TIME OF COMPLETION___SOLUTION____________            NAME_____________________________

 

 

                                           DEPARTMENT OF NATURAL SCIENCES

 

PHYS 1111, Exam 3                                                                                                      Section 1

Version 1                                                                                                                  July22, 2003

Total Weight: 100 points

 

1.         Check your examination for completeness prior to starting.  There are a total of ten (10) problems on seven (7) pages.

 

2.         Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet  (provided by your instructor).

 

3.         You will have 80 minutes to complete the examination.

 

4.         The total weight of the examination is 100 points.

 

5.         There are six (6) multiple choice and four (4) calculation problems. Work all problems.  Show all work; partial credit will be given for correct work shown.

 

6.         If you have any questions during the examination, see your instructor who will

be located in the classroom.

 

7.         Start:               11:00 a.m.

Stop:                12:20 p.m

 

 

 

              PROBLEM

 

                 POINTS

 

                CREDIT

 

                     1-6

 

                     30

 

 

 

                      7

 

                     20

 

 

 

8

 

                     15

 

 

 

                      9

 

                     20

 

 

 

                     10

 

                     15

 

 

 

                 TOTAL

 

                    100

 

 

 

 

 

           PERCENTAGE

 

 

 

 

CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT.

 

1. The mass on a spring vibrates 14.6 times in 19 s. Its frequency is

 

a.      2.10 Hz.

 

b.     0.650 Hz.    F = (# of oscillations)/(time) = 14.6/19 Hz

(5)

c.      0.768 Hz.

 

d.     1.30 Hz.

 

 

 

2.  The center of gravity

 

e.      Is always at the geometric center of a body.

 

f.       Always has some material at it.

(5)

g.      Is always outside of a body.

 

h.      None of the above.

 

 

3. A cylinder with its mass concentrated toward the center has a moment of inertia I = 0.100 M R2.

If this cylinder is rolling without slipping along a level surface with a linear velocity v, what is the ratio of its rotational kinetic energy to its linear kinetic energy?

 

a.      1/10.     KER = ˝ Iw2 = ˝ (0.100 MR2) w2       KEL = ˝ mV2

 

b.     1/5.      KER / KEL = ˝ (0.100 MR2) w2 / (˝ mV2 )= 0.100, since  Rw = V

(5)

c.      1/2.

 

d.     1/1.

 

 

 

4.  The Earth moves about the Sun in an elliptical orbit. As the Earth moves closer to the Sun, which of the following best describes the Earth-Sun system’s moment of inertia?

 

a.      Decreases.

 

b.     Increases.

(5)

c.      Remains constant.

 

d.     None of the above.

 

 

  5.   A large spring requires a work of 150 J to be done on it in order to compress it by 0.100 m.

What is the spring constant of the spring? 

 

a.      125,000 N/m.   W = ˝ kx2                    150 J = ˝ k (0.100 m)2

 

b.     30,000 N/m.

            (5)

c.      10.0 N/m.

 

d.     1.25 N/m.

 

 

 

6.  A mass is attached to a spring and oscillates on a frictionless horizontal surface.

 

a.      The displacement and the acceleration of the object are always in the opposite direction.

(5)

b.     The acceleration and the velocity of the object are always in the same direction.

 

c.      The acceleration and the velocity of the object are always in the opposite direction.

 

d.     The displacement, the velocity, and the acceleration are always in the same direction.

 

 

 

7. A cat named Feynman  (of 4.50 kg mass) is sitting on a horizontal bar supported by two vertical cables. The length of the bar is 2.00 m and its mass is 10.0 kg. The right cable exerts a force of 66.5 N on the bar.

 

 

a.       What is the tension in the left cable?

 

(wC)x = MC g cos (- 90o) = 0

(wC)y = MC g sin (- 90o) = - (4.50 kg) (9.80 m/s2) = - 44.1 N

 

(wB)x = MB g cos (- 90o) = 0

(wB)y = MB g sin (- 90o) = - (10.0 kg) (9.80 m/s2) = - 98.0 N

 

(TR)x =  TR cos (90o) = 0

(TR)y =  TR sin (90o) = 66.5 N

 

(TL)x =  TL cos (90o) = 0

(TL)y =  TL sin (90o) = TL

 

SFx = 0

SFy = 0                     - 44.1 N – 98.0 N + 66.5 N  + TL  = 0    TL = 75.6 N

 

 

b.   How far from the left cable does Feynman sit?

