Discussion Class Solutions
Week 14: 28 October – 1 November
HRW 9th ed. Ch 11: Q7, Q9, P41, P43
HRW 9th ed. Ch 16: Q1a, Q2, Q3, Q5, Q8, Q10, P1, P13, P31
Chapter 11
Q7.
(a) Remains the same. The beetle and disk form a closed isolated system and hence must have a
constant total angular momentum
(b) Increases. The beetle is now moving around the centre of the disk faster than it was before,
hence its angular momentum must increase. 𝑣𝑏𝑒𝑓𝑜𝑟𝑒 = 𝜔𝑟 but 𝑣𝑎𝑓𝑡𝑒𝑟 = 𝑤𝑟 + 𝑣𝑤𝑎𝑙𝑘𝑖𝑛𝑔 >
𝑣𝑏𝑒𝑓𝑜𝑟𝑒 and 𝑙 = 𝑟𝑚𝑣 sin 𝜃.
(c) Decreases. Since 𝑙𝑏𝑒𝑒𝑡𝑙𝑒 increases and 𝐿must be a constant then 𝑙𝑑𝑖𝑠𝑘 decreases (𝐿 = 𝑙𝑑𝑖𝑠𝑘 +
𝑙𝑏𝑒𝑒𝑡𝑙𝑒 )
(d) Remains the same. Decreases. Increases. The beetle and disk form a closed isolated system and
hence must have a constant total angular momentum. The beetle is now moving around the centre
of the disk slower than it was before, hence its angular momentum must decrease. 𝑣𝑏𝑒𝑓𝑜𝑟𝑒 = 𝜔𝑟
but 𝑣𝑎𝑓𝑡𝑒𝑟 = 𝑤𝑟 − 𝑣𝑤𝑎𝑙𝑘𝑖𝑛𝑔 > 𝑣𝑏𝑒𝑓𝑜𝑟𝑒 and 𝑙 = 𝑟𝑚𝑣 sin 𝜃. Since 𝑙𝑏𝑒𝑒𝑡𝑙𝑒 decreases and 𝐿must be a
constant then 𝑙𝑑𝑖𝑠𝑘 increases (𝐿 = 𝑙𝑑𝑖𝑠𝑘 + 𝑙𝑏𝑒𝑒𝑡𝑙𝑒 )
Q9.
D, B, A=C. 𝜏 =
𝑑𝐿
𝑑𝑡
thus the steeper the gradient the grater the magnitude of the torque acting on the
wheel.
P41.
(a) For the hoop, we use Table 10-2(h) and the parallel-axis theorem to obtain
I1  I com  mh 2 
1
3
mR 2  mR 2  mR 2 .
2
2
Of the thin bars (in the form of a square), the member along the rotation axis has (approximately) no
rotational inertia about that axis (since it is thin), and the member farthest from it is very much like it
(by being parallel to it) except that it is displaced by a distance h; it has rotational inertia given by the
parallel axis theorem:
I 2  I com  mh2  0  mR 2  mR 2 .
Now the two members of the square perpendicular to the axis have the same rotational inertia (that
is I3 = I4). We find I3 using Table 10-2(e) and the parallel-axis theorem:
1
𝑅 2 1
2
𝐼3 = 𝐼𝑐𝑜𝑚 + 𝑚𝑅 =
𝑚𝑅 + 𝑚 ( ) = 𝑚𝑅 2
12
2
3
2
Therefore, the total rotational inertia is
I1  I 2  I 3  I 4 
19
mR 2  16
. kg  m2 .
6
(b) The angular speed is constant:

 2 

 2.5 rad s.
t 2.5
Thus, L  I total  4.0 kg  m2 s.
P43.
We assume that from the moment of grabbing the stick onward, they maintain rigid postures so that
the system can be analyzed as a symmetrical rigid body with center of mass midway between the
skaters.
(a) The total linear momentum is zero (the skaters have the same mass and equal and opposite
velocities). Thus, their center of mass (the middle of the 3.0 m long stick) remains fixed and
they execute circular motion (of radius r = 1.5 m) about it.
(b) Using Eq. 10-18, their angular velocity (counterclockwise as seen in Fig. 11-47) is

