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Physics H7A, Fall 2011
Homework 6 Solutions
1. (Hooke’s law)
(a) What are the dimensions (in terms of M , L, and T ) of the spring constant k in
Hooke’s law?
(b) If it takes 4.00 J of work to stretch a Hooke’s-law spring 10.0 cm from its unstressed
length, determine the extra work required to stretch it an additional 10.0 cm.
(a) [k] = [F/x] = M/T 2
(b) The work done by your hand in stretching a spring from x1 to x2 is
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Wx1 →x2 = U (x2 ) − U (x1 ) = k(x22 − x21 ).
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So if W0→10 = 21 k(102 ) = 4 J, then W10→20 = 21 k(202 − 102 ) = 12 J.
2. (K&K Problem 2.31) Find the frequency of oscillation of mass m suspendended by
two springs having constants k1 and k2 , in each of the configurations
shown. p
p
Ans. clue. If k1 = k2 = k, ωa = k/2m, ωb = 2k/m
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Two springs in series behave as a single spring with spring constant ka = (1/k1 + 1/k2 )−1 ,
while two springs in parallel behave as a single spring with kb = k1 + k2 . Thus
r
r
r
1
ka
kb
k1 + k2
=p
=
.
ωa =
,
ωb =
m
m
m
(1/k1 + 1/k2 )m
3. (Pencil dive) A 70-kg diver steps off a 10-m tower and drops straight down into the
water. If he comes to rest 5 m beneath the surface of the water, determine the average
resistance force exerted by the water on the diver.
√
The diver hits the water with speed v = 2gh = 14 m/s. To come to rest after ∆x = 5 m,
he feels an average upward acceleration aavg = v 2 /2∆x = 19.6 m/s2 = 2g. Considering that
gravity still pulls him down with force mg while in the water, the average resistance force from
the water must be Fw = 3mg = 2058 N.
4. (Pendulum in the wind) A ball having mass m is connected by a strong string of
length L to a pivot point and held in place in a vertical position. A wind exerting a
constant force of magnitude F is blowing from left to right as in the figure below.
(a) If the ball is released from rest, show that the maximum height H reached by the
ball, as measured from its initial height, is
H=
2L
.
1 + (mg/F )2
Check that this result is valid both for cases when 0 ≤ H ≤ L and for L ≤ H ≤ 2L.
(b) Compute the equilibrium height Heq of the ball (the height at which the net force
on the ball is zero).
(c) Could the equilibrium height ever be larger than L? Explain.
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(a) The wind force F here is constant, so by analogy with the gravitational force mg we can
assign a potential energy Uw = −F x to the ball, where x is the horizontal displacement
of the ball relative to some arbitrary origin (which we will take to be starting position
of the ball). Then the total initial energy is E0 = Uw0 + Ug0 = −F x0 + mgy0 = 0, while
the final energy at its maximum height is E = −F x + mgH, and energy conservation
implies F x = mgH. From the geometry of the figure we have
2
2
2
L = x + (L − H) =
mgH
F
2
2
2
+ L − 2LH + H = 1 +
mg 2 F
H 2 − 2LH + L2 ,
hence H = 2L/[1 + (mg/F )2 ].
(b) At equilibrium the net force on the ball is zero, so
Fnet = F x̂ − mgŷ + T (− sin θx̂ + cos θŷ) = 0,
where T is the tension in the string and θ is the angle the string makes with respect to
the vertical. Setting the x and y components separately to zero, we get F = T sin θ and
mg = T cos θ, hence tan θ = F/mg. The equilibrium height is therefore
!
1
1
1
Heq = L(1−cos θ) = L 1 −
=L 1− √
=L 1− p
.
sec θ
1 + (F/mg)2
1 + tan2 θ
(c) The equilibrium height can never be greater than L, because if the ball is higher than
L then there is no upward force to balance the downward force of gravity.
5. (K&K Problem 3.8) A 50-kg woman jumps straight into the air, rising 0.8 m from
the ground. What impulse does she receive from the ground to attain this height?
v0 =
√
mgh,
√
∆p = mv0 = m mgh = 990 kg m/s
6. (Tennis volley) A tennis player receives a shot with the ball (0.060 kg) traveling
horizontally at 50.0 m/s and returns the shot with the ball traveling horizontally at
40.0 m/s in the opposite direction. (a) What is the impulse delivered to the ball by
the racquet? (b) What work does the racquet do on the ball?
(a) ∆p = mvf − mvi = (0.060 kg)(40.0 m/s) − (0.060 kg)(−50.0 m/s) = 5.4 kg m/s
(b) W = ∆K = 12 mvf2 − 12 mvi2 = −27.0 J
7. *(Leaky bucket) At t = 0, a massless bucket contains a mass M of sand. It is
connected to a wall by a massless spring with constant tension T (that is, independent
of length). The ground is frictionless, and the initial distance to the wall is L. At
later times, let x be the distance from the wall, and let m be the mass of sand in
the bucket. The bucket is released, and on its way to the wall, it leaks sand at a rate
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dm/dx = M/L. In other words, the rate is constant with respect to distance, not time;
and it ends up empty right when it reaches the wall. Note that dx is negative, so dm
is also.
