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TIME OF COMPLETION_______________  &nb= sp;         NAME____SOLUTION= _________________________

 

 

<= 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;            =       December 6, 2005

Total Weight: 100 points

 

1.         = Check your examination for completeness prior to starting.  There are a total of nine (9) 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.         = You will have 50 minutes to complete the examination.

 

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

 

5.         = There are five (5) multiple choice and four (4) calculation problems. Work all multiple choice problems and 4 calculation 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:        &= nbsp;      10:30 a.m.

Stop:            &= nbsp;   11:45 a.m

 

 

 

&= nbsp;           &nbs= p; PROBLEM

 

&= nbsp;           &nbs= p;    POINTS

 

&= nbsp;           &nbs= p;   CREDIT

 

   &nbs= p;            &= nbsp;    1-5

 

   &nbs= p;            &= nbsp;    20

 

&= nbsp;

 

   &nbs= p;            &= nbsp;     6

 

   &nbs= p;            &= nbsp;    20

 

&= nbsp;

 

7

 

   &nbs= p;            &= nbsp;    20

 

&= nbsp;

 

   &nbs= p;            &= nbsp;     8

 

   &nbs= p;            &= nbsp;    20

 

&= nbsp;

 

&= nbsp;           &nbs= p;         9

 

&= 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 MULTIPL= E CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT.

 

1.  Rolling without slipping depends on  

  &nbs= p;        

  &nbs= p;         a. Kinetic friction between the rolling object and the ground.

  &nbs= p;        

  &nbs= p;         b. Static friction between t= he rolling object and the ground.

(4)  &= nbsp;   

  &nbs= p;         c. Tension between the rolling object and the ground.

  &nbs= p;        

  &nbs= p;         d. The force of gravity between the rolling object and the Earth.

 

 

 

 

2.  A car is negotiating a flat circular curve of radius 50 m with a speed of 20 m/s.  The maximum centripetal = force (provided by static friction) is 1.2 × 104 N. What is the centripetal acceleration of the car?

 

  &nbs= p;         a. 0.40 m/s2

 

  &nbs= p;         b. 0.80 m/s2

(4)

  &nbs= p;         c. 4.0 m/s2

 

  &nbs= p;         d. 8.0 m/s2    acp =3D Vt2/R =3D (20.0 m/s)2/(50.0 m)

  

 

 

 

3. The gravitational force between two objects i= s proportional to  

 

  &nbs= p;         a. The distance between the two objects.

 

  &nbs= p;         b. The square of the distance between the two objects.

(4)

  &nbs= p;         c. The product of the masses= of the two objects. <= o:p>

 

  &nbs= p;         d. The square of the product of the masses of the two objects.

 

 

 

 

4.  Two equal forces are applied to a door at the doorknob.  The first force is applied perpend= icular to the door; the second force is applied at 30° to the plane of the door.  Which force exerts the greater torque?

 

  &nbs= p;         a. The first applied perpend= icular to the door.

 

  &nbs= p;         b. The second applied at an angle.

(4)

  &nbs= p;         c. Both exert equal non-zero torques.

 

d. Both exert zero torques.

 

 

 

 

5. Two uniform solid spheres have the same mass,= but one has twice the radius of the other.&nbs= p; The ratio of the larger sphere's moment of inertia to that of the smaller sphere is =

 

  &nbs= p;         a. 4/5.

 

  &nbs= p;         b. 8/5.

(4)

  &nbs= p;         c. 2.

 

  &nbs= p;         d. 4.        =        (2/5 M (2R)2)/(2/5 M R2) =3D 4

=  <= /span>

6.       Three children are trying to balance on a seesaw, which consists of a fulcrum roc= k, acting as a pivot at the center, and a very light board 3.6 m long. Two playmates = are already on either end. Boy A has a mass of 50 kg, and girl B a mass of 35 k= g.

a. Whe= re should girl C, whose mass is 25 kg, place herself so as to balance the sees= aw?

 

 

tA =3D (50.0kg)(9.81 m/s2)(1.80 m) =3D 883 N-m

 

tB =3D -  (35.0kg)(9.81 m/s2)(1.80 m) =3D - 618 N-m

 

tC =3D - (25.0kg)(9.81 m/s2)(x) =3D - (245N)(x)

 

St =3D 0

 

883 N-m - 618 N-m – (245 N)(x) =3D 0=

 

x =3D 1.08 m

 

b. What is the magnitude of the nor= mal force that the rock exerts on the board?

