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Physics H7A, Fall 2011
Homework 4 Solutions
1. (K&K Problem 2.11)
A mass m is connected to a vertical revolving axle by two
strings of length l, each making an angle of 45◦ with the axle,
as shown. Both the axle and mass are revolving with angular
velocity ω. Gravity is directed downward.
(a) Draw a clear force diagram for m.
(b) Find the tension in the upper string, Tup , and the lower
string, Tlow .
√
√
Ans. clue. If lω 2 = 2g, Tup = 2mg.
(a) (Insert figure here.)
(b) From the figure, we have
X
√
√
√
Fx = −Tup / 2 − Tlow / 2 = max = −mω 2 l/ 2,
X
√
√
Fy = Tup / 2 − Tlow / 2 − mg = may = 0.
Solving these equations gives
√
Tup = m(ω 2 l/2 + g/ 2),
√
Tlow = m(ω 2 l/2 − g/ 2).
2. (K&K Problem 2.13) Masses M1 and M2 are connected to a system of strings
and pulleys as shown. The strings are massless and inextensible, and the pulleys are
massless and frictionless. Find the acceleration of M1 .
Ans. clue. If M1 = M2 , ẍ1 = g/5.
1
After drawing free body diagrams for M1 , M2 , and the lower pulley, we obtain the equations
T1 − M1 g = M1 a1 , T2 − M2 g = M2 a2 , T1 − 2T2 = 0.
This provides three equations for four unknowns; the remaining equation comes from
relating the accelerations of the two masses: a2 = 2apulley = −2a1 . Solving the four
equations gives
2M2 − M1
a1 =
g.
M1 + 4M2
3. *(K&K Problem 2.34) A mass m whirls around on a string which passes through a
ring, as shown. Neglect gravity. Initially the mass is distance r0 from the center and is
revolving at angular velocity ω0 . The string is pulled with constant velocity V starting
at t = 0 so that the radial distance to the mass decreases. Draw a force diagram and
obtain a differential equation for ω. This equation is quite simple and can be solved
either by inspection or by formal integration. Find
(a) ω(t).
Ans. clue. For V t = r0 /2, ω = 4ω0
(b) The force needed to pull the string.
Hint: It might be convenient to use the expression for the acceleration vector a in polar
coordinates, as on page 36 of K&K. In this context, ω(t) = θ̇(t).
2
The string is pulled downward with constant velocity V , so the radius of the mass must
likewise decrease steadily at the rate ṙ = −V . Since r(0) = r0 , we have r(t) = r0 − V t.
Since the only force acting on the mass is from the tension in the string (remember we’re
neglecting gravity), we have F = −Tr̂ (the −r̂ is our way of saying that the force acts
radially inward). Recalling the expression for the acceleration vector in polar coordinates,
a = (r̈ − rθ̇2 )r̂ + (rθ̈ + 2ṙθ̇)θ̂,
we see that F = ma implies the two equations
−T = m(r̈ − rθ̇2 ),
rθ̈ + 2ṙθ̇ = 0.
The first equations tells us that the applied force must be T = rω 2 , while the second
gives us a differential equation for ω:
(r0 − V t)
dω
= 2V ω.
dt
This is a separable differential equation, which we can solve directly in just a few steps:
dω
2V dt
=
ω
r0 − V t
Z ω(t)
Z t
dω
2V
=
dt
ω
0 r0 − V t
ω0
ω(t)
r0 − V t
ln
= −2 ln
ω0
r0
2
ω0 r0
ω(t) =
(r0 − V t)2
The force pulling on the string must therefore be
F (t) = mω(t)2 r(t) =
mω02 r0
.
(r0 − V t)3
As an aside, note that the equation rω̇ + 2ṙω = 0 is equivalent to
d
(ωr2 ) = 0.
dt
3
Thus, it is satisfied if and only if ωr2 is constant, i.e. ω(t) ∝ r(t)−2 . This is a special
case of the law of conservation of angular momentum: in the absence of tangential forces
acting on an point mass (torques), the angular momentum L = mωr2 is conserved.
4. (K&K Problem 4.5) Mass m whirls on a frictionless table, held to circular motion by
a string which passes through a hole in the table. The string is slowly pulled through
the hole so that the radius of the circle changes from l1 to l2 . Show that the work done
in pulling the string equals the increase in kinetic energy of the mass.
Note: The previous problem should tell you that ω ∝ r−2 . This also follows directly
from angular momentum conservation, which we will discuss later in the course.
Since ω ∝ r−2 , we can say without loss of generality that ω = ω0 r02 r−2 for some values
of ω0 and r0 . Since the speed of a rotating object is v = ωr = ω0 r02 r−1 , the change in
kinetic energy of the mass is
1
1
K2 − K1 = m(v22 − v12 ) = mω02 r04 (l2−2 − l12 ).
2
2
From the previous problem, the force needed to pull the string at constant velocity is
F = mω 2 r = mω02 r04 r−3 . The work done is therefore
Z
2
Z
l1
F · dr =
W21 =
1
l2
1
mω02 r04 r−3 dr = mω02 r04 (l2−2 − l1−2 ) = K2 − K1 .
2
Notice that I swapped the integration limits in going from the second to the third expression above. Understanding the reason for this will prevent a lot of minus sign errors,
R 2 and
save you a lot of headaches. From a physicist’s perspective, the line integral W = 1 F·dr
represents the following: we take our path from 1 to 2, break it up into a connected series of path elements dr, compute the dot product F · dr for each element, and sum the
results. In this picture, it’s clear that if F · dr is positive for each path element, then
W is obviously positive as well, since we just added up a bunch of positive numbers.
However,
R b for ordinary
R a1-D integrals, mathematicians like to define things in such a way
that a f (x)dx = − b f (x)dx, so that if a > b the integral of a positive function f (x)
will come out negative. So to correct for this effect, when I express my line integral as an
ordinary 1-D integral, I manually swap the integration limits so that my answer comes
out positive. (An alternative and more robust approach is to use the formal procedure
for evaluating line integrals, as explained in K&K pp. 167-168. This will always get you
the right answer, but may take more lines of work.)
5. (K&K Problem 4.6) A small block slides from rest from the top of a frictionless
sphere of radius R. How far below the top x does it lose contact with the sphere? The
sphere does not move.
Ans. R/3
4
Let θ represent the angular position of the block with respect to the vertical when it
separates from the globe. From the figure, x = R(1 − cos θ). Since the block starts from
rest (or very close to it), the kinetic energy it picks up is equal to the change in potential
energy: 21 mv 2 = mgx = mgR(1 − cos θ). With a free body diagram, we see that the
forces acting on the block in the radial direction are the normal force N (outward), and
the a component mg cos θ of the gravitational force (inward). So long as the block is
sliding along the surface of the sphere, the radial component of the acceleration must be
v 2 /R inward (it will obviously have tangential acceleration, since it’s speeding up, but
ar = −v / R still holds). Thus
N − mg cos θ = −mv 2 /R = −2mg(1 − cos θ).
At the point where the block separates from the sphere, the normal force vanishes (N = 0),
which gives
2
−mg cos θ = −2mg(1 − cos θ)
=⇒
cos θ = .
3
Hence x = R/3.
5