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GE Lab 9


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10.14. A bucket of water of mass 15.6 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder
with diameter 0.340 m with mass 11.0 kg . The cylinder pivots on a frictionless axle through its center. The
bucket is released from rest at the top of a well and falls a distance 10.1 m to the water. You can ignore the
weight of the rope.
(a) What is the tension in the rope while the bucket is falling?
(b) With what speed does the bucket strike the water?
(c) What is the time of fall?
(d) While the bucket is falling, what is the force exerted on the cylinder by the axle?


10.16. A 12.0-kg box resting on a horizontal, frictionless surface is attached to a 5.00-kg weight by a thin, light wire
that passes without slippage over a frictionless pulley. The pulley has the shape of a uniform solid disk of
mass 2.50 kg and diameter 0.520 m .
(a) After the system is released, find the horizontal tension in the wire.
(b) After the system is released, find the vertical tension in the wire.
(c) After the system is released, find the acceleration of the box.
(d) After the system is released, find magnitude of the horizontal and vertical components of the force that
the axle exerts on the pulley.




(e)

10.20. A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg .
The free end of the string is held in place and the hoop is released from rest (the figure (Figure 1)). After the
hoop has descended 95.0 cm , calculate:
(a) the angular speed of the rotating hoop and
(b) the speed of its center.




10.57. A thin, uniform 3.90-kg bar, 90.0 cm long, has very small 2.50-kg balls glued on at either end. It is
supported horizontally by a thin, horizontal, frictionless axle passing through its center and perpendicular to
the bar. Suddenly the right-hand ball becomes detached and falls off, but the other ball remains glued to the
bar.
(a) Find the angular acceleration of the bar just after the ball falls off.
(b) Will the angular acceleration remain constant as the bar continues to swing? If not, will it increase or
decrease?
(c) Find the angular velocity of the bar just as it swings through its vertical position.




10.80. A large turntable with radius 6.00 m rotates about a fixed vertical axis, making one revolution in 8.00 s. The
moment of inertia of the turntable about this axis is 1200 kg⋅m2. You stand, barefooted, at the rim of the
turntable and very slowly walk toward the center, along a radial line painted on the surface of the turntable.
Your mass is 71.0 kg . Since the radius of the turntable is large, it is a good approximation to treat yourself
as a point mass. Assume that you can maintain your balance by adjusting the positions of your feet. You find
that you can reach a point 3.00 m from the center of the turntable before your feet begin to slip. What is the
coefficient of static friction between the bottoms of your feet and the surface of the turntable?
10.81. In your job as a mechanical engineer you are designing a flywheel and clutch-plate system like the one
in figure. Disk A is made of a lighter material than disk B, and the moment of inertia of disk A about the shaft
is one-third that of disk B. The moment of inertia of the shaft is negligible. With the clutch disconnected, A is
brought up to an angular speed ω0; B is initially at rest. The accelerating torque is then removed from A,
and A is coupled to B. (Ignore bearing friction.) The design specifications allow for a maximum of 3600 J of
thermal energy to be developed when the connection is made. What can be the maximum value of the
original kinetic energy of disk A so as not to exceed the maximum allowed value of the thermal energy?




28.16. A very long, straight horizontal wire carries a current such that 8.25×1018 electrons per second pass any
given point going from west to east.
(a) What is the magnitude of the magnetic field this wire produces at a point 4.50 cm directly above it?
(b) What is direction of the magnetic field?

28.31. The wires in a household lamp cord are typically 2.5 mm apart center to center and carry equal currents in
opposite directions.
(a) If the cord carries current to a 100−W light bulb connected across a 120−V potential difference, what
force per meter does each wire of the cord exert on the other? (Model the lamp cord as a very long
straight wire.)
(b) Is the force attractive or repulsive?
(c) Is this force large enough so it should be considered in the design of lamp cord?




2
28.43. Coaxial Cable. A solid conductor with radius a is supported by insulating disks on the axis of a conducting
tube with inner radius b and outer radius c in the following figure. The central conductor and tube carry
equal currents I in opposite directions. The currents are distributed uniformly over the cross sections of each
conductor.
(a) Derive an expression for the magnitude of the magnetic field at points outside the central, solid
conductor but inside the tube (a < r < b).
(b) Derive an expression for the magnitude of the magnetic field at points outside the tube (r > c).




