PHY–302 K. Solutions for problem set #9.
Textbook problem 7.10:
For linear motion at constant acceleration a, average velocity during some time interval from
t1 to t2 is the average of the velocities v(t1 ) and v(t2 ) at the beginning and at the end of the
interval, vavg = 12 (v(t1 ) + v(t2)). Likewise, for rotation at constant angular acceleration α, the
average angular velocity during some time interval is
ωavg [between t1 and t2 ] =
ω(t1 ) + ω(t2 )
.
2
(S.1)
Hence, the angular displacement during this interval is
∆ϕ = ωavg × ∆t =
ω(t1 ) + ω(t2 )
× (t2 − t1 ).
2
(S.2)
The washer’s tub in question spins up from zero to ω = 2πf = 2π × 5.0 radians per second
in ∆t1 = 8.0 s, so during this time it rotates through angle
∆ϕ1 =
0, + (2π × 5.0 rad/s)
× 8.0 s = 2π × 20 rad,
2
(S.3)
or 20 complete revolutions. After this spin-up, the tub decelerates from ω = 2π × 5.0 rad/s
to zero in δt2 = 12.0 s, and during this deceleration, the tub rotates through further
∆ϕ2 =
(2π × 5.0 rad/s) + 0
× 12.0 s = 2π × 30 rad,
2
or 30 complete revolutions.
Altogether, the tub rotates through 20 + 30 = 50 compete revolutions.
1
(S.4)
Textbook problem 8.2:
~ acting on it. The point B where the
Figure P8.2 shows the tooth in question and the force F
force is acting is at distance r = 1.20 cm from the pivot point A at the root of the tooth.
~ has magnitude F = 80.0 N and direction θ = 180◦ − 48◦ = 132◦ from the radius
The force F
vector ~r from A to B. Hence, the force has lever arm
ℓ = r × sin θ = 1.20 cm × sin 132◦ = 0.892 cm
(S.5)
τ = F × ℓ = 80.0 N × 0.892 cm = 71.3 N · cm = 0.713 N · m.
(S.6)
and the torque
Modified textbook problem 8.7:
The arm on figure P8.7 is in static equilibrium, so the net force and the net torque on the
arm must vanish,
X
Fx = 0,
X
Fy = 0,
X
τ = 0.
(S.7)
~ g = m~g, the tendon pull F
~ t , and the
There are 3 forces acting on the arm: the weight F
~ s at the shoulder joint. We know the direction and magnitude mg = 50.0 N (note
force F
modification) of the weight force, the direction — φ = 12◦ above the humerus bone — but
~ t , and neither direction θ nor magnitude of the force
not the magnitude of the tendon force F
~ s at the joint.
F
To solve the problem one unknown at a time, let start with the torque equation
P
τ = 0.
We may use any pivot point we like to calculate the torques (as long as it’s the same point
~ s — regardless
for all the forces), so let’s use shoulder joint O. With this choice, the force F
of its magnitude or direction — has no lever arm and no torque, hence the torque equation
becomes
X
~ t ) + τ (F
~ g ) = 0.
τ = 0 + τ (F
(S.8)
~ t acts at the point where the tendon is attached to the humerus bone.
The tendon pull F
This point is not labeled on the figure, but we know that it lies at distance rt = 8.0 cm from
2
~ t has lever arm
the shoulder joint O. Consequently, F
ℓt = rt × sin φ = 8.0 cm × sin 12◦ = 1.66 cm
(S.9)
~ t ) = Ft × ℓt .
and upward torque τ (F
~ g is distributed all over the arm, but for the purpose of calculating
The force of gravity F
its torque, we may treat it as acting at the arm’s center of gravity (point A on the figure).
This center of gravity lies at distance rg = 29.0 cm from the pivot point O, so assuming the
arm is horizontal (it looks that way on the figure, and we are not told otherwise), the gravity
~ g ) = −mg × ℓg .
force has lever arm ℓg = rg = 29.0 cm and downward torque τ (F
Altogether, the net torque about the shoulder joint is
X
τ = 0 + Ft × ℓt − mg × ℓg .
