NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES

TIME

OF COMPLETION_______________ NAME____SOLUTION_________________________

DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 3 Version 1 Total Weight: 100 points

Section 1 December 6, 2005

1. Check your examination for completeness prior to starting. There are a total of nine (9) 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 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: Stop:

10:30 a.m. 11:45 a.m

PROBLEM

POINTS

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100 PERCENTAGE

CREDIT

CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. Rolling without slipping depends on a. Kinetic friction between the rolling object and the ground. b. Static friction between the rolling object and the ground. (4) c. Tension between the rolling object and the ground. 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? a. 0.40 m/s2 b. 0.80 m/s2 (4) c. 4.0 m/s2 d. 8.0 m/s2

acp = Vt2/R = (20.0 m/s)2/(50.0 m)

3. The gravitational force between two objects is proportional to a. The distance between the two objects. b. The square of the distance between the two objects. (4) c. The product of the masses of the two objects. 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

perpendicular to the door; the second force is applied at 30° to the plane of the door. Which force exerts the greater torque? a. The first applied perpendicular to the door. b. The second applied at an angle. (4) 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. The ratio of the larger sphere's moment of inertia to that of the smaller sphere is a. 4/5. b. 8/5. (4) c. 2. d. 4.

6.

(2/5 M (2R)2)/(2/5 M R2) = 4

Three children are trying to balance on a seesaw, which consists of a fulcrum rock, 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 kg. a. Where should girl C, whose mass is 25 kg, place herself so as to balance the seesaw?

= (50.0kg)(9.81 m/s2)(1.80 m) = 883 N-m = - (35.0kg)(9.81 m/s2)(1.80 m) = - 618 N-m  C= - (25.0kg)(9.81 m/s2)(x) = - (245N)(x) = 0 883 N-m - 618 N-m – (245 N)(x) = 0

x = 1.08 m b. What is the magnitude of the normal force that the rock exerts on the board? Nx = 0 Ny = N wAx = 0 wAy = - mA g wBx = 0 wBy = - mB g wCx = 0 wCy = - mC g Fx = 0 Fy = 0 N - mA g – mB g – mC g = 0 N = mA g + mB g + mC g = (mA + mB + mC)g = 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 angular momentum, what would the Sun’s new rotation rate be? (Take the Sun’s current period to be about 30 days.) II I = IF F II = 2/5 M0 R02 II = 2/5 (M0/2)(R0/100)2 (2/5 M0 R02) I = (2/5 (M0/2)(R0/100)2) F

I = (1/2)(1/100)2) F I = (1/20000) F F = (20000) I I = (2)/T = (2)/(30 days x (24 hours/1 day)(3600 s/1hour)) = 2.42 x 10-6 rad/s F = 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 = ½ I 2 KI = ½ (2/5 M0 R02I2 KF = ½ (2/5 (M0/2)(R0/100)2 F2 KF /KI = [½ (2/5 (M0/2)(R0/100)2 ) F2] / [½ (2/5 (M0)(R0)2 ) I2] KF /KI = [(1/2)(1/100)2 ) F2] / [I2] = (1/20000)(F2/ I2) = 20000

8.

The rings of Saturn 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 chunk of ice at the inner radius and the period of a chunk at the outer radius. Compare your numbers with 26 Saturn’s mean rotation period of 10 hours and 39 minutes. The mass of Saturn is 5.7 10 kg.

TO2 = ((42)/(GMS))RO3 TO2 = ((42)/(6.673 x 10-11 N-m2/kg2)(5.70 x 1026 kg)))(170,000,000 m)3 = 5.06 x 109 s2 TO = 7.11 x 104 s = 19.8 hours TI2 = ((4 2)/(GMS))RI3 TI2 = ((4 2)/(( 6.673 x 10-11 N-m2/kg2)(5.70 x 1026 kg)))(73,000,000 m)3 = 4.00 x 108 s2

TI = 2.00 x 104 s = 5.56 hours

TIME

OF COMPLETION_______________ NAME_______SOLUTION______________________


Section 1 December 4, 2006



4.


