Physics 125 Winter 2008: Exam #3 Solution
Form #1
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Instructions
! Fill in your name above.! Each exam will have a different version number which will need to be written in the
area of the scantron marked “Form”.
! Record all your answers on the answer sheet provided.
! This will be a 1.5 hour, closed book exam. The exam includes 20 questions.
! You may use three 3x5 note cards containing information you think might be useful.
You can write on both sides of the cards if you like.
! This is a closed-book exam. You may use a calculator which is for numerical calculations only and not to refer to stored information.
! There is only ONE correct answer for each multiple-choice question. No partial credit
will be awarded.
! You may be asked to show your University student ID card when you turn in your exam.
Discussion section reminder:
Moments of inertia for some
geometric shapes
Data:
g = 9.8 m/s
2G = 6.673
x
10
-11N m
2/ kg
21. A wheel is attached to a fixed shaft through its center and the system is free to rotate without friction. A tape of negligible mass wrapped around the shaft is pulled, starting from rest, with a known constant force, F = 13 N. When a length L = 5m of tape has unwound, the system is rotating with angular speed ! = 0.5 rad/s. What is the moment of inertia of the system, I ?
The tape comes off the wheel without slipping so
Now use kinematics to find the angular acceleration :
Now apply Newton's 2 Law for rotation:
2. A thin rod of length L and mass M rotates with angular speed ! about an axis through
one of its ends and perpendicular to its length. The rotating rod has kinetic energy K. If the rod rotates instead at angular speed 2! about an axis through its center and
perpendicular to its length, the kinetic energy will be:
2
end end end ctr ctr ctr
end
end end
ctr
ctr ctr ctr ctr
ctr end
end end end end
3. A 5.0 m long strut of negligible weight is hinged to a wall and used to support an 800-N block as shown. The tension in the horizontal rope is:
.
3 3
(4 m) - (3 m) = 0 (800
4 4
V
H V H V
From the free body diagram of the hanging weight: Recognizing that the strut is the hypotenuse of a 3-4-5 right triangle and summing the
torques about the hinge:
T W
T T T T
#
( # # N) = 600 N
A) 400 N B) **600 N C) 800 N D) 1000 N E) 1200 N
4. The drawing shows a rectangular block of wood. The length from A to D is 1.5 m and the length from A to B is 0.75 m. The force applied to corners B and D have the same magnitude of 6.5 N and are directed parallel to the long and short sides of the rectangle, as shown. An axis of rotation is shown perpendicular to the plane of the rectangle at its center. What are the magnitude and direction of the net torque on the block?
0.75 m 1.50 m
(6.5 N) 2.43 N m 2.4 N m
Taking torques about the given axis: AB AD
2 2 2 2
F F
' # , # *- , +.# , (
/ 0 clockwise.
A) 4.9 N m; clockwise B) 9.8 N m; clockwise C) **2.4 N m; clockwise D) 0 N m
5. In class we demonstrated that the torque due to the weight of a gyroscope causes it to precess (rotate) in a horizontal circle and that the angular speed 1 for precession about a vertical axis through the pivot point at the end of the gyroscope can be expressed as
L
'
1 # , where ' and Lare the magnitudes of the torque and angular momentum vectors respectively, as shown in the diagram below. Neglecting the mass of the support frame, treat the gyroscope as spinning wheel of mass m with rotational inertia I spinning about the x axis with angular speed&. The weight of the spinning wheel acts at its center which is a distance r from the support pivot as shown. The axis of the spinning wheel (the x axis) is parallel to the table top. The precession angular speed is therefore:
The magnitude of the angular momentum for the spinning wheel is and the torque due to the wheel's weight acting downward at a horizontal distance from the pivot axis has magnitude so the
pr
I mg
r rmg
&
' #
ecession angular speed is rmg
I&
1 #
A) I
rmg
&
1 #
B) None of the given expressions is correct C)
** rmg
I&
1 #
D) rIg
m&
1 #
E) rm
Ig
&
6. The Earth has a radius of 6.38 x 106 m and rotates on its axis once every 24 hours. Suppose that the Earth suddenly collapsed to a radius of 137 m (giving it approximately the density of a neutron star) under the action of internal forces. What would the rotation period be for the collapsed Earth with its new 137 m radius? Assume that the Earth can be treated as a uniform sphere in both cases. Possibly useful information: The moment of inertia of a uniform sphere rotated about an axis passing through its center is
7. Galileo is said to have discovered the laws governing the motion of a pendulum by observing lamps suspended by long cords from the ceiling swinging back and forth in the cathedral of Pisa. If the period of a swinging lamp was 6.5 s, what was the length of the rope from which the lamp was suspended?