(S.10)
Demanding that this net torque should vanish, we find that the tendon pull must be
Ft =
mg × ℓg
50.0 N × 29.0 cm
=
= 872 N.
ℓt
1.66 cm
(S.11)
~ t of the tendon, we may find the force at the shoulder from
Now that we know the force F
the force balance equations
X
X
Fx = Fs × cos θ − Ft × cos φ = 0,
Fy = −Fs × sin θ + Ft × sin φ − Fg = 0.
(S.12)
Solving these equations, we find
Fs × cos θ = Ft × cos φ = 877 N × cos 12◦ = 853 N,
Fs × sin θ = Ft × sin φ − Fg = 877 N × sin 12◦ − 50.0 N = 131 N,
q
Fs =
(853 N)2 + (131 N)2 = 863 N,
Fs × sin θ
131 N
=
= 0.154,
Fs × cos θ
853 N
θ = arctan 0.154 = 8.75◦ .
tan θ =
3
(S.13)
Modified textbook problem 8.29:
A system of several point-like particles has net moment of inertia
I =
X
i
mi × ri2
(S.14)
where ri is the distance between particle #i and the axis of rotation.
• For rotation around the x axis, ri2 = yi2 + zi2 .
• For rotation around the y axis, ri2 = x2i + zi2 .
• For rotation around the z axis, ri2 = x2i + yi2.
The rectangle in question has width 6 m, height 8 m (note modification) and no depth. The
rectangle is symmetric with respected to all 3 coordinate axes, so the 4 particles at its vertices
are located at
(x, y, z)1,2,3,4 = (±3 m, ±′ 4 m, 0).
(S.15)
p
Consequently, each of the four particles is at distance rx = y 2 + z 2 = 4 m from the x axis,
p
√
at distance ry = x2 + z 2 = 3 m from the y axis, and at distance rz = x2 + y 2 = 5 m from
the z axis. Therefore:
(a) Moment of inertia around the x axis is
Ix =
X
i
mi (yi2 + zi2 ) = (y 2 + z 2 ) ×
X
mi
i
2
= (4 m) × (3 kg + 2 kg + 2 kg + 4 kg = 11 kg)
(S.16)
= 176 kg · m2 .
(b) Moment of inertia around the y axis is
Iy =
X
i
mi (x2i + zi2 ) = (x2 + z 2 ) ×
2
X
mi
i
= (3 m) × (3 kg + 2 kg + 2 kg + 4 kg = 11 kg)
= 99 kg · m2 .
4
(S.17)
(c) Moment of inertia around the z axis is
Iz =
X
i
mi (x2i + yi2 ) = (x2 + y 2 ) ×
X
mi
i
2
= (5 m) × (3 kg + 2 kg + 2 kg + 4 kg = 11 kg)
(S.18)
= 275 kg · m2 .
Textbook problem 8.32:
The potter’s wheel decelerates from f = 50 RPM (i.e.,
50
60
= 0.833 Hz) to stop in ∆t = 6.0 s
at constant rate
α =
∆ω
0 − 2πf
=
= −0.87 rad/s2 .
∆t
∆t
(S.19)
This angular acceleration requires net torque
τ net = Iα = 12 kg · m2 × −0.87 rad/s2 = −10.5 N · m.
(S.20)
This net torque is due to all forces acting on the wheel. Altogether, there are 4 forces: its
weight W , the force Fa from the axis of the wheel, the normal force N from the wet rag in the
potter’s hand, and the kinetic friction force f from the same rag. However, Fa and W act at the
axis of rotation, so they have zero lever arms and hence zero torques. (I assume a symmetric
wheel so its center of gravity is on the axis.) The normal force N acts at the rim of the wheel
but in a radial direction, so the line of this force goes through the axis. Consequently, N also
has zero lever arm and zero torque. Only the friction force does contribute to the torque, thus
τ net = τ (f )
(S.21)
Similar to N, the friction force f acts at the rim of the wheel but in a tangential rather
than the radial direction. Consequently, its lever arm is equal to the wheel’s outer radius,
ℓ = R = 0.50 m, and the torque is
τ (f ) = −R × f
(S.22)
where the ‘−’ sign indicates the direction of this torque being against the wheel’s rotation.