5. There are six (6) multiple choice and four (4) calculation problems. Work five (5) multiple choice problems and four (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: Stop:

6:00 p.m. 7:15 p.m

PROBLEM

POINTS

1-6

20

7

20

8

20

9

20

10

20

CREDIT

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. Tripling the mass of the bob on a simple pendulum will cause a change in the frequency of the pendulum swing by what factor? a. 0.330. b. 1.00. (4) c. 3.00. d. 9.00.

2. What condition or conditions are necessary for static equilibrium? a. ΣFx = 0 b. ΣFx = 0, ΣFy = 0, Σ = 0 (4) c. Σ = 0 d. ΣFx = 0, ΣFy = 0

3. Two equal forces are applied to a door at the doorknob. The first force is applied perpendicular to the door; the second force is applied at 30° to the plane of the door. Which force exerts the greater torque? a. The first applied perpendicular to the door. b. The second applied at an angle. (4) c. Both exert equal non-zero torques. d. Both exert zero torques.

4. A solid cylinder of mass 10 kg is pivoted about a frictionless axis thought the center O. A rope wrapped around the outer radius R1 = 1.0 m, exerts a force F1 = 5.0 N to the right. A second rope wrapped around another section of radius R2 = 0.50 m exerts a force F2 = 6.0 N downward. What is the angular acceleration of the cylinder? a. 1.0 rad/s2 b. 0.60 rad/s2 (4) c. 0.40 rad/s2 d. 0.80 rad/s2 5. A mass on a spring undergoes SHM. When the mass is at its maximum displacement from equilibrium, its instantaneous velocity a. Is maximum. b. Is less than maximum, but not zero. (4) c. Is zero. d. Cannot be determined from the information given.

2. The moment of inertia of a rigid body a. Depends on the axis of rotation. b. Cannot be negative. (4) c. Depends on mass distribution. d. All of the above.

7. A centrifuge rotor rotating at 10,300 rpm is shut off and is eventually brought uniformly to rest by a frictional torque of 1.20 m  N. If the mass of the rotor is 4.80 kg and it can be approximated as a solid cylinder of radius 0.0710 m, a. What is the rotor’s moment of inertia? I = ½ MR2 = ½ (4.80 kg)(0.0710 m)2 = 0.0121 kgm2

b. What is the angular acceleration of the rotor as it is brought to rest? /I = (1.20 N-m)/(0.0121 kgm2) = 99.2 rad/s2 c. Through how many revolutions will the rotor turn before coming to rest? 2 = 2 + 2 )  = 10,300 rpm = 1079 rad/s =0 )= - 2 / (2 ) = (1079 rad/s)2/(-2 x 99.2 rad/s2) = 5868 rad = 934 rev 8. A uniform disk turns at 2.4 rev s around a frictionless spindle. A nonrotating rod, of the same mass as the disk and length equal to the disk’s diameter, is dropped onto the freely spinning disk. They then both turn around the spindle with their centers superposed. What is the angular frequency in rev s of the combination?

Ii = ½ MR2

If = ½ MR2 + (1/12) ML2 = ½ MR2 + (1/12) M(2R)2 = ½ MR2 + (1/3) MR2 = (5/6) MR2 Iii = Iff f = Iii / If f = (2.40 rev/s) (½ MR2) /( 5/6 MR2) = 1.44 rev/s

9. A 0.150-kg toy is undergoing simple harmonic motion on the end of a horizontal spring with spring constant k = 300 N/m. The amplitude of the motion is 0.200 m.

a. Find the total mechanical energy of the toy at any point of its motion. E = ½ kA2 = ½ (300 N/m) (0.200 m)2 = 6.00 J

c. Find the maximum speed attained by the toy. ½ m Vmax2 = ½ kA2 m Vmax2 = kA2 Vmax2 = (k/m)A2 Vmax = ((300 N/m)/(0.150 kg))½ (0.200 m) = 8.94 m/s d. How many oscillations per second does the toy undergo? T = 2  (m/k)½ T = 2  ((0.150 kg)/(300 N/m))½ T = 0.140 s F = 1/T = 7.11 Hz e. What is the speed of the toy at 10.0 cm to the right from the equilibrium position? V = +- (k/m(A2 – x2)) ½ V = +- ((300 N/m)/(0.150 kg)((0.200 m)2 – (0.100 m)2)) ½ = +- 7.75 m/s

V = 7.75 m/s 7. Pictured below is a very light wooden plank with two masses, 10.0 kg each, on top of it. Find the reaction forces at points A and B.

