TIME OF COMPLETION______________

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; = November 28, 2007

Total Weight: 100 points

1. = 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. = You will have 75 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 calculation problems and 5 (five) multiple choice. 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; 6:00 p.m.

Stop: &= nbsp; 7:15 p.m

&= nbsp; &nbs= p; PROBLEM

&= nbsp; &nbs= p; POINTS

&= nbsp; &nbs= p; CREDIT

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

&nbs= p; &= nbsp; 20

&= nbsp;

&nbs= p; &= nbsp; 7

&nbs= p; &= nbsp; 20

&= nbsp;

&nbs= p; &= nbsp; 20

&= nbsp;

&nbs= p; &= nbsp; 9

&nbs= p; &= nbsp; 20

&= 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. A solid disk rolls down the slope as shown below. What is the direction of = the disc’s angular momentum?

(A) Down the slope.<= /span>

(B) Up the slope.

(D) Into the page.

(C) Out of the page.=

(E) Clockwise.

2. A very light rod with two identical masses attached to it is rotating in the counterclockwise direction around = the axis which goes through one of the rod’s ends as shown below. Which one of the masses has the mo= st kinetic energy?

(A) A.

(B) B.

(C) Both masses have the same a= mount of kinetic energy.

(D) Neither one of the masses has any kinetic energy.

3. Given the two masses from Problem 2, as the mass A undergoes the angular displacement of 1.40 rad, the mass B undergoes the angular displacement of =

(A) 0 rad.

(B) 0.700 rad.

(D) 2.80 rad.

(E) 5.60 rad.

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4. A merry-go-around spins feely when Janice moved quickly to the center along t= he radial direction. Is it true to say that

= <= /span>(A) The moment of iner= tia of the system decreases and the angular speed increases.

= (= B) The moment of inertia of the system decreases and the angular speed decreas= es.

= (= C) The moment of inertia of the system decreases and the angular speed remains= the same.

= (= D) The moment of inertia of the system increases and the angular speed increas= es.

&= nbsp; (E) The moment of inertia of the system increases and the angu= lar speed decreases.

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5. What are the dimensions of the mom= ent of inertia?

(A) M·L.

(B) M·L^2.

(D) M/L².

= (= E) L/T².

6. What is the correct expression for torque, in= terms of the magnitude of the force, F, the radial distance from the axis of rotation, r, and the angle between the force and the radial direction, q?

  &nbs= p;         (A) t =3D F r sin <= span style=3D'font-size:12.0pt;font-family:Symbol;color:red'>q.

  &nbs= p;         (B) t =3D F r cos q.

  &nbs= p;         (C) t =3D F r tan q.

  &nbs= p;         (D) t =3D F r = cot q.

  &nbs= p;         (E) t =3D F r = q.

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7. Two d= isks of identical masses but different radii (r a= nd 2r) are spinning on frictionless be= arings at the same angular speed of 1.00 rad/s but in opposite direction. The two disks are brought slowly together. The frictional force between the surfaces eventually brings them to a common angular velocity. What is the magnitude = of the final angular velocity?

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L_I =3D L_F

L_I=3D (1/2 m r² ) w_I+ (1/2 m (2r)² ) (-= w_I)

L_F=3D (1/2 m r² ) wF_{+ (1/2 m (2r)² ) wF}

(1/2 m r² ) w_I+ (1/2 m (2r)² ) (-= w_I) =3D (1/2 m r² ) <= /span>wF_{+ (1/2 m (2r)² ) wF}

r² w_I+ (2r)²(-w_{I) =3D r² wF}_{+ (2r)² wF}

r² w_I+ 4r²(-w_I) =3D r² wF_{+ 4r² wF}

w_I+ 4(-w_I) =3D w_F_{+ 4w_F}

w_F_{=3D - 3/5 wI}

w_F =3D - 0.600 rad/s=

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8.   &n= bsp; A uniform beam of = mass m and length l is hung from two cables, one at the end of the beam and the ot= her 5l/8 of the way to the other end as shown below. If m =3D 10.0 kg and l =3D 2.00= m, determine the magnitudes of the forces the cable exerts on the beam. <= /o:p>



Note: we will use standard x-y coordinate system with the positive x-axis pointin= g to the right and positive y-axis pointing upward.