5
Hence, in light of eqs. (S.20) and (S.21), we must have
−R × f = τ net = −10.5 N · m
(S.23)
−τ net
+10.5 N · m
f =
=
= 21 N.
R
0.50 m
(S.24)
and
Finally, the kinetic friction coefficient µk obtains as the ratio of this friction force and the
normal force N = 70 N, thus
µk =
21 N
f
=
= 0.30.
N
70 N
(S.25)
Textbook problem 8.34:
Approximating the bicycle wheel as a hoop — all the mass is at the rim of the wheel — we
obtain its moment of inertia as
I = MR2 = 1.80 kg × (0.320 m)2 = 0.184 kg · m2 .
(S.26)
(Note the diameter 64.0 cm is 2R.) To give this wheel angular acceleration α = 4.50 rad/s2 ,
we need net torque
τ net = Iα = (0.182 kg · m2 ) × (4.50 rad/s2 ) = 0.829 N · m.
(S.27)
Now consider the forces acting at the wheel and their torques. The wheel’s weight W = mg
acts at the wheel’s center of gravity; assuming the wheel is balanced, the center of gravity lies
on the wheel’s axis, so W has zero lever arm and zero torque. Likewise, the force Fa from the
axis of the wheel (this force is needed to keep the wheel from moving linearly) has zero lever
arm and zero torque. The resistive force Fr = 120 N acts at the rim of the tire in a tangential
6
direction, so its lever arm is ℓr = R = 0.32 m and the torque is
τ (Fr ) = −Fr × R = −120 N × 0.32 m = −38.4 N · m.
(S.28)
Finally, the force Fc of the bicycle’s chain acts at the sprocket. The direction of this force is
tangent to the sprocket, so the lever arm is equal to the sprocket’s radius rs and the torque is
τ (Fc ) = +Fc × rs .
(S.29)
Altogether, the net torque on the wheel is
τ net = τ (W ) + τ (Fa ) + τ (Fr ) + τ (Fc ) = 0 + 0 − 38.4 N · m + Fc × rs .
(S.30)
In light of eq. (S.27), this means that we need
Fc × rc = 38.4 N · m + 0.83 N · m = 39.2 N · m.
(S.31)
(a) For the sprocket radius rs = 4.50 cm, this formula requires chain force
Fc =
39.2 N · m
= 872 N = 196 lb.
0.045 m
(S.32)
(b) For the sprocket radius rs = 2.8 cm, we need a stronger chain force
Fc =
39.2 N · m
= 1401 N = 315 lb.
0.028 m
(S.33)
Sanity check: The chain forces we have just computed are way to strong for a bicycle. I’ve
double checked the above calculations and found no mistakes, so there has to be an error in
the problem’s data. Since we ended up with τ (Fr ) ≫ τ net , which is suspicious by itself, the
7
resistive force Fr = 120 N ought to be wrong. Most likely, it is a typo for Ff = 12.0 N. In
this case,
τ (Fr ) = −Fr × R = −12.0 N × 0.32 m = −3.84 N · m
(S.34)
Fc × rs = τ (Fc ) = +3.84 N · m + 0.83 N · m = 4.67 N · m.
(S.35)
and
Consequently, (a) for the sprocket radius rs = 4.5 cm, the chain force is Fc = 104 N (or 23.4
pounds), and (b) for the sprocket radius rs = 2.8 cm, the chain force is Fc = 167 N (37.5
pounds).
PS: Students who used the problem data as it’s written in the textbook and correctly calculated
the chain forces for this situation will get full credit. Students who calculated those forces
and then realized that they are way to strong for a bicycle will get extra credit.
