MIME-Version: 1.0 Content-Type: multipart/related; boundary="----=_NextPart_01C5975A.3F91AE90" 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_01C5975A.3F91AE90 Content-Location: file:///C:/D9286902/Summer05.Phys1111.Exam3.SolutionWeb.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" TIME OF COMPLETION_______________

TIME OF COMPLETION_______________  &nb= sp;         NAME_____SOLUTIO= N________________________

 

 

<= span style=3D'font-family:"Times New Roman"'>&nb= sp;            =             &nb= sp;            =      DEPARTMENT OF NATURAL SCIENCES

 

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

Version 1           &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =            August 1, 2005

Total Weight: 100 points

 

1.         = Check your examination for completeness prior to starting.  There are a total of ten (10) prob= lems on eight (8) pages.

 

2.         = Authorized references include your calculator with calculator handbook, and the Refere= nce 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; part= ial 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:        &= nbsp;      11:30 a.m.

Stop:            &= nbsp;   12:50 p.m

 

 

 

&= nbsp;           &nbs= p; PROBLEM

 

&= nbsp;           &nbs= p;    POINTS

 

&= nbsp;           &nbs= p;   CREDIT

 

   &nbs= p;            &= nbsp;    1-6

 

   &nbs= p;            &= nbsp;    30

 

&= nbsp;

 

   &nbs= p;            &= nbsp;     7

 

   &nbs= p;            &= nbsp;    20

 

&= nbsp;

 

8

 

   &nbs= p;            &= nbsp;    15

 

&= nbsp;

 

   &nbs= p;            &= nbsp;     9

 

   &nbs= p;            &= nbsp;    15

 

&= nbsp;

 

&= nbsp;           &nbs= p;        10

 

&= nbsp;           &nbs= p;        20

 

 

 

&= nbsp;           &nbs= p;    TOTAL

 

&= nbsp;           &nbs= p;       100

 

 

 

 

 

&= nbsp;          PERCENTA= GE

 

 

 

 

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

 

  1. Whi= ch one of the following quantities is zero when an object in simple harmonic = motion is at its maximum displacement?

 

= a.<= span style=3D'font:7.0pt "Times New Roman"'>      Potential energy.=

(5)

= b.<= span style=3D'font:7.0pt "Times New Roman"'>     Acceleration.

 

= c.      Kinetic energy.

 <= /p>

= d.<= span style=3D'font:7.0pt "Times New Roman"'>     Frequency.

 

 

      2.   Two equal forces are applied to a door.  The first force is appl= ied at the midpoint of the         =        door; the second force is applied at the doorknob.  Both forces are applied perpendicu= lar         =      to the door.  Which force exerts = the greater torque?

 

  &nbs= p;            &= nbsp;           a. The first at the midpoint.

(5)

  &nbs= p;            &= nbsp;           b. The second at the doorkno= b.

 = ;

            &n= bsp;            = ;  c. Both exert equal non-zero torques.<= o:p>

 

            &n= bsp;            = ;  d. Both exert zero torques.

 

 

3. The quantity “angular momentum” (in terms of the fundamental quantities of mass, length, and time) is equivalent to: <= /p>

 

a.<= span style=3D'font:7.0pt "Times New Roman"'>      MLT -2.

 

b.     ML2T -1.

(5)

c.<= span style=3D'font:7.0pt "Times New Roman"'>      ML2T -3.=

 

d.<= span style=3D'font:7.0pt "Times New Roman"'>     ML3T.<= /p>

 

4.  Two moons orbit a planet in nearly circular orbits.  Moon A has orbital radius r, and m= oon B has orbital radius 4r.  Moon A= takes 20 days to complete one orbit.  How long does it take moon B to complete an orbit?

 

= a.<= span style=3D'font:7.0pt "Times New Roman"'>      20 days.

 

= b.<= span style=3D'font:7.0pt "Times New Roman"'>     80 days.

(5)

= c.<= span style=3D'font:7.0pt "Times New Roman"'>      160 days.

 

= d.<= span style=3D'font:7.0pt "Times New Roman"'>     320 days.

 

 

 

5.  The moment of inertia of a spinnin= g body about its spin axis depends on its:

 

= a.<= span style=3D'font:7.0pt "Times New Roman"'>       Angular speed, shape, and mass.

 

= b.<= span style=3D'font:7.0pt "Times New Roman"'>     Angular acceleration, mass, and axis position.

(5)

= c.<= span style=3D'font:7.0pt "Times New Roman"'>      Mass, shape, axis positio= n, and size.

 

= d.<= span style=3D'font:7.0pt "Times New Roman"'>     Mass, size, shape, and speed.<= /o:p>

 

 

 

 

6. A boy and a girl are riding on a merry-go-rou= nd which is turning at a constant rate.  The boy is near the outer edge, and the girl is closer to the center.  Who has the greater l= inear speed? =

 

    1. T= he boy.

 

    1. T= he girl.

(5)

    1. B= oth have the same non-zero translational velocity.

 

    1. B= oth have zero translational velocity.

 

 

 

 

7.  A 64.0-kg person stands on a weigh= tless diving board supported by two pillars, one at the enof the board, the other 1.10-m away. The pillar at the end of the board exerts a downward force of = 828 N. The board has a length of 3.00.

