# CHAPTER 15 SOLUTION FOR PROBLEM 15 The maximum force ```CHAPTER 15
SOLUTION FOR PROBLEM 15
The maximum force that can be exerted by the surface must be less than &micro;s N or else the block
will not follow the surface in its motion. Here, &micro;s is the coefficient of static friction and N is the
normal force exerted by the surface on the block. Since the block does not accelerate vertically,
you know that N = mg, where m is the mass of the block. If the block follows the table and
moves in simple harmonic motion, the magnitude of the maximum force exerted on it is given by
F = mam = mω 2 xm = m(2πf )2 xm , where am is the magnitude of the maximum acceleration,
ω is the angular frequency, and f is the frequency. The relationship ω = 2πf was used to obtain
the last form.
Substitute F = m(2πf )2 xm and N = mg into F &lt; &micro;s N to obtain m(2πf )2 xm &lt; &micro;s mg. The
largest amplitude for which the block does not slip is
2
xm
&micro;s g
(0.50)(9.8 m/s )
=
=
= 0.031 m .
2
(2πf )
(2π &times; 2.0 Hz)2
A larger amplitude requires a larger force at the end points of the motion. The surface cannot
supply the larger force and the block slips.
CHAPTER 15
SOLUTION FOR PROBLEM 39
(a) Take the angular displacement of the wheel to be θ = θm cos(2πt/T ), where θm is the
amplitude and T is the period. Differentiate with respect to time to find the angular velocity:
Ω = −(2π/T )θm sin(2πt/T ). The symbol Ω is used for the angular velocity of the wheel so it is
not confused with the angular frequency. The maximum angular velocity is
Ωm =
=
T
0.500 s
(b) When θ = π/2, then θ/θm = 1/2, cos(2πt/T ) = 1/2, and
√
sin(2πt/T ) = 1 − cos2 (2πt/T ) = 1 − (1/2)2 = 3/2 ,
where the trigonometric identity cos2 A + sin2 A = 1 was used. Thus
X√ ~
w
W
w
W
2πt
2π
3
2π
Ω = − θm sin
=−
T
T
0.500 s
2
The minus sign is not significant. During another portion of the cycle its angular speed is
(c) The angular acceleration is
d2 θ
α= 2 =−
dt
When θ = π/4,
α=−
w
w
2π
T
W2
θm cos(2πt/T ) = −
2π
0.500 s
Again the minus sign is not significant.
w
2π
T
W2 p Q
π
2
4
W2
θ.
CHAPTER 15
SOLUTION FOR PROBLEM 55
0
(a) The period of the pendulum is given by T = 2π I/mgd, where I is its rotational inertia,
m is its mass, and d is the distance from the center of mass to the pivot point. The rotational
inertia of a rod pivoted at its center is mL2 /12 and, according to the parallel-axis theorem, its
rotational inertia when it is pivoted a distance d from the center is I = mL2 /12 + md2 . Thus
2
2
m(L /12 + d )
L2 + 12d2
= 2π
.
T = 2π
mgd
12gd
The square of the period is 4π2 (L2 + 12d2 )/12gd) and the derivative of this with respect to d is
]
}
dT 2 4π 2
L2 + 12d2
.
=
24 −
dd
12g
d2
√
Set this equal to zero and solve for d. The result is d = L/ 12. Substitute this back into the
expression for T and obtain
L
2.20 m
= 2.26 s .
T = 2π √ = 2π √
2
3g
3(9.8 m/s )
(b) According √
to the expression obtained in part (a) for the minimum period this period is
proportional to L, so it increases as L increases.
(c) According to the expression obtained in part (a) for the minimum period this period is
independent of m. T does not change when m increases.
CHAPTER 15
HINT FOR PROBLEM 17
(a) The displacement x and acceleration a at any instant of time are related by x = −ω2 a, where
ω is the angular frequency. The frequency is f = ω/2π.
(b) The angular frequency, mass m, and spring constant k are related by ω2 = k/m.
(c) If xm is the amplitude then we may take the displacement to be x = xm cos(ωt). The velocity
is v = −ωxm sin(ωt). Use the trigonometric identity sin2 (ωt) + cos2 (ωt) = 1 to find an expression
for xm .
J
ans: (a) 5.58 Hz; (b) 0.325 kg; (c) 0.400 m
o
CHAPTER 15
HINT FOR PROBLEM 25
Suppose the smaller block is on the verge of slipping when the acceleration of the blocks has
its maximum value. This occurs when the displacement of the blocks is equal to the amplitude
xm of their oscillation and the value is am = ω 2 xm , where ω is the angular frequency of the
oscillation. The magnitude of the force of friction is f = mam = mω 2 xm . Since the block is on
normal force of
the verge of slipping f = &micro;s N = &micro;s mg, where N (= mg) is the magnitude of the 0
either block on the other. Solve for xm . The angular frequency is given by ω = k/(m + M ).
J
ans: 22 cm
o
CHAPTER 15
HINT FOR PROBLEM 33
0
(a) Use f = (1/2π) k/m, where k is the spring constant and m is the mass of the object.
(b) The potential energy is given by U = 12 kx2 .
(c) The kinetic energy is given by K = 12 mv 2 , where v is the speed of the object.
(d) The mechanical energy, which is the sum of the kinetic and potential energies, is 12 kx2m ,
where xm is the amplitude of the oscillation.
J
ans: (a) 2.25 Hz; (b) 1125 J; (c) 250 J; (d) 86.6 cm
o
CHAPTER 15
HINT FOR PROBLEM 41
The period of a physical pendulum with rotational inertia
0 I and mass m, suspended from a point
that is a distance h from its center of mass, is T = 2π I/mgh. The distance between the point
of suspension and the center of oscillation is the same as the length of a 0
simple pendulum with
the same period. The period of a simple pendulum of length L0 is T = 2π L0 /g. Set these two
expressions for the period equal to each other and solve for L0 .
CHAPTER 15
HINT FOR PROBLEM 53
At time t = 0 the angle is θ0 = θm cos(φ) and its rate of change is (dθ/dt)0 = −ωθm sin(φ), where
ω (= 4.44 rad/s) is the angular frequency of oscillation. Solve the first equation for cos(φ) and
the second for sin(φ). Use the results to find an expression for tan(φ) in terms of θ0 , (dθ/dt)0 ,
and ω. Then find θ itself. Be sure you obtain the correct result for φ. Both the sine and cosine
should be positive. Now add the expressions for the square of the sine and the square of the
cosine to find an expression for θm .
J