Textbook problem 8.40:
Treating the spool as a solid disk of radius R = 0.600 m, mass M = 5.00 kg, and uniform
density and thickness, we have its moment of inertia as
I =
2
1
2 MR .
(S.36)
The kinetic energy of the rotating spool is
Kspool =
1
2I
× ω2 =
2 2
1
4 MR ω ,
(S.37)
so the net kinetic energy of the spool and the bucket (of mass m = 3.00 kg) is
Knet = Kspool + Kbucket =
2 2
1
4 MR ω
+
2
1
2 mv .
(S.38)
If the string on which the bucket is suspended does not slip off the spool, the bucket’s linear
8
displacement is related to the spool’s angular displacement as
∆y bucket = −R × ∆ϕspool .
Consequently,
vybucket = −R × ωspool
(S.39)
and hence
Knet =
2 2
1
4 MR ω
+
2
1
2 mv
=
2 2
1
4 MR ω
+
2
1
2 m(Rω)
=
M + 2m
× R2 ω 2 .
4
(S.40)
The potential energy of the spool remains constant — it rotates but does not move up or
down as a whole — while the bucket’s potential energy is
U = mgy.
(S.41)
As a bucket goes down by ∆y = −4.00 m, its potential energy changes by ∆U = mg∆y < 0,
and since the net mechanical energy is conserved, the released potential energy becomes the
kinetic energy of the bucket and the spool:
E = Unet + Knet = const
=⇒
∆Knet = −∆Unet = −∆Ubucket = −mg∆y. (S.42)
(0)
Initially, the spool and the bucket are at rest, so Knet = 0 and
∆Knet = Knet =
M + 2m
× R2 ω 2 .
4
(S.43)
Comparing this formula with eq. (S.42), we obtain
M + 2m
× R2 ω 2 = mg × (−∆y)
4
9
(S.44)
and hence
4m
× g × (−∆y)
M + 2m
4 × 3.00 kg
=
× 9.8 m/s2 × (+4.00 m)
5.00 kg + 2 × 3.00 kg
= 42.8 m2 /s2 .
R2 ω 2 =
q
Rω =
42.8 m2 /s2 = 6.54 m/s.
(S.45)
6.54 m/s
Rω
=
R
0.600 m
= 10.9 rad/s.
ω =
Textbook problem 8.42 (a):
The kinetic energy of a rotating flywheel is
K =
2
1
2 Iω
(S.46)
The rotation rate f = 5000 rev/min = 83.3 rev/s corresponds to the angular velocity ω =
2πf = 524 rad/s. As to the flywheel’s moment of inertia, the general formula is
I = CMR2
(S.47)
where C is the numeric coefficient depending on the mass distribution over the flywheel.
Typically, C is between
1
2
(a solid disk) and 1 (all mass at the rim). Since the problem does
not tell us the mass distribution, I am going to assume C = 1 — all mass at the rim — to
maximize the energy capacity. Consequently,
I = MR2 = 500 kg × (2.00 m)2 = 2, 000 kg · m2
(S.48)
and
K =
2
1
2 Iω
=
1
2 (2, 000
kg · m2 ) × (524 s−1 )2 = 2.74 · 108 J.
(S.49)
(b) Power is the rate at which the work is done or energy is transferred. Energy store E (for
10
the flywheel) used to produce power P would last for time
t =
E
.
P
(S.50)
For the flywheel, E = K = 2.74 · 108 J. Expending it at the rate P = 10.0 hp = 7.46 kW to
power a car, the energy would last for
t =
2.74 · 108 J
= 36, 700 s ≈ 10 hours.
7.46 · 103 W
(S.51)
PS: Students who have guessed — and written down — that the flywheel is a solid disk and has
I = 21 MR2 will get full credit if the subsequent calculations are correct for this assumption.
Ditto for any other spelled out assumption about the flywheel’s geometry and the appropriate
formula for I in terms of M and R.
