MIME-Version: 1.0 Content-Type: multipart/related; boundary="----=_NextPart_01C5500E.AFA5F3A0" This document is a Single File Web Page, also known as a Web Archive file. If you are seeing this message, your browser or editor doesn't support Web Archive files. Please download a browser that supports Web Archive, such as Microsoft Internet Explorer. ------=_NextPart_01C5500E.AFA5F3A0 Content-Location: file:///C:/D9286902/Spring05.Phys1111.exam3.SolutionWeb.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" TIME OF COMPLETION_______________

TIME OF COMPLETION_______________&n= bsp;           NAM= E____SOLUTION_________________________

 

 

<= span style=3D'mso-bidi-font-size:10.0pt'> &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;    DEPARTMENT OF NATURAL SCIENCES

 

PHYS 1111, Exam 3            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;  Section 1

Version 1            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;          April 27, 2005

Total Weight: 100 points

 

1.     &nb= sp;   Check your examination for completeness prior to starting.  There are a total of ten (10) prob= lems on seven (7) pages.

 

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

 

3.     &nb= sp;   You will have 75 minutes to complete the examination.

 

4.     &nb= sp;   The total weight of the examination is 100 points.

 

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

 

6.     &nb= sp;   If you have any questions during the examination, see your instructor who will=

be located in the classroom.

 

7.     &nb= sp;   Start:        &= nbsp;      6:00 p.m.

Stop:&n= bsp;            = ;   7:15 p.m

 

 

 

 = ;             <= /span>PROBLEM

 

 = ;            &n= bsp;   POINTS

 

 = ;            &n= bsp;  CREDIT

 

     &nb= sp;            =    1-6

 

     &nb= sp;            =    20

 

 = ;

 

     &nb= sp;            =     7

 

     &nb= sp;            =    23

 

 = ;

 

8

 

     &nb= sp;            =    23

 

 = ;

 

     &nb= sp;            =     9

 

     &nb= sp;            =    17

 

 = ;

 

 = ;            &n= bsp;       10

 

 = ;            &n= bsp;       17

 

 

 

 = ;            &n= bsp;   TOTAL

 

 = ;            &n= bsp;      100

 

 

 

 

 

 = ;          PERCENTAGE

 

 

 

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

 

1.   The moment of inertia of a r= igid body 

 

a.&n= bsp;     Depends on the axis of rotation.

&n= bsp;

b.&n= bsp;    Cannot be negative.

   (4)

c.&n= bsp;     Depends on mass distribution.

&n= bsp;

d.     All of the above.<= o:p>

 <= /p>

 <= /p>

 

2.  If the period of a system undergoi= ng a SHM is doubled, the frequency of the system is 

 

a.&n= bsp;     Doubled.

&nb= sp;

b.     Halved.=

  (4)

c.&n= bsp;     For times as large.

 

d.&n= bsp;    One-quarter as large.

 

   <= /p>

 

 

 

3.  = Which one of the following quantities is zero when an object in simple harmonic motio= n is at its maximum displacement?

 

a.&n= bsp;     Potential energy.

 

b.&n= bsp;    Acceleration.

 (4)

c.&n= bsp;     Kinetic energy.

&n= bsp;

d.&n= bsp;    Frequency.

 

 

 

 

 

 

 

 

4. What condition or conditions are necessary for rotat= ional equilibrium? =

 

a= . ΣFx =3D 0.

 =

b= . ΣFx =3D 0, ΣΤ =3D 0.

(4)=

c. ΣΤ =3D 0.

 =

d= . ΣFx =3D 0, Σ= ;Fy =3D 0. =

 

 

5. What is the quantity used to measure an objec= t's resistance to changes in rotation?

 

  &nbs= p;         a. Mass.=

 

  &nbs= p;         b. Moment of inertia.=

(4)

  &nbs= p;         c. Torque.

 

  &nbs= p;         d. Angular velocity. =  

 

 

 

6.  If a net torque is applied to an o= bject, that object will experience:

 

    1. A constant angular velocity.

 

    1. An angular acceleratio= n.

(4)

    1. A constant moment of inertia.

 

    1. An increasing moment of inertia.

 

 

7. A person of mass 75.0 kg stands at the center of a rotating merry-go-round of radius 3.00 m and moment of inertia of 920 kgm2.  The merry-go-round rotates without friction with angular velocity of 2.00 rad/s. T= he person walks radially to the edge of the merry-go-round.

 

a. Calculate the angular ve= locity of the merry-go-round when the person reaches the edge.