v 1.4 m/s

 0.93 rad/s.
r 1.5 m
(c) Their rotational inertia is that of two particles in circular motion at r = 1.5 m, so Eq. 10-33 yields
I   mr 2 2  50 kg 1.5 m   225 kg  m 2 .
2
Therefore, Eq. 10-34 leads to
K
1 2 1
2
I    225 kg  m 2   0.93rad/s   98 J.
2
2
(d) Angular momentum is conserved in this process. If we label the angular velocity found in part (a)
 i and the rotational inertia of part (b) as Ii, we have
I ii   225 kg  m 2   0.93rad/s   I f  f .
The final rotational inertia is
 mr
2
f
where rf = 0.5 m so I f  25 kg  m2 . Using this value, the
above expression gives  f  8.4 rad s.
(e) We find
Kf 
1
1
2
I f  2f   25 kg  m 2   8.4 rad/s   8.8 102 J.
2
2
(f) We account for the large increase in kinetic energy (part (e) minus part (c)) by noting that the
skaters do a great deal of work (converting their internal energy into mechanical energy) as they pull
themselves closer — “fighting” what appears to them to be large “centrifugal forces” trying to keep
them apart.
Chapter 16
Q1.
(a) 1, 4, 2, 3. 𝑣 =
𝜔
𝑘
where ωis simply the coefficient of t and k the coefficient of x.
Q2.
(a) 2. The greatest valley corresponds to when the two waves would interfere constructively in the
negative y direction. This means that a valley in 1 must overlap will a valley in the left going wave.
The wave 2 has the greatest of all the left going waves meaning that as the two waves overlap there
will be a stage where the valley is larger than any valley produced by any other wave.
(b) 4. Overlapping these two waves completely results in completely destructive interference. This is
not possible with any of the other waves.
(c) 3. This will require partial cancelation of the peaks of each wave with a valley in the other wave.
The only wave in which this overlap would be possible is 3 and the results of other overlaps is either
completely constructive or completely destructive.
Q3.
a and b move upwards while c and d move downwards. As the wave moves to the right, a and b are
approached by a crest (peak) and hence are pushed upwards, whereas c and d are approached by a
valley and move downwards.
e and f are moving downwards, g and h are moving upwards. The direction motion for the particle is
given by the velocity of the particle which is
𝑑𝑦
𝑑𝑡
hence the sign of the slope dictates direction of travel
Q5.
Intermediate (almost destructive). Any shift by a multiple of a full wavelength means that the two
waves would be out of phase by 0 rad and the waves would undergo constructive interference. Any
shift by half a wavelength will mean that they are out of phase by π rad and they would undergo
completely destructive interference. Any shift between these two produces intermediate
interference but since we know that it is almost a half multiple shift then the interference is very
close to destructive interference
Q8
(a) A node. sin 5(0) = sin 0 = 0 hence y’=0 regardless of time.
𝜋
𝜋
(b) An antinode. sin(5(0) + 2 ) = sin( 2 ) = 1 hence 𝑦 ′ (𝑡) = 3 cos 4𝑡.
Q10
(a) 8 nodes. For the nth harmonic there are n+1 nodes as there are n half wavelengths in the given
length
(b) An antinode. If n is even then there exists a node at the midpoint while if n is odd then there
exists an antinode
(c) Longer. Since L is the same and 𝜆 =
2𝐿
𝑛
then the higher n the smaller the wavelength.
(d) Higher. Since 𝑓 = 𝑣/𝜆 and v has remained the same (the same string is used) then f is higher.
P1.
Let y1 = 2.0 mm (corresponding to time t1) and y2 = –2.0 mm (corresponding to time t2). Then we
find
kx + 600t1 +  = sin1(2.0/5.0)
and
kx + 600t2 +  = sin1(–2.0/5.0) .
Subtracting equations gives
600(t1 – t2) = sin1(2.0/5.0) – sin1(–2.0/5.0).
Thus we find t1 – t2 = 0.0014 s (or 1.4 ms).
P13.
Using v = f, we find the length of one cycle of the wave is
 = 320/500 = 0.640 m = 640 mm.
From f = 1/T, we find the time for one cycle of oscillation is T = 1/500 = 2.00  10–3 s = 2.00 ms.
(a) A cycle is equivalent to 2 radians, so that /3 rad corresponds to one-sixth of a cycle. The
corresponding length, therefore, is /6 = 640/6 = 107 mm.
(b) The interval 1.00 ms is half of T and thus corresponds to half of one cycle, or half of 2 rad. Thus,
the phase difference is (1/2)2 =  rad.
P31.
The displacement of the string is given by
y  ym sin(kx  t )  ym sin(kx  t   )  2 ym cos  12   sin  kx  t  12   ,
where  = /2. The amplitude is
A  2 ym cos  12    2 ym cos( / 4)  1.41ym .