Textbook problem 8.44 (a):
The net kinetic energy of a rolling sphere is a sum of kinetic energies due to its linear motion
(i.e., speed v of the center of mass) and due to its spin about axis through the center of mass,
linear
K = Kcm
+ K spin =
2
1
2 Mv
+
2
1
2 Iω
(S.52)
If the sphere rolls without slipping, its linear velocity v is related to the angular velocity ω of
its spin as
v = R × ω.
(S.53)
Consequently, the net kinetic energy of the sphere is
K =
2
1
2 m(Rω)
+
2
1
2 Iω
=
2
1
2 (MR
+ I) × ω 2 .
(S.54)
While the sphere rolls down the ramp through distance L = 6.0 m, its potential energy
11
changes by
∆U = Mg∆y = Mg × −L sin θ.
(S.55)
Since the net kinetic + potential energy of the sphere is conserved, the released potential
energy becomes kinetic,
E = K + U = const
=⇒
∆K = −∆U = +MgL sin θ.
(S.56)
Initially, the sphere is at rest and K0 = 0, so after it has rolled down, it has
K = K0 + ∆K = 0 + MgL sin θ.
(S.57)
Comparing this formula to eq. (S.54) for the kinetic energy, we obtain
2
1
2 (MR
+ I) × ω 2 = MgL sin θ
(S.58)
2MgL sin θ
.
MR2 + I
(S.59)
and hence
ω2 =
The problem does not specify if the sphere is hollow or solid, and this leads to ambiguity
of the moment of inertial I. Since a solid sphere is properly called a ball, I assume the sphere
is question is hollow spherical shell with a thin wall (like an inflated basketball). Such spheres
have
I = 23 MR2 ,
(S.60)
so eq. (S.59) becomes
ω2 =
2MgL sin θ
6 gL sin θ
=
×
.
2
2
2
5
R2
MR + 3 MR
Note that the sphere’s mass M (or weight Mg) cancels out of this formula.
12
(S.61)
Numerically
ω2 =
6 (9.8 m/s2 )(6.0 m) sin 37◦
×
= 1062 s−2
5
(0.20 m)2
(S.62)
and ω = 32.6 inverse seconds (i.e., 32.6 rad/s which corresponds to 5.15 rev/s or 311 rev/min).
PS: Students who have assumed that the sphere is solid and has
I =
2
2
5 MR
(S.63)
will get full credit if they have correctly derived eq. (S.59) and hence
ω2 =
10 gL sin θ
×
= 1264 s−2
7
R2
(S.64)
and ω = 35.6 inverse seconds (i.e., 35.6 rad/s which corresponds to 5.66 rev/s or 340 rev/min).
Non-textbook problem #1
Fp
pivot
CM θ
free end
mg
α
a
Let’s start by calculating the net torque on the rod with respect to the pivot point at its left
end. There are two forces acting on the rod: its weight Mg, and some force Fp at the pivot
which makes sure the pivot does not move as the rod swings down. We do not know the
magnitude or the direction of the Fp , but fortunately we do not need to: Because this force
act at the pivot point, it has zero lever arm and zero torque. As to the gravity force Mg, it’s
distributed all over the rod, but it has the same torque as if it were acting at the rod’s center
of mass. For a rod of uniform density and thickness, the center of mass is in the middle of the
13
rod, at the distance Lcm = 21 L from the pivot end. The horizontal coordinate of the center of
mass (counting from the pivot) is
Xcm = Lcm × cos θ =
1
2 L cos θ,
(S.65)
and that’s the lever arm of the weight force. Therefore, net torque is
τ net = τ (Fp ) + τ (Mg) = 0 + Mg × 12 L cos θ.
(S.66)
Given this net torque, we can find the rod’s angular acceleration α from
τ net = Iα
(S.67)
where I is the rod’s moment of inertia with respect to the pivot. The pivot is at the end of
the rod, so according to the bottom right picture in Table 8.1 on page 241 of the textbook —
and also according to the top item on page 3 of my notes — the moment of inertia is
2
1
3 ML .
I =
(S.68)
Substituting this formula into eq. (S.67) and comparing to eq. (S.66), we arrive at
2
1
3 ML
× α = τ net = Mg × 21 L cos θ.
