Q11.1.a:
What is the direction of
< 0, 0, 3> x < 0, 4, 0>?
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
1)
2)
3)
4)
5)
6)
7)
+x
–x
+y
–y
+z
–z
zero magnitude
Q11.1.c:
1)
2)
3)
4)
5)
6)
7)
What is the direction of
< 0, 0, 6> x < 0, 0, -3>?
+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?
1)
2)
3)
4)
5)
6)
7)
A
+x
–x
+y
–y
+z
–z
zero magnitude
(z-axis points
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
You can do this with the parallelogram method. Using geometry,
r sin (theta) = r (perp). r (perp) * p = 40 kg m/s gives the magnitude. Using RHR we get the direction to be –z.
(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?
1) same direction as momentum
2) opposite direction as momentum
3) into the page
4) out of the page
3) zero magnitude
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
zero magnitude
(z-axis points
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? 8
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?
1)
2)
3)
4)
3)
same direction as momentum
opposite direction as momentum
into the page
out of the page
zero magnitude
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
zero magnitude
(z-axis points
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
zero magnitude
(z-axis points
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?
1) L increases.
2)
L
decreases.
3)
L
stays the same, but the direction of
4) The direction and magnitude of
L
L
changes.
remain the same.
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
2Md 2
1
Md 2
3) 2
1
Md 2
4) 4
2
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.
2 pi radians per year
or 2 pi radians per pi e 7 seconds
or 2 e -7 radians per second
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
(z-axis points
6) –z
out of page)
7) zero magnitude
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-go-round. For the system of the child + disk
(excluding the axle and the Earth), which statement is true from just before to just after impact?
( K = total kinetic energy, p = total linear momentum, L = total angular momentum about the
axle)
Q11.5.b
What is the initial angular momentum of the child + disk about the
axle?
K , p , and L do not change
2) p and L do not change
3) L does not change
4) K and p do not change
5) K and L do not change
1)
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
Q11.7.d
After the collision, what is the speed (in m/s)
of the child?
(definition of radian measure, theta = arclength/radius so R=s, divide by time to get v = R)
1) < 0, 0, 0 >
2) < 0, –I, 0 >
3) < 0, I, 0 >
4) < 0, 0, –I >
5) < 0, 0, I >
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?
Q11.7.f (11.P.56)
What principle should we use to find the final velocity of the space
station?
Q11.7.g (11.P.56)
What principle should we use to find the final angular speed of the
space station?
1) < 0, 0, 0 >
2) < 0, –Rm, 0 >
3) < 0, Rm, 0 >
4) < 0, –Rm(R), 0 >
5) < 0, Rm(R), 0 >
1) The momentum principle
2) The energy principle
3) The angular momentum principle
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? 2 & 3
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 2
1 e2
mv 
mv 
4 0 r 2
3) 2
4 0 r
1) 2
1 2
1 e2
mv 
4 0 r
2) 2
Q11.10.d: Starting from the idea that the angular momentum of the
electron is quantized, Bohr found the following for the radius of the
circular orbit:
1 2
1 e2
mv 
4 0 r 2
4) 2
5) 0
 N 2 2 


m 
r 
 1 2

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
2) r = (5.0E+23 meter)N
5) r = (1.2E-38 meter)N2
2
3) r = (4.8E-1 meter)N
  1.05E - 34 J  s , m  9E - 31 kg , e  1.6E - 19 C ,
1
4 0
 9E9 N  m 2 / C 2
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  1 2  .

e 
 4 0 