2)
3)
4)
5)
6)
7)
+x
–x
+y
–y
+z
–z
zero magnitude
Q11.1.d: A ball falls straight down in the xy plane. Its momentum is shown by the
red arrow. What is the direction of the ball's angular momentum about location A?
A
1)
2)
3)
4)
5)
6)
7)
+x
–x
+y
–y
+z
–z
(z-axis points
zero magnitude out of page)
Q11.1.e: A ball falls straight down in the xy plane. Its momentum is shown by the
red arrow. What is the ball's z component of angular momentum about location A?
10 kg m/s
A
4m
1)
2)
3)
4)
5)
+10
–10
+40
–40
0
(z-axis points
out of page)
Q11.1.f: A planet orbits a star, in a circular orbit in the xy plane. Its momentum is
shown by the red arrow.
What is the direction of the angular
momentum of the planet?
Q11.1.g: A planet orbits a star, in a circular orbit in the xy plane. Its momentum is
shown by the red arrow.
1)
2)
3)
4)
5)
6)
7)
+x
–x
+y
–y
+z
–z
(z-axis points
zero magnitude out of page)
Q11.1.h: A comet orbits the Sun, in an elliptical orbit in the xy plane. The red arrow
indicates its momentum. Which arrow best shows the direction of the vector
where A is at the center of the Sun?
Q11.1.i: A comet orbits the Sun, in the xy plane. Its momentum is shown by the red
arrow. What is the direction of the comet's angular momentum about the Sun?
Q11.1.j: A comet orbits the Sun, in the xy plane. Its momentum is shown by the red
arrow. What is the direction of the comet's angular momentum about the Sun?
1)
2)
3)
4)
5)
6)
7)
+x
–x
+y
–y
+z
–z
(z-axis points
zero magnitude out of page)
Q11.1.k: A comet orbits the Sun, in the xy plane. Its momentum is shown by the red
arrow. What is the direction of the comet's angular momentum about the Sun?
1)
2)
3)
4)
5)
6)
7)
+x
–x
+y
–y
+z
–z
(z-axis points
zero magnitude out of page)
Q11.1.l: A comet orbits the Sun in the xz plane. Its momentum is shown by the red
arrow. What is the direction of the comet's angular momentum about the Sun?
1)
2)
3)
4)
5)
6)
7)
+x
–x
+y
–y
+z
–z
zero magnitude
Q11.1.m: If an object is traveling at a constant speed in a vertical circle, how does
the object's angular momentum change as the object goes from the top of the circle
to the bottom of the circle?
Q11.2.a: A diatomic molecule such as molecular nitrogen (N2) consists of two
atoms each of mass M, whose nuclei are a distance d apart. What is the moment of
inertia of the molecule about its center of mass?
1) Md 2
2) 2Md 2
1
3) Md 2
2
1
4) Md 2
4
5) 4Md 2
d
Q11.2.b: The spokes of a bicycle wheel have low mass, so almost all of the mass of
the wheel is concentrated in the rim. What is the moment of inertia of a bicycle
wheel of radius R and mass M?
1) MR2
2) 2  MR2
3) 2  RM
4) (1/2) MR2
5)  MR2
Q11.2.c: The Earth rotates on its axis once every 24 hours. What is its angular
speed? Radius: 6.4e6 m Mass: 6e24 kg
1)  = 2  / (24*60*60)
2)  = 2  * 6.4e6 / (24*60*60)
3)  = (6e24) * 2  * 6.4e6 / (24*60*60)
4)  = (6e25) * (6.4e6)2 * 2  / (24*60*60)
Q11.2.d: The Earth orbits the Sun in a nearly circular orbit. What is its angular
speed?
Click any button when you have
calculated the answer.