T_Lx =3D 0

T_Ly =3D T_L

T_Rx =3D 0

T_Ry =3D T_R

w_x =3D 0

w_y =3D - m g

SF_x =3D 0

SF_y =3D 0

T_L + T_R – mg =3D 0

T_L + T_R =3D (10.0 kg)(9.81 m/s²) =3D 98.1 N

t_TL =3D 0

t_TR =3D T_{R <=
/sub>(1.25
m)}

t_w =3D - = (10.0 kg)(9.81 m/s²)(1.0= 0 m) =3D - 98.1 N-m

St =3D 0

T_R(1.25 m) – 98.1 N-m =3D 0

T_R =3D 78.5 N

T_L =3D 19.6 N

An automotive engine shaft revs up from 3.00 x 10³ rev/min to = 4.50 x 10³ rev/min during a 1.20 s interval. <= /li>

(a) Calculate the magnitude of the = angular acceleration of the shaft assuming it is constant during the interval.=

w₀=3D (3.00 x 10³ rev/min)(2 p rad/1 rev)(1 min/60 s) =3D 0.314 x= 10³ rad/s

w=3D (4.50 x 10³ rev/min)(2 p rad/1 rev)(1 min/60 s) =3D 0.471 x= 10³ rad/s

w=3D w₀+ a t

a =3D (w- w₀) / <= /span>t =3D 131= rad/s²

(b)=     Determine the number of revolutions= of the shaft during the interval.

w²=3D w₀²⁼+ 2 a (q - q₀)<= /o:p>

(q - q₀) =3D (w²- w₀²)/ (2 a) =3D (469 rad)(1 rev/2 p rad) =3D 74.6 rev

(c)=     Assuming that the shaft can be approximated as a solid cylinder with a radius of 5.00 cm and mass of 10.0 = kg, what is the change in the shaft’s kinetic energy during this motion?<= span style=3D'mso-spacerun:yes'>

I =3D ½ m R²=

I =3D ½ (10.0 kg) (0.0500 m)= ² =3D 0.0125 kg-m²

K_i =3D ½ I= w₀^=
2=3D ½ (0.0125 kg-m²)(0.314 rad/s)² =3D = 616 J

K_f =3D ½ I= w²₌=3D &fra= c12; (0.0125 kg-m^{2)(0.471 rad/s)² =3D =
1387 J}

DK =3D 771 J

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10. Thre= e forces are applied to a wheel of radius 0.350 m, as shown below. One force is perpendicular to the rim, one is tangent to it, and the third makes a 40.0<= sup>o angle with the radius.

(a) What is the net torque on the wheel due to these three forces for the axis perpendicular to the wheel and passing through its center?

t₁ =3D (11.9 N) (0.350 m)= sin(0^o) =3D 0

t₂ =3D (14.6 N) (0.350 m) sin(40.0^o) =3D -3.28 N-m

t₃ =3D (8.50 N) (0.350 m)= sin(90^o) =3D 2.98 N-m

St =3D -0.310 N= -m

(b) Given that the mass of the wheel is 5.00 kg, what is its moment of inertia?=

I =3D ½ m R²

I =3D ½ (5.00 kg) (0.350 m)² =3D = 0.306 kg-m²

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(c) Find the angular acceleration that the wheel has as a result of these three forces.

St =3D I a

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a =3D St / <= span style=3D'font-family:"Times New Roman"'>I =3D -1.= 01 rad/s²

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(d) Does the wheel rotate clockwise or counterclockwise?

Clockwis= e.

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LI =3D LF

LI =3D (1/2 m r2 ) wI + (1/2 m (2r)2 ) (-= wI)

LF =3D (1/2 m r2 ) wF + (1/2 m (2r)2 ) wF

(1/2 m r2 ) wI + (1/2 m (2r)2 ) (-= wI) =3D (1/2 m r2 ) <= /span>wF + (1/2 m (2r)2 ) wF

r2 wI + (2r)2(-wI) =3D r2 wF + (2r)2 wF

r2 wI + 4r2(-wI) =3D r2 wF + 4r2 wF

wI + 4(-wI) =3D wF + 4wF

wF =3D - 3/5 wI

wF =3D - 0.600 rad/s=

L_I =3D L_F

L_I=3D (1/2 m r² ) w_I+ (1/2 m (2r)² ) (-= w_I)

L_F=3D (1/2 m r² ) wF_{+ (1/2 m (2r)² ) wF}

(1/2 m r² ) w_I+ (1/2 m (2r)² ) (-= w_I) =3D (1/2 m r² ) <= /span>wF_{+ (1/2 m (2r)² ) wF}

r² w_I+ (2r)²(-w_{I) =3D r² wF}_{+ (2r)² wF}

r² w_I+ 4r²(-w_I) =3D r² wF_{+ 4r² wF}

w_I+ 4(-w_I) =3D w_F_{+ 4w_F}

w_F_{=3D - 3/5 wI}

w_F =3D - 0.600 rad/s=