2 2
8. The hull of a WW II German U-boat submarine could withstand a gauge pressure of 20 atmospheres (2.026 x 106 Pa) before crushing. At what depth under water would the hull be crushed? Use"water #10 kg/m3 3.
gauge-max water crush crush
gauge-max crush
water
where is the depth under the surface of the water
at which the hull is crushed
9. A figure skater spins on frictionless ice with her arms extended. She then slowly pulls in her arms. Which statement best describes the situation while she is pulling in her arms?
Since only internal forces (and hence internal torques) act, her angular momentum is conserved. Pulling in her arms reduces her moment of inertia so her angular velocity must increase so that angular momentum is conserved.
A) Her angular velocity and her angular momentum both increase.
B) Her angular velocity and her angular momentum both remain the same.
C) **Her angular velocity increases and her angular momentum remains the same. D) Her angular velocity remains the same and her angular momentum increases. E) Her angular velocity increases and her angular momentum decreases.
10. Which one of the following quantities would you need to increase in order to increase the periodof simple harmonic motion for a mass attached to a spring oscillating on a horizontal frictionless surface?
2
2 2
SHM
To the , the .
k m
T T
m k
2
& & 2 2
&
# # ( # # (
increase period increase mass
A) The amplitude of the motion. B) The frequency.
C) The spring constant. D) **The mass
11. A stoplight with a mass of 24 kg is hung vertically from a steel wire with a
cross-1.0210 m 1 millimeter (2.0 10 N/m )(3.0 10 m )
12. A uniform solid ball floats in ethyl alcohol ( 3
806 kg/m
"# ) with 20% of its volume above the surface of the liquid. What is the specific gravity of the ball?
in Archam edes' Principle: The m agnitude of the buoyant force equals the weight of the displaced fluid. Since the ball is floating in equilibrium the buoyant force equals the ball's weight. Call the v
V
B fluid ball fluid alcohol in ball ball ball
ball
in in
alcohol in ball ball ball alcohol
ball alcohol ball
olum e of ball floating subm erged in the alcohol
t submerged in
13. A mass is undergoing simple harmonic motion according to the equationx#1.2 cos(7.4 )t , where x is in meters and t is in seconds. What is the maximum acceleration experienced by the mass?
2
( ) cos( ).
1.2 m and 7.4 rad/s
The general form for SHM is Since is in meters and is in seconds, the units are correct and we identify
. The general form for the acceleration in SHM is
x t A t x
14. A wooden beam is being acted upon by various external forces. The sum of the external forces is zero. What can you conclude?
0 0
For the beam to be in equilibrium the must be and the about any chosen axis must be
The
F # ' #
7
!7
!
net exteral force zero
net external torque zero :
net external f is stated to be so it cannot have a acceleration, but the due to these external forces
is stated to be , so on the beam
and i
orce zero, linear
net external torque
not zero a net external torque may exist
t may have an angular acceleration.
A) The beam has no linear acceleration and no angular acceleration.
B) The beam has no angular acceleration but may have a linear acceleration. C) **The beam has no linear acceleration but may have an angular acceleration. D) The beam has both a linear and an angular acceleration.