Q11.4.a: A yo-yo is in the xy plane. You pull up on the string with a force of
magnitude 0.6 N. What is the direction of the torque you exert on the yo-yo?
r = 0.005 m, R= 0.035 m
1) +x
2) –x
3) +y
4) –y
5) +z
6) –z
(z-axis points
7) zero magnitude out of page)
Q11.4.b: A yo-yo is in the xy plane. You pull up on the string with a force of
magnitude 0.6 N. What is the magnitude of the torque you exert?
1) 0.005 N· m
r = 0.005 m, R= 0.035 m
2) 0.003 N· m
3) 0.021 N· m
4) 0.035 N· m
5) 0.6 N· m
6) cannot be determined without
knowing the length of the string
Q11.5.a
Child runs and jumps on playground merry-goround. For the system of the child + disk
(excluding the axle and the Earth), which
statement is true from just before to just after
impact?
Q11.5.b
What is the initial angular momentum
of the child + disk about the axle?
1) < 0, 0, 0 >
2) < 0, –Rmv, 0 >
3) < 0, Rmv, 0 >
4) < 0, 0, –Rmv >
5) < 0, 0, Rmv >
Q11.7.c
The disk has moment of inertia I, and after the
collision it is rotating with angular speed . The
rotational angular momentum of the disk alone (not
counting the child) is
1) < 0, 0, 0 >
2) < 0, –I, 0 >
3) < 0, I, 0 >
4) < 0, 0, –I >
5) < 0, 0, I >
Q11.7.d
After the collision, what is the speed (in m/s)
of the child?
1) R
2) 
3) R2
4) R
5) 2R
Q11.7.e
After the collision, what is the translational angular
momentum of the child about the axle?
1) < 0, 0, 0 >
2) < 0, –Rm, 0 >
3) < 0, Rm, 0 >
4) < 0, –Rm(R), 0 >
5) < 0, Rm(R), 0 >
Q11.7.f (11.P.56)
What principle should we use to find 1) The momentum principle
the final velocity of the space station? 2) The energy principle
3) The angular momentum principle
Q11.7.g (11.P.56)
What principle should we use to find
the final angular speed of the space
station?
1) The momentum principle
2) The energy principle
3) The angular momentum principle
Q11.10.a
In the original Bohr model of the hydrogen atom, the electron moves in circular
orbits around the proton. Apply the Momentum Principle to this model. Which of
these equations is the result? )
Q11.10.b
For the bound states of the hydrogen
atom, which statement is true?
1) K+U is positive
2) K+U is negative
3) K+U is zero
Q11.10.c: Which is the correct expression for K+U for the hydrogen atom?
e = +1.6 e -19 C
1 2
1 e2
1) mv 
2
4 0 r
1 2
1 e2
2) mv 
2
4 0 r
1 2
1 e2
3) mv 
2
4 0 r 2
1 2
1 e2
4) mv 
2
4 0 r 2
5) 0
Q11.10.d: Starting from the idea that the
 N 2 2 


angular momentum of the electron is
m 

quantized, Bohr found the following for
r
 1 2
the radius of the circular orbit:

e 
 4 0 
What does this predict for the numerical value of r ? (Leave N2 as a factor.)
1) r = (8.5E-30 meter)N2
4) r = (5.3E-11 meter)N2
2) r = (5.0E+23 meter)N2
5) r = (1.2E-38 meter)N2
3) r = (4.8E-1 meter)N2
  1.05E - 34 J  s , m  9E - 31 kg , e  1.6E - 19 C ,
1
 9E9 N  m 2 / C 2
4 0
Notes on answers
1) r = (8.5E-30 meter)N2 Didn’t square the electron charge e.
2) r = (5.0E+23 meter)N2 Didn’t square  .
3) r = (4.8E-1 meter)N2 Didn’t divide by 1 .
4 0
4) r = (5.3E-11 meter)N2 Correct. Useful to write as r  0.53E - 10 meter N 2 .
5) r = (1.2E-38 meter)N2 Didn’t divide by 

e 2  .
 4 0 
1