Download Physics 140 HOMEWORK Chapter 10B Q7. Figure 10

Physics 140
HOMEWORK Chapter 10B
Q7. Figure 10-24a is an overhead view of a horizontal bar that can pivot; two horizontal forces act on
the bar, but it is stationary. If the angle between the bar and is now decreased from 90◦ and the bar
is still not to turn, should F2 be made larger, made smaller, or left the same?
———
made greater. The torque F2 exerts about the pivot must be maintained. The factor are |~r| doesn’t
~ | has to increase.
change, so if sin θ is to decrease, |F
Q9. Figure 10-25 shows a uniform metal plate that had been square before 25% of it was snipped off.
Three lettered points are indicated. Rank them according to the rotational inertia of the plate around
a perpendicular axis through them, greatest first.
———
Ic > Ia > Ib . This is an application of the parallel axis theorem, I// = Icom + M h2 . By inspection, the
com is between b and a, but closer to b. (If you don’t see that, work it out in a coordinate system with
the origin at b and the x axis pointed at a.) The h in the formula above is the distance from the com
to the axis.
P43. The uniform solid block in Fig. 10-35 has mass 0.172 kg and edge lengths a = 3.5 cm, b = 8.4 cm,
and c = 1.4 cm. Calculate its rotational inertia about an axis through one corner and perpendicular to
the large faces.
———
Parallel axis theorem: First, look up and compute Icom :
Icom = (1/12)M (a2 + b2 ) = (1/12)(0.172 kg)[(0.035 m)2 + (0.084 m)2 ] = 1.187 × 10−4 kg m2 .
Next, use parallel axis theorem, noting that h2 = (a2 + b2 )/4:
Icorner = Icom + M h2 = 1.21 × 10−4 + (0.172)(0.0352 + 0.0842 )/4 = 4.75 × 10−4 kgm2 .
P48. The length of a bicycle pedal arm is 0.152 m, and a downward force of 111 N is applied to the
pedal by the rider. What is the magnitude of the torque about the pedal arm’s pivot when the arm is
at angle (a) 30◦ , (b) 90◦ , and (c) 180◦ with the vertical?
———
~ | sin θ:
Very basic problem: τ = |~r| kF
(a) τ = (0.152 m)(111 N)(0.5) = 8.44 N m.
(b) τ = (0.152 m)(111 N)(1.0) = 16.9 N m.
(c) τ = (0.152 m)(111 N)(0.0) = 0 N m.
P51. In Fig. 10-38, block 1 has mass m1 = 460 g, block 2 has mass m2 = 500 g, and the pulley,
which is mounted on a horizontal axle with negligible friction, has radius R = 5.00 cm. When released
from rest, block 2 falls 75.0 cm in 5.00 s without the cord slipping on the pulley. (a) What is the
magnitude of the acceleration of the blocks? What are (b) tension T2 and (c) tension T1 ? (d) What is
the magnitude of the pulley’s angular acceleration? (e) What is its rotational inertia?
———
Recognize a Newton’s second law problem with torques. Since m2 > m1 , m1 will accelerate upward.
XFBD for circular pulley: T1 downward at left edge; T2 downward at right edge; Mp downward at
center; n upward at center. (n is the normal, or support, force from the axle.) ~a = 0; α is CW. With
only one object rotating, we can make CW positive, the opposite of the standard convention.
FBD for m1 : T1 up, m1 g down. ~a upward, so make +x upward.
FBD for m2 : T2 up, m2 g down. ~a downward, so make +x downward.
Note that the coordinate systems make the accelerations of m1 and m2 both positive, and that they
are equal to each other. We have also switched the usual angular convention so that α is positive.
We are given data from which the acceleration can be calculated. The steps lead us through the problem
in a logical order.
(a) ∆x = 0.75 m = (1/2)at2 ⇒ a = 2∆x/t2 = 2(0.75 m)/(5 s)2 = 0.0600 m/s2 .
(b) Eq for m2 : m2 g − T2 = m2 a ⇒ T2 = m2 (g − a) = (0.50 kg)(9.74 m/s2 ) = 4.8700 N
(c) Eq for m1 : T1 − m1 g = m1 a ⇒ T1 = m1 (g + a) = (0.46 kg)(9.86 m/s2 ) = 4.5356 N
(d) α = a/R = (0.06 m/s2 )/(0.05 m) = 1.20 rad/s2
(e) Torque Eq for pulley: T2 R − T1 R = Iα ⇒
I = (T2 − T1 )R/α = (0.3344 N)(0.05 m)/(1.2 rad/s2 ) = 0.0139 kg m2 .