28.47. A solenoid is designed to produce a magnetic field of 3.50×10−2 T at its center. It has a radius of 1.50 cm and
a length of 30.0 cm , and the wire can carry a maximum current of 13.5 A .
(a) What minimum number of turns per unit length must the solenoid have?
(b) What total length of wire is required?


28.64. The long, straight wire AB shown in the figure carries a current of 14.0 A. The rectangular loop whose long
edges are parallel to the wire carries a current of 5.00 A.
(a) Find the magnitude of the net force exerted on the loop by the magnetic field of the wire.
(b) Find the direction of the net force exerted on the loop by the magnetic field of the wire.




28.65. Two long, parallel wires hang by 4.00-cm-long cords from a common axis. The wires have a mass per unit
length of 1.80×10−2 kg/m and carry the same current in opposite directions. What is the current in each wire
if the cords hang at an angle of 6.00 ∘ with the vertical?




3
Solutions

10.14. IDENTIFY: Apply å Fy = ma y to the bucket, with +y downward. Apply å t z = I a z to the cylinder, with
the direction the cylinder rotates positive.
SET UP: The free-body diagram for the bucket is given in Figure 10.14a and the free-body diagram for the
cylinder is given in Figure 10.14b. I = 12 MR 2 . a(bucket) = Ra (cylinder)

EXECUTE: (a) For the bucket, mg - T = ma. For the cylinder, åt z = Ia z gives TR = 12 MR 2a . a = a /R
then gives T = 12 Ma. Combining these two equations gives mg - 12 Ma = ma and
mg æ 15.0 kg ö 2 2
a= =ç ÷ (9.80 m/s ) = 7.00 m/s .
m + M /2 è 15.0 kg + 6.0 kg ø
T = m( g - a ) = (15.0 kg)(9.80 m/s 2 - 7.00 m/s 2 ) = 42.0 N.

(b) v 2y = v02y + 2a y ( y - y0 ) gives v y = 2(7.00 m/s 2 )(10.0 m) = 11.8 m/s.
(c) a y = 7.00 m/s 2 , v0 y = 0, y - y0 = 10.0 m. y - y0 = v0 yt + 12 a yt 2 gives
2( y - y0 ) 2(10.0 m)
t= = = 1.69 s
ay 7.00 m/s 2
(d) å Fy = ma y applied to the cylinder gives n - T - Mg = 0 and
n = T + mg = 42.0 N + (12.0 kg)(9.80 m/s 2 ) = 160 N.
EVALUATE: The tension in the rope is less than the weight of the bucket, because the bucket has a
downward acceleration. If the rope were cut, so the bucket would be in free fall, the bucket would strike the
2(10.0 m)
water in t = = 1.43 s and would have a final speed of 14.0 m/s. The presence of the cylinder
9.80 m/s 2
slows the fall of the bucket.




Figure 10.14
! !
10.16. IDENTIFY: Apply å F = ma to each box and åt z = Ia z to the pulley. The magnitude a of the acceleration
of each box is related to the magnitude of the angular acceleration a of the pulley by a = Ra .
SET UP: The free-body diagrams for each object are shown in Figure 10.16. For the pulley, R = 0 . 280 m
and I = 12 MR 2 . T1 and T2 are the tensions in the wire on either side of the pulley. m1 = 12.0 kg,
!
m2 = 5.00 kg and M = 2.10 kg. F is the force that the axle exerts on the pulley. For the pulley, let
clockwise rotation be positive.
EXECUTE: (a) å Fx = max for the 12.0 kg box gives T1 = m1a. å Fy = ma y for the 5.00 kg weight gives
m2 g - T2 = m2a. åt z = Ia z for the pulley gives (T2 - T1 ) R = ( 12 MR 2 )a . a = Ra and T2 - T1 = 12 Ma.
Adding these three equations gives m2 g = (m1 + m2 + 12 M )a and
æ m2 ö æ 5.00 kg ö 2 2
a=ç ÷g = ç ÷ (9.80 m/s ) = 2.71 m/s . Then
ç m1 + m2 + 1 M ÷ è 12. 0 kg + 5.00 kg + 1 .05 kg ø
è 2 ø




4
T1 = m1a = (12.0 kg)(2.71 m/s 2 ) = 32.6 N. m2 g - T2 = m2a gives
T2 = m2 ( g - a ) = (5.00 kg)(9.80 m/s 2 - 2.71 m/s 2 ) = 35.4 N. The tension to the left of the pulley is 32.6 N
and below the pulley it is 35.4 N.
(b) a = 2.72 m/s 2
...

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