15. The spring in a mechanical wristwatch produces a clockwise torque on the balance wheel of the watch when the balance wheel undergoes a counterclockwise rotation through an angle $ according to the relationship ' # ,8$ where 8is a constant. The balance wheel is a uniform disk of mass m and radius r pivoted about an axis through its center ( 1 2
2
I # mr ). If the balance wheel is rotated to an initial angle $0and then released, the period for the resulting angular oscillations of the balance wheel is:
The torque has the form of Hooke's Law:
where the minus sign indicates that it is a restoring torque, that is, a torque is produced for a
Torque = - (constant)(angular displacement)
clockwise counterclock
angular displacement. Simple Harmonic Moition (SHM) is possible
for this system apply Newton's 2 Law and solve for the acceleration:
16. A bowling ball rolls without slipping and encounters a 0.760-m vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore rolling friction or air resistance and assume that ball is a uniform solid sphere. The translational speed of the ball is 3.50 m/s at the bottom of the rise. Find the translational speed at the top.
W e ig nore rolling friction and air resistance, so th ere is no nonconservative w ork . D efine the ze ro for gravitational potential energy at th e botto m of the rack and apply the W ork-E ne rgy T he orem , includ
ing both ro ta tional and transla tional k inetic energy te rm s:
17. What force must be applied to the small piston on the right hand side of the hydraulic lift to support the car? The weight of the car and lift combined is 47,200 N. The piston is circular in cross-section with a radius of 2 cm. The lift is also circular in cross section with a radius of 30 cm.
small
A force applied to the small piston produces a pressure which
by Pascal's Principle is applied uniformly to the fluid and the vessle walls. The applied pressure produces an upward force on t
F
he large piston
which equals the weight of the car plus lift since
the car and lift are in equilibrium
18. In the figure below a light rope is wrapped around a wheel of radius R= 2.0 m. The wheel is free to rotate about its center on frictionless bearings. A 14 kg block is suspended from the end of the rope. When the block is released from rest the block descends 10 m in 2.0 s. The rope does not slip relative to the wheel as the mass falls. What is the moment of inertia of the wheel?
nd
nd
Apply Newton's 2 Law to the falling mass taking downward as positive: Apply Newton's 2 Law to the accelerating wheel:
Since the rope comes off the pully without slipping
mg T ma I RT
Substitute into Find the
acceleration from the given information using kinematics:
19. Three identical 1kg balls are held in a triangle with a 2m edge length by massless rods. What is the rotational inertia of this system around an axis through the center of the solid colored ball, perpendicular to the paper? Treat the balls as point masses.
2m
2m
2m
2 1
2 m
(1
The axis passes through the black ball, which therfore contributes zero to the rotational inertia since it is at zero radius. The remaining two balls are each
from the rotation axis: N
i i i
I m R
#
#
7
# 2 2 2 2kg)(2 m) )(1 kg)(2 m) #2(4 kg m )#8 kg m
20. The wonderful thing about Tiggers, is that Tiggers are wonderful things. Their heads are made out of rubber, and their tails are made out of springs. If Tigger's tail is a spring with force constant k = 5000 N/mand he has a mass m = 20 kg, how high does Tigger rise above the ground before coming momentarily to rest if he compresses his tail a distance of 0.5 m and then launches himself vertically into the air with his tail? Neglect any friction or air resistance.
Apply the Work-Energy Theorem to Tigger. The only forces that act are the conservative Hooke's Law spring force and gravity. Tigger has no kinetic energy at launch and none when he comes momentarily t
2
o rest at apogee (the highest point of his bounce, errrr uh... ). Define the zero for gravitational potential energy to be at Tigger's launch elevation:
kx
“What will we do about poor little Tigger? If he never eats nothing, he’ll never get bigger. He doesn’t like honey or haycorns or thistles, Because of the taste or because of the bristles. And all the good things that an animal likes,
Have the wrong sort of swallow or too many spikes. But whatever his weight in pounds, shillings or ounces, He’ll always seem bigger because of his bounces.”