 

Ii =3D 920 kgm2

 

If =3D  920 kg= m2 + (75.0 kg)(3.00 m)2  =3D 1595 kgm2

 

wi =3D 2.00 rad/s

 

Li  =3D Lf 

 

Ii  wi =3D  If  wf

 

wf  =3D Ii  wi / If

 

wf  =3D (920 kgm2) (2.00 = rad/s) / (1595 kgm2) =3D 1.15 rad/s

 

b. Calculate the rotational kinetic energy of the system before and after the person’s walk.=

 

KEi =3D  ½ Ii  wi 2 =3D  ½ (920 kgm2) (2.0= 0 rad/s)2 =3D 1840= J=

&nb= sp;

KEf =3D  ½ If  wf 2 =3D  ½ (1595 kgm2) (1.= 15 rad/s)2 =3D 1055= J=

 

 

d. If the merry-go-rou= nd rotates in a clockwise direction as viewed from above, what is the directio= n of its angular momentum?

 

 

Inward. (Right Hand Rul= e!)

 

 

8. A shop sign weighing 245 N is supported by a unifor= m beam as shown below. Find the tension in the wire and the horizontal and vertica= l forces exerted by the hinge on the beam. &nb= sp;

Remember to draw a free-b= ody diagram first!

(w)x =3D w cos(- 90o) =3D 0

(w)y =3D w sin= (- 90o) =3D - 245 N

 

(FV)x =3D FV cos(90o) =3D 0

(FV)y =3D FV sin(90o) =3D FV

 

(FH)x =3D FH cos(0o) =3D FH

(FH)y =3D FH sin(0o) =3D 0

 

 

(T)x =3D  T cos(= 145o)

(T)y =3D  T sin(145o)

 

SFx =3D 0        =        FH + T cos(145o) = =3D 0

      

 

SFy =3D 0        =       -245 N + FV + T sin(145o)  =3D 0 

 

 

 

tw =3D - (245 N)(1.70 m) =3D - 416.5 Nm

 

tT=3D + (T)(1.350 m) sin(35o)

 

tFH =3D 0   <= /o:p>

 

tFV =3D 0

 

St =3D 0

 

 -416.5 Nm + (T)(1.350 m) sin(= 35o) =3D 0        =     T =3D 538 N

 

FH =3D 441 N        =      

 

FV =3D -63.6 N

 

9. Neptune is an average distance of 4.50 x 109 km from the Sun. Estimate t= he length of the Neptunian year given that the Earth is 1.50 x 108 = km from the Sun on the average.

 

TN2= =3D (4p2/GMS )RN3

 

TE2= =3D (4p2/GMS )RE3

 

TN2= /TE2 =3D RN3/RE3

 

TN/TE =3D (RN3/RE3)1/2

 

TN/TE =3D ((4.50 x 109)3/(1.50 x 108)3)1/2  =3D 164

 

TN =3D 164 years

 

What is the Neptune’s orbital speed?

 

 

V =3D 2pR/T

 

V =3D 2p(4.50 x 1012m)/(164x365x24x3= 600 s) =3D 5467 m/s

 

 

10. A mass-spring system undergoes SHM in horizontal directi= on. If the mass is 0.250 kg, the spring constant is 12.0 N/m, and the amplitude= is 15.0 cm,

 

<= span style=3D'mso-list:Ignore'>a.&n= bsp;     What would be the maximum speed of the mass?

 

Vmax =3D {(k/m) A2}½

 

Vmax =3D {[(12.0 N/m)/(0.250 kg)](0.150 m)2}½ =3D 1.03 m/s

 

 

<= span style=3D'mso-list:Ignore'>b.&n= bsp;    Where would it occur?

 

@ x=3D 0

 

<= span style=3D'mso-list:Ignore'>c.&n= bsp;     What would be the speed at a half-amplitude positio= n?

 

V =3D {(k/m)(A2 – x2)}½

 

V =3D {[(12.0 N/m)/(0.250 kg)][(0.150 m)2 - (0.0750 m)2 ]}½ <= span style=3D'color:red'>=3D 0.900 m/s

 

 

<= span style=3D'mso-list:Ignore'>d.&n= bsp;    How many oscillations per second does the system go through?

 

f =3D (1/2p) (k/m)½=   =3D (1/2p) {(12.0 N/m)/(0.= 250 kg)}½  =3D 1.10 Hz<= /o:p>

 

 

<= span style=3D'mso-list:Ignore'>e.&n= bsp;     What would be the magnitude of the mass acceleratio= n at the half-amplitude position?

 

a =3D F/m

 

F =3D k x

 

a =3D k x/m

 

a =3D (12.0 N/m)(0.0750 m)/(0.250 kg) =3D 3.60 m/s2=

 

 

 

 

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PHYS 1111 Exam 3, Version 1

Spring 2005        =             &nb= sp;            =             &nb= sp;            =             &nb= sp;    7

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