(S.69)
Solving this equation for α gives us the rod’s angular acceleration
α =
Mg × 12 L cos θ
1
2
3 ML
=
3g cos α
.
2L
(S.70)
It remains to work out the linear motion of the rod’s free end. As the rod swings on a
pivot at its left end, the right end moves in circular arc of radius r = L. The linear motion of
14
the free end is related to the angular motion of the rod according to
v = L × ω,
ac = L × ω 2 ,
at = L × α.
(S.71)
Immediately after the release, the rod has non-zero angular acceleration α according to
eq. (S.70), but it has not yet acquired an angular velocity, ω ≈ ω0 = 0. Consequently, the free
end has zero speed and zero centripetal acceleration, but a non-zero tangential acceleration,
v = 0,
ac = 0,
at = L × α 6= 0.
(S.72)
Hence, the net linear acceleration a is the tangential acceleration,
a = at = L × α = L ×
3g cos α
= g × 32 cos θ.
2L
(S.73)
◦
PS: Note that for θ <
∼ 48 , the acceleration of the rod’s free end is faster than g. And for
< 35◦ , even the vertical component of the free end’s motion is faster than free fall,
θ∼
|ay | = a × cos θ = g × 32 cos2 θ > g.
(S.74)
So if you place a coin on top of the rod’s free end, the coin would not be able to follow the
rod when it’s released. Instead, it will separate from the rod and fall down at a slower rate.
Non-textbook problem #2
LB
LC
L
FC
mg
FB
The equilibrium conditions for any rigid body are
X
Fx = 0,
X
Fy = 0,
X
τ = 0.
(S.75)
The stretcher in question is subject to three forces: The patient’s weight Mg, force FB of
P
Bob’s hands, and force FC from Charlie’s hands. All these forces are vertical, so
Fx = 0 is
15
trivially true, and there is only one non-trivial balance-of-forces equation,
X
Fy = FB + FC − Mg = 0.
In the balance-of-torques equation
P
(S.76)
τ = 0, we may treat any point P we like as a pivot,
as long as we calculate toques of all forces with respect to the same pivot point P. For the
problem at hand, it’s convenient to put P at one end of the stretcher, for example at the front
end where Bob holds the stretcher. For this choice of a pivot, the force FB has zero lever
arm and hence zero torque. On the other hand, the force FC acts at the other end of the
stretcher, so its lever arm is L = 8 ft — the full length of the stretcher — and the torque is
τ (FC ) = −FC × L, where the ‘−’ sign indicates the counterclockwise direction of this torque.
Finally, the patient’s weight Mg is distribute all over the patient’s body, but for the
purpose of calculating the torque we may treat it as acting at the patient’s center of gravity
(which is at his center of mass). This center of mass lies at LB = 3 ft from the pivot point
(Bob’s hands), so the lever arm of Mg is LB and the torque is τ (Mg) = +Mg × LB , where
the ‘+’ sign indicates the clockwise direction of this torque.
Altogether, the net torque around out chosen pivot point is
X
τ = τ (FB ) + τ (FC ) + τ (Mg) = 0 + FC × L − Mg × LB .
(S.77)
Demanding that this net torque cancels out, we have
FC × L − Mg × LB = 0
(S.78)
and therefore
FC = Mg ×
LB
3 ft
= 200 lb ×
= 80 lb.
L
8 ft
(S.79)
Consequently, according to the balance-of-forces equation (S.76),
FB = Mg − FC
5 ft
LB
LB
LC
= 200 lb×
= Mg − Mg×
= Mg× 1 −
=
= 120 lb.
L
L
L
8 ft
(S.80)
Thus, Charlie carries 80 pounds of patient’s weight and Bob carries the remaining 120 pounds.
16