NAx = NA cos (90o) = 0 NAy = NA sin (90o) = NA NBx = NB cos (90o) = 0 NBy =NB sin (90o) = NB w1x = w1 cos (-90o) = 0 w1y = w1 sin (-90o) = - m1 g = - (10.0 kg)(9.81 m/s2) = - 98.1 N w2x = w2 cos (-90o) = 0 w2y = w2 sin (-90o) = - m2 g = - (10.0 kg)(9.81 m/s2) = - 98.1 N Fx = 0 => 0 = 0 Fy = 0 => NA + NB – 98.1 N – 98.1 N = 0 NA + NB= 196 N  Assuming that the axis of rotation is at point A:

NA = 0 NB = NB (3.00 m) w1 = - (98.1 N) (0.750 m) = - 73.6 N-m sp = - (98.1 N) (2.50 m) = - 245 N-m  = 0 NB (3.00 m) – 73.6 N-m – 245 N-m = 0 NB = 106 N NA= 196 N – 106 N = 90.0 N

TIME

OF COMPLETION_______________ NAME___SOLUTION__________________________


Section 1 November 28, 2007

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.


4.


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: Stop:

6:00 p.m. 7:15 p.m

PROBLEM

POINTS

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

CREDIT

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. A solid disk rolls down the slope as shown below. What is the direction of the disc’s angular momentum? (A) Down the slope. (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 most kinetic energy?

(A) A. (B) B. (C) Both masses have the same amount 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. (C) 1.40 rad. (D) 2.80 rad. (E) 5.60 rad.

4. A merry-go-around spins feely when Janice moved quickly to the center along the radial direction. Is it true to say that (A) The moment of inertia of the system decreases and the angular speed increases. (B) The moment of inertia of the system decreases and the angular speed decreases. same.

(C) The moment of inertia of the system decreases and the angular speed remains the (D) The moment of inertia of the system increases and the angular speed increases. (E) The moment of inertia of the system increases and the angular speed decreases.

5. What are the dimensions of the moment of inertia? (A) M·L. (B) M·L2.

(C) M·L3. (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, ? (A) = F r sin . (B) = F r cos . (C) = F r tan . (D) = F r cot . (E) = F r .

7. Two disks of identical masses but different radii (r and 2r) are spinning on frictionless bearings 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? LI = LF LI = (1/2 m r2 ) + (1/2 m (2r)2 ) (-) LF = (1/2 m r2 ) F+ (1/2 m (2r)2 ) F (1/2 m r2 ) + (1/2 m (2r)2 ) (-) = (1/2 m r2 ) F+ (1/2 m (2r)2 ) F r2 + (2r)2(-) = r2 F+ (2r)2 F r2 + 4r2(-) = r2 F+ 4r2 F + 4(-) = F+ 4F F= - 3/5 I

F = - 0.600 rad/s

8. A uniform beam of mass m and length l is hung from two cables, one at the end of the beam and the other 5l/8 of the way to the other end as shown below. If m = 10.0 kg and l = 2.00 m, determine the magnitudes of the forces the cable exerts on the beam.