P56. Figure 10-43 shows particles 1 and 2, each of mass m, attached to the ends of a rigid massless rod
of length L1 + L2 , with L1 = 20 cm and L2 = 80 cm. The rod is held horizontally on the fulcrum and
then released. What are the magnitudes of the initial accelerations of (a) particle 1 and (b) particle 2?
———
Notice how hard this would be without rotations. The rigid rod exerts an upward force, but how much?
With rotations, it’s just a matter of τ = Iα, along with a = Rα.
The rod will obviously rotate CW. Since the problem is so simple, reverse the usual sign convention
and take angular quantities as positive when CW. We need the torque τ and the rotational inertia I:
τ = mgL2 − mg(L1 ) = mg(L2 − L1 ).
I = Σmi ri2 = mL21 + mL22 .
α = g(L2 − L1 )/(L21 + L22 ) = (9.8 m/s2 )(0.6 m)/(0.22 + 0.82 ) m2 = 8.65 rad/s2
(a) a1 = r1 α = (0.2 m)(8.65 rad/s) = 1.73 m/s2 upward.
(b) a2 = r2 α = (0.8 m)(8.65 rad/s) = 6.92 m/s2 downward.
P73. A uniform helicopter rotor blade is 7.80 m long, has a mass of 110 kg, and is attached to the
rotor axle by a single bolt. (a) What is the magnitude of the force on the bolt from the axle when the
rotor is turning at 320 rev/min? (Hint: For this calculation the blade can be considered to be a point
mass at its center of mass. Why?) (b) Calculate the torque that must be applied to the rotor to bring
it to full speed from rest in 6.70 s. Ignore air resistance. (The blade cannot be considered to be a point
mass for this calculation. Why not? Assume the mass distribution of a uniform thin rod.) (c) How
much work does the torque do on the blade in order for the blade to reach a speed of 320 rev/min?
———The rotor is held to the center axle at one end.
(a) Fcent = ma = mv 2 /r. We will need v = rω, where ω = 320 rev/min = 33.5 rad/s, and r = L/2 =
3.90 m. So:
v = (3.90 m)(33.5 rad/s) = 130.7 m/s.
F = (110 kg)(130.7 m/s)2 /(3.9 m) = 481 kN.
Why can we only consider the com here? It’s easier if we use the alternate form of acent = rω 2 . Then
the force on each element is proportional to r, and the total force is the total of mi ri , which is the com.
(b) τ = Iα. We need I = (1/3)M L2 = (1/3)(110 kg)(7.8 m)2 = 2231 kg m2 . From the definition,
α = ∆ω/∆t = (33.5 rad/s)/(6.7 s) = 5.00 rad/s2 .
τ = (2231 kg m2 )(5.000 rad/s2 ) = 11.2 kN · m
The distribution must be considered because the contribution to the kinetic energy scales as r2 .
(c) W = τ θ. We first need θ = θ0 + ω0 t + (1/2)αt2 = 0 + 0 + 112.2 = 112.2 rad.
W = (1.12 × 104 N m)(1.122 × 102 rad = 1.26 MJ.
You might check that this equals the K = (1/2)Iω 2 .
P87. In Fig. 10-52, a wheel of radius 0.20 m is mounted on a frictionless horizontal axle. A massless
cord is wrapped around the wheel and attached to a 2.0 kg box that slides on a frictionless surface
inclined at angle θ = 20◦ with the horizontal. The box accelerates down the surface at 2.0 m/s2 . What
is the rotational inertia of the wheel about the axle?
———
Recognize a Newton’s second law problem with rotation.
XFBD for wheel: mwheel g at com, down; T , at edge, angled down/left; naxle on wheel at com, angled up
and right. ~a = 0, which is how we determine the general direction of naxle on wheel , the normal (support)
force from the axle.
The x- and y-equations for the wheel would give us the components of naxle on wheel . These would be
necessary if we were designing the axle and bearings, but are not needed in the present problem.