  1. How far from that pillar is the person standing?

&nb= sp;

&nb= sp;

N= 1x =3D N1 cos (-90o) =3D 0

N= 1y =3D N1 sin (-90o) =3D - N1 =3D -828 N=

N= 2x =3D N2 cos (90o) =3D 0

N= 2y =3D N2 sin (90o) =3D N2<= /p>

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

w= y =3D w sin (-90o) =3D - w =3D - m g =3D  - (64.0 kg)(9.80 m/s2) = =3D - 627 N

SFx =3D 0  =3D>   0 =3D 0

SFy =3D 0  =3D>    (-828 N) + N2 – (627 N)  =3D 0

N= 2 =3D  1455 N

tN2 =3D   N2 (1.10 m) =3D (1455 N)= (1.10 m) =3D 1601 Nm

tN1 =3D  0

tw =3D   - (627 N) (x)

St =3D 0 

- (627= N) x + 1601 N m =3D 0

x =3D = 2.55 m

&nb= sp;

&nb= sp;

  1. Find the force exerted by the = second pillar. 

 

N= 2 =3D  1455 N

 

8.      A potter= 217;s wheel is rotating around a vertical axis through its center at a frequency = of  The wheel c= an be considered a uniform disk of mass 5.00 kg and diameter 0.400 m. The potter = then throws a 3.10-kg chunk of clay, approximately shaped as a flat disk of radi= us 8.00 cm, onto the center of the rotating wheel.

&nb= sp;

    1. What is the frequency of the = wheel after the clay sticks to it?

<= /span>

Ii<= /sub> =3D ½ mw R2 =3D ½ (5.00 kg)(0.200 m)2 =3D= 0.100 kg m2

 

If<= /sub> =3D ½ mw R2 + ½ mc R2 =3D &fra= c12; (5.00 kg)(0.200 m)2 + ½ (3.10 kg)(0.0800 m)2 =3D 0.1= 0992 kg m2

 

Ii<= /sub> wi =3D If wf

 

wf =3D (Ii /If) wi =3D 1.36= rev/s

<= /span>

 angular momentum?of the wheel'e dir= ection as viewed from above, what is the direction of the system'

    1. (Bonus) If the= wheel rotates in clockwise direction as viewed from above, what is the direction of the wheel’s angular momentum?

&nb= sp;

&nb= sp;

 

 

 

 

 

 

9. A 0.500-kg m= ass resting on a horizontal frictionless surface is connected to a fixed spring with spring constant of 80.0 N/m. The mass is displaced 16.0 cm from its equilibrium position and released.

 

  1. What is the amplitude of oscillations?

 

A =3D 0.160 m

 

  1. What is the maximum speed of = the mass?

 

Vmax= =3D (k/m)1/2 A =3D ((80.0 N/m)/(0.500 kg)) 1/2 (0.160 m) =3D 2.02 m/s

 

 

  1. What is the total mechanical energy of the system?

 

TE =3D ½= m Vmax2 =3D ½ (0.500 kg) (2.02 m/s)2 =3D 1.02 J

 

  1. What is the frequency of oscillations?

 

f =3D (1/(2p))(k/m)1/2 =3D 2.01= Hz

 

  1. What is the period of oscillations?

 

T =3D 1/f =3D 0.497 s

 

10. A uniform disk of mass 1.00 kg and radius 10.0 m has four symmetrically pla= ced masses, each of mass 0.250 kg fastened at positions 5.00 m from the center.= The disk is free to rotate around its

center.

 

a.<= span style=3D'font:7.0pt "Times New Roman"'>      What is the moment of inertia of th= e disk relative to its center?

 

 

I =3D &f= rac12; mdisk R2 + 4 m r2 =3D ½ (1.00 kg)(1= 0.0 m)2 + 4 (0.250 kg)(5.00 m)2 =3D 75.0 kg m<= sup>2

 

 

b.<= span style=3D'font:7.0pt "Times New Roman"'>     Two tangential forces are applied t= o the disk as shown in the figure.

What is the net torque = acting on the disk?

 

 

 

t1 =3D F1 d1 =3D (40.0N)(10.0m) =3D 400 Nm=

 

t2 =3D - F2 d2 =3D - (20.0N)(5.00m) =3D -100 Nm

 

St =3D 400 Nm – 100 Nm =3D 300 Nm

 

c.<= span style=3D'font:7.0pt "Times New Roman"'>      What is the angular acceleration of= the disk?

 

 

St =3D I a <= /span>

 

a =3D St / I =3D (300 Nm)/ (75.0 kgm2<= /sup>) =3D 4.00 rad/s2  

 

 

d.<= span style=3D'font:7.0pt "Times New Roman"'>     Assuming that initially the system = was at rest, what is the kinetic energy of the disk 10.0 s after the start of the motion?

 

 

 

 

KE =3D ½ I w2 =3D ½ I (w0 + a t)2 =3D= ½ (75.0 kgm2)((4.00 rad/s2)(10.0 s))2 =3D 60,000 J

 

 

 

 

 

 

 

 

 

 

 

 

 

 

------=_NextPart_01C5975A.3F91AE90 Content-Location: file:///C:/D9286902/Summer05.Phys1111.Exam3.SolutionWeb_files/header.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii"

 

 

= 8

PHYS 1111 Exam 3, Version 1

Summer 2005

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