(a) What is the kinetic energy of the (sand in the) bucket, as a function of x? What
is its maximum value?
(b) What is the magnitude of the momentum of the bucket, as a function of x? What
is its maximum value?
(a) The leaky bucket can be considered as a simple example of a rocket, in which the
“exhaust” is expelled with relative velocity u = 0 (because the sand leaks out, instead of
being propelled backward). Using Equation 3.19 from K&K, we have T = m(t)(dv/dt),
where m(t) is the mass of the sand left in the bucket at time t. We can now change
independent variables from t to x, using the given relation m(x) = M x/L along with
the identity
dv
dv dx
dv
=
=v ,
dt
dx dt
dx
to obtain the differential equation
T =
M x dv
v
L dx
=⇒
dx
M
=
vdv.
x
LT
Integrating from the initial point (x0 = L, v0 = 0) to a final point (x, v), we have
ln
x
M 2
=
v .
L
2LT
Therefore the kinetic energy as a function of position is
1
x
K(x) = m(x)v(x)2 = T x ln ,
2
L
with maximum Kmax = T L/e.
(b) The momentum of the bucket as a function of position is
r
2T M
x
p(x) = m(x)v(x) = x −
ln
L
L
(notice the minus sign since ln Lx < 0 for x < L), with maximum pmax =
4
p
M T L/e.
8. *(Speedy travel) A straight tube is drilled between two points on the earth, as shown
in the figure below. An object is dropped into the tube. What is the resulting motion?
How long does it take to reach the other end? Ignore friction, and assume (incorrectly)
that the density of the earth is constant (ρ = 5.5 g/cm3 ).
Let a be the distance from the center of the earth to the tube, and let x be the horizontal
displacement of the object along the tube (measured
√ from the midpoint) at any given time.
At the position x, the object lies a distance r = a2 + x2 from the center of the earth. The
mass that lies within this radius is m(r) = ME (r/RE )3 , so the strength of the gravitational
force is
Gm(r)mobj
GME mobj
F = |F| =
=
r.
3
r2
RE
But this gravitational force on the object is directed radially inward, and since the object is
constrained to travel within the tube, only the horizontal component of the gravitational force
contributes to the motion. This component is
Fx = −F sin θ = −F
GME mobj
x
=−
x.
3
r
RE
(The minus sign tells us the force acts to the left when x is to the right, and vice versa.) Thus,
the x component of Newton’s second law (F = ma) gives
−
GME mobj
x = mobj ax
3
RE
=⇒
GME
3 x = 0.
RE
ẍ +
This is the generic equation of motion for simple harmonic oscillation (ẍ + ω 2 x = 0), so the
object will undergo SHO with period
2π
T =
= 2π
ω
3
RE
GME
1/2
.
9. *(Mine shaft)
(a) If the earth had constant density, the gravitational force would decrease linearly
with radius as you descend in a mine shaft. However, the density of the earth is
not constant, and in fact the gravitational force increases as you descend. Show
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that the general condition under which this is true is ρc < (2/3)ρavg , where ρavg is
the average density of the earth, and ρc is the density of the crust at the surface.
(The values for the earth are ρc ≈ 3 g/cm3 and ρavg ≈ 5.5 g/cm3 .)
(b) Assuming that the density of a planet is a function of radius only, what should
ρ(r) look like if you want the gravitational force to be independent of the depth
in a mine shaft, all the way down to the center of the planet?
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(a) At the surface of the earth (r = R), the strength of the gravitational force on an object
of mass mobj is
GM mobj
F (R) =
,
R2
3
where M = 4π
3 R ρavg is the total mass of the earth. If you descend a small vertical
distance ∆R down a mine shaft, then the strength of the gravitational force will be
F (R − ∆R) =
G(M − ∆M )mobj
,
(R − ∆R)2
3
3
where ∆M = 4π
3 [R − (R − ∆R) ]ρc is the mass of spherical shell between radii R − ∆R
and R (which lies within the crust). Using the general result that
(1 + x)n ≈ 1 + nx
for x 1,
we can expand the above expression for F (R − ∆R) to first order in the small quantity
∆R/R. In the end we find
GM mobj
ρc
∆R
3
.
F (R) − F (R − ∆R) =
−
2
R2
ρavg
R
Thus, for ρc < (2/3)ρavg , we have F (R) < F (R − ∆R), i.e. the force increases as you go
down the mine shaft.
(b) The gravitational force at a radius r from the center of the planet is proportional to
m(r)/r2 , where
Z r
m(r) = 4π
ρ(r0 )r02 dr0
0
is the mass within radius r. To make this force independent of r, we must have m(r) =
Ar2 for some constant A. If the radius of the planet is R and its total mass M , then
m(R) = M implies A = M/R2 . To find the density, we differentiate the above expression
for m(r), using the fundamental theorem of calculus:
dm
= 4πρ(r)r2 .
dr
On the other hand dm/dr = (M/R2 )2r, so
ρ(r) =
M 1
.
2πR2 r
(Note that even though the density goes to infinity at the center, the total mass is still
finite.)
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