FBD for box: mg down, n angled up and left. ~a downplane, given as 2 m/s2 . +x downplane.
x-eq for box: mg sin θ − T = ma ⇒
T = m(g sin θ − a) = (2 kg)[(9.8 m/s2 )(0.342) − 2.0 m/s2 ] = 2.70 N.
For the torque eq, if we take the axle as the reference point, only T exerts a non-zero torque
Torque Eq. τNET = Iα: T R = Iα = Ia/R ⇒
I = T R2 /a = (2.70 N)(0.2 m)2 /(2 m/s2 ) = 0.0541 kg m2 .
P98. A yo-yo-shaped device mounted on a horizontal frictionless axis is used to lift a 30 kg box
as shown in Fig. 10-56. The outer radius R of the device is 0.50 m, and the radius r of the hub is
0.20 m. When a constant horizontal force of magnitude 140 N is applied to a rope wrapped around
the outside of the device, the box, which is suspended from a rope wrapped around the hub, has an
upward acceleration of magnitude 0.80 m/s2 . What is the rotational inertia of the device about its axis
of rotation?
———
Newton’s second law with rotation.
XFBD for double-pulley (“device”): T downward at r on right; Fappl at top at R to left; n at com
upward; mdevice g downward at com. ~a = 0. α CCW.
FBD for box: T upward; mg down. ~a = 0.8 m/s2 upward. Make +x upward.
Eq for box: T − mg = ma ⇒ T = m(g + a) = (30 kg)(9.8 + 0.8) m/s2 = 318 N.
Torque Eq: Fappl R − T r = Iα. Note that rα = a. Hence
I = (r/a)(Fappl R − T r) = (0.2 m/0.8 m/s2 )(70 N m − 63.6 N m) = 1.6 kg m2 .
P65 Opt. A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin
rod of length 55.0 m. At the instant it makes an angle of 35.0◦ with the vertical as it falls, what are
(a) the radial acceleration of the top, and (b) the tangential acceleration of the top. (Hint: Use energy
considerations, not a torque.) (c) At what angle θ0 is the tangential acceleration equal to g?
———(a) For the radial accceleration, this is the centripetal acceleration: acent = v 2 /r = rω 2 . Get ω 2 via
conservation of energy.
In applying Ug to an extended object, the y value is simply the y value of the com.
mg(L/2) = mg(L/2)(cos θ) + (1/2)Iω 2 . Since I = (1/3)mL2 ,
ω 2 = 3g(1 − cos θ)/L ⇒ ar = 3g(1 − cos θ) = 5.32 m/s2 .
(b) For this part, we do need the torque (I think).
In calculating torque by a gravitational force, we treat the force as applied at the com of the object.
α = τ /I = [mgL sin θ /2]/[(1/3)M L2 ] = (3/2)g sin θ/L =
(3/2)(9.8 m/s2 )(0.5736)/(55 m) = 0.1533 rad/s2 .
At the tip of the chimney, at = (3/2)g sin θ = 8.43 m/s2 .
(c) In the previous eq, set at = g ⇒ 1 = (3/2) sin θ0 ⇒ θ0 = 41.8◦ .
The point is this: at some angle, the tangential acceleration of the upper part of the chimney would
exceed g if the chimney remained intact. Meanwhile, the radial acceleration at points in the upper
part of the chimney would increase to the point that the chimney is under tension, not compression. In
actual collapses, the chimney frequently breaks maybe 70% of the way up while falling, and the lower
part hits the ground first.
P81 Opt. The thin uniform rod in Fig. 10-50 has length 2.0 m and can pivot about a horizontal,
frictionless pin through one end. It is released from rest at angle θ = 40◦ bove the horizontal. Use the
principle of conservation of energy to determine the angular speed of the rod as it passes through the
horizontal position.
———We will solve with the release point at θ0 and the final point at θ1 .
Apply conservation of energy with Ug = mg(L/2) sin θ and K = Krot = (1/2)Iω 2 , with I = (1/3)mL2 .
This chooses Ug to be zero when the rod is horizontal.
mg(L/2) sin θ0 = mg(L/2) sin θ1 + (1/2)(mL2 /3)(ω 2 ) solve for ω:
p
ω = (3g/L)(sin θ0 − sin θ1 ) = 3.07 rad/s.