 

Nx =3D 0=

 

Ny =3D N

 

wAx =3D 0

 

wAy =3D - mA g

 

wBx =3D 0

 

wBy =3D - mB g

 

wCx =3D 0

 

wCy =3D - mC g

 

 

SFx =3D 0

 

SFy =3D 0

 

N - mA g – m<= sub>B g – mC g =3D 0

 

N =3D mA g + m= B g + mC g =3D (mA + mB + mC)g = =3D 1079 N

 

 

7.    Suppose our Sun eventually collapses into a white dwarf, losing about half its mass= in the process, and winding up with a radius 1.0% of its existing radius.

a. Assuming the lost mass carries away no angul= ar momentum, what would the Sun’s new   rotation rate be? (Take the Sun’s current period to be about 30 days.)

II wI<= /sub> =3D IF wF<= /span>

II =3D 2/5 M0 R02

II =3D 2/5 (M0/2)(R0= /100)2

(2/5 M0 R02) wI =3D (2/5 (M0/2)(R0/100)2) wF

 <= /span>wI =3D (1/2)(1/100)2) wF<= /sub>

wI =3D (1/20000) wF

wF =3D (20000) wI

      = ;      wI =3D (2p)/T =3D (2p)/(30 days x (24 hours/1 day)(3600 s/1hour)) = =3D 2.42 x 10-6 rad/s

wF =3D 4.84 x 10-2 rad/s

(To practice, convert the new angular velocity = into the new period)

b. What would be its final ke in terms of its initial ke of today?

K =3D ½ I w2

 

KI =3D ½ (2/5 M0 R02) wI2

KF =3D ½ (2/5 (M0/2)(R0/100= )2 ) w= F2

KF /KI =3D [½ (2/5 (M0/2)(R0/100= )2 ) wF2] /  [½ (2/5 (M0)(R0)2 ) <= span style=3D'font-family:Symbol;mso-bidi-font-style:italic'>wI2]

 

KF /KI =3D [(1/2)(1/100)2 ) wF2] / [wI2] = =3D (1/20000)(wF2/ wI2) = =3D 20000

 

 

8.      The rings of Satur= n are composed of chunks of ice that orbit the planet. The inner radius of the rings is 73= ,000 km, while the outer radius is 170,000 km. Find the period of an orbiting ch= unk of ice at the inner radius and the period of a chunk at the outer radius. Compare your numbers with Saturn’s mean rotation period of 10 hours a= nd 39 minutes. The mass of Saturn is

 

 

 

TO2 =3D ((4p2)/(GMS))RO3

 

TO2 =3D ((4p2)/(6.673 x 10-11 N-m2/kg2)(5.70 x 1026 kg)))(170,000,000 m)3 =3D 5.06 x 109 s2

 

TO  =3D 7.11 x 104 s = =3D 19.8 hours

 

TI2 =3D ((4 p2)/(GMS))RI3

 

TI2 =3D ((4 p2)/(( 6.673 x 10-11 N-m2/kg2)(5.70 x 1026 kg)))(73,000,000 m)3 =3D 4.00 x 108 s2

 

TI =3D 2.00 x 104 s =3D 5.56 hours=

 

 

 

 

 

 

 

 

 

 

 

 

 

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9. After you pick up a spare, your bowling ball rolls without slipping back to= ward the ball rack with a linear speed of 2.85 m/s. To reach the rack, the ball = rolls up the ramp that gives the ball a 0.530-m vertical rise. What is the speed = of the ball when it reaches the top of the ramp?

 

Ei =3D m g yi= +  ½ mVi2  +  ½ Iwi2&= nbsp; =3D 0 +  ½ mVi2  +  ½ Iwi2&= nbsp; =3D  ½ mVi2  +  ½ (2/5mR2)w2   =3D  ½ mVi2  +  &fr= ac12; (2/5)mVi2&= nbsp;

 

Ef =3D m g yf= +  ½ mVf2  +  ½ Iwf2   =3D  m g yf +  &f= rac12; mVf2  +<= sup>  ½ (2/5mR2)w2   =3D m g yf +  ½ mVf2  +  ½ (2/5)mVf2&= nbsp;

 

Ei =3D Ei

 

        =     Vi2  + (2/5)Vi2&= nbsp; =3D 2 g y= f + Vf2  +=  (2/5= )Vf2&= nbsp;

 

(7/5= )Vi2&= nbsp; =3D 2 g y= f + (7/5)Vf2&= nbsp;

 

Vi2&= nbsp; =3D (5/7)= (2 g yf)+ Vf2&= nbsp;

 

Vf2&= nbsp; =3D - (5/= 7)(2 g yf)+ Vi2&= nbsp;

 

Vf&n= bsp; =3D (- (5= /7)(2 g yf)+ Vi2)  1/2=

 

Vf&n= bsp; =3D (- (5= /7)(2 (9.81 m/s2) (0.530 m))+ (2.= 85 m/s)2)  1/2

 

Vf&n= bsp; =3D  0.834 m/= s

 

 

 

 

 

 

 

 

 

 

 

Bonus (5 points): Wh= at is the direction of the ball’s angular momentum?

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7

PHYS 1111 Exam 3, Version 1

Fall 2005

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