 

tTL= 0

tTR = + (66.5 N)(2.00 m) = 133 Nm

tWC= - (44.1 N) (x)

tWB= - (98.0 N) (1.00 m) = -98.0 Nm

St = 0

 133 Nm – (44.1 N)(x) – 98.0 Nm = 0          x = 0.793 m (from the left)

 

 

 

 

 

 

 

 

 

 

 

8. A cat named Feynman is playing with his toy mouse which happens to be attached to the wall by a light spring. Amazingly enough the horizontal floor is frictionless (don’t ask how). Feynman discovers that by applying the 1.00-N force he is able to stretch the spring by 12.0 cm relative to its equilibrium position. The mass of the toy is 0.200 kg.

 

    1. What is the spring constant of the spring?

 

F = k x 

k = F/x

k = 8.33 N/m

 

    1. Feynman stretches the spring by 12.0 cm and lets the toy go. The mouse starts to oscillate. What is the frequency of the resulting oscillations?

f = 1/2p sqrt(k/m) = 1/2p sqrt {(8.33 N/m)/(0.200 kg)} = 1.03 Hz

 

    1. What is the maximum kinetic energy of the toy mouse?

 

KEmax = PEmax = ˝ k A2 = ˝ (8.33 N/m) (0.120 m)2 = 0.0600 J

 

    1. What is the position of the mouse when its velocity is 0.200 m/s?

 

V = sqrt{k/m(A2 – x2)}

 

V2 = k/m (A2 – x2)

 

mV2/k  = A2 – x2

 

- mV2/k  + A2  =  x2

 

x = 0.116 m

    1. How long does it take the mouse to go from being 12.0 cm away from the equilibrium to the equilibrium position? (Hint: think about it in terms of a  period.)

T = 1/f

T = 0.973 s

 

t = T/4 = 0.243 s

 

 

 

9. A thin hollow sphere of mass 1.00 kg is filled with a liquid of mass 0.500 kg. The filled sphere is spinning freely with no friction at angular speed 1.00 rad/s about an axis through the center of mass of the sphere. The hollow sphere and all parts of the fluid have the same angular speed. A leak develops along the axis of the sphere and all the fluid leaks out. What is the angular speed of the now hollow sphere? (The moment of inertia of a thin spherical shell I = 2/3 mR2 , a sphere I = 2/5 mR2 ).

(Hint: I did promise you a problem on the conservation of angular momentum.)

 

.

 Ii  = 2/3 mSR2 + 2/5 mLR2

 If = 2/3 mSR2

 

  Li  = Lf 

 

 (2/3 mSR2 + 2/5 mLR2) wI = (2/3 mSR2 ) wf

 

 (2/3 mS + 2/5 mL) wI = (2/3 mS) wf

 

 wf = 1.30 rad/s

 

10. Find the center of mass of the collection of mass points in the Figure below. All masses sit on the light (neglect its mass) rigid rod.

 

 

 

XCM = (m1x1+  m2x2 + m3x3) / (m1 +  m2 + m3) = ((1.00 kg)(0.00 m) + (2.00 kg)(4.00 m) + (3.00 kg)(6.00 m))/(1.00 kg + 2.00 kg + 3.00 kg)

 

XCM = 4.33 m

 

Now we start to rotate the rod with the masses on it around a vertical axis going through the 1.00 kg mass. What is the moment of inertia of the system relative to this axis?

 

I = m1x12 +  m2x22  + m3x32 = (1.00 kg)(0.00 m)2 + (2.00 kg)(4.00 m)2 + (3.00 kg)(6.00 m)2 = 140 kg m2

 

 

Eventually the system reaches the angular velocity of 2.50 rad/s. What is the rotational kinetic energy of the system at this velocity?

 

KER = ˝ Iw2 = ˝ (140 kg m2) (2.50 rad/s)2 = 438 J

 

 

 What amount of work needed to be done on the system to reach this angular velocity?

 

W = KEF – KEI = 438 J – 0 J = 438 J