Note: we will use standard x-y coordinate system with the positive x-axis pointing to the right and positive y-axis pointing upward. TLx = 0 TLy = TL TRx = 0 TRy = TR wx = 0 wy = - m g Fx = 0

Fy = 0 TL + TR – m g = 0 TL + TR = (10.0 kg)(9.81 m/s2) = 98.1 N TL= 0 TR= TR (1.25 m)  w= - (10.0 kg)(9.81 m/s2)(1.00 m) = - 98.1 N-m = 0 TR (1.25 m) – 98.1 N-m = 0 TR = 78.5 N TL = 19.6 N

9. 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. (a) Calculate the magnitude of the angular acceleration of the shaft assuming it is constant during the interval. = (3.00 x 103 rev/min)(2  rad/1 rev)(1 min/60 s) = 0.314 x 103 rad/s  = (4.50 x 103 rev/min)(2  rad/1 rev)(1 min/60 s) = 0.471 x 103 rad/s = +  t  = (- ) t = 131 rad/s2 (b) Determine the number of revolutions of the shaft during the interval. = + 2  () () = (- )/ (2  = (469 rad)(1 rev/2  rad) = 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?

I = ½ m R2 I = ½ (10.0 kg) (0.0500 m)2 = 0.0125 kg-m2 Ki = ½ I= ½ (0.0125 kg-m2)(0.314 rad/s)2 = 616 J Kf = ½ I= ½ (0.0125 kg-m2)(0.471 rad/s)2 = 1387 J K = 771 J

10. Three 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 40.0o 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? = (11.9 N) (0.350 m) sin(0o) = 0 = (14.6 N) (0.350 m) sin(40.0o) = -3.28 N-m  = (8.50 N) (0.350 m) sin(90o) = 2.98 N-m  = -0.310 N-m (b) Given that the mass of the wheel is 5.00 kg, what is its moment of inertia? I = ½ m R2 I = ½ (5.00 kg) (0.350 m)2 = 0.306 kg-m2

(c) Find the angular acceleration that the wheel has as a result of these three forces. = I  I = -1.01 rad/s2

(d) Does the wheel rotate clockwise or counterclockwise?

Clockwise.

TIME

OF COMPLETION_______________ NAME____SOLUTION_________________________


Section 1 April 27, 2005



4.


5. 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. If you have any questions during the examination, see your instructor who will be located in the classroom. 7.

Start: Stop:

6:00 p.m. 7:15 p.m

PROBLEM

POINTS

1-6

20

7

23

CREDIT

8

23

9

17

10

17

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 moment of inertia of a rigid body e. Depends on the axis of rotation. f. Cannot be negative. (4) g. Depends on mass distribution. h. All of the above.

2. If the period of a system undergoing a SHM is doubled, the frequency of the system is a. Doubled. b. Halved. (4) c. For times as large. d. One-quarter as large.

3. Which one of the following quantities is zero when an object in simple harmonic motion is at its maximum displacement? a. Potential energy. b. Acceleration. (4) c. Kinetic energy. d. Frequency.

4. What condition or conditions are necessary for rotational equilibrium? a. ΣFx = 0. b. ΣFx = 0, ΣΤ = 0. (4) c. ΣΤ = 0. d. ΣFx = 0, ΣFy = 0.

5. What is the quantity used to measure an object's resistance to changes in rotation? a. Mass. b. Moment of inertia. (4) c. Torque. d. Angular velocity.

6. If a net torque is applied to an object, that object will experience: a. A constant angular velocity. b. An angular acceleration. (4) c. A constant moment of inertia. d. 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. The person walks radially to the edge of the merry-go-round. a. Calculate the angular velocity of the merry-go-round when the person reaches the edge. Ii = 920 kgm2 If = 920 kgm2 + (75.0 kg)(3.00 m)2 = 1595 kgm2

 i rad/s Li = Lf Ii i  If f f = Ii i If f = (920 kgm2) 2.00 rad/s 1595 kgm2) = 1.15 rad/s b. Calculate the rotational kinetic energy of the system before and after the person’s walk. KEi = ½ Ii i 2  ½ (920 kgm2) (2.00 rad/s)2 = 1840 J KEf = ½ If f 2  ½ (1595 kgm2) (1.15 rad/s)2 = 1055 J

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

Inward. (Right Hand Rule!)

8. A shop sign weighing 245 N is supported by a uniform beam as shown below. Find the tension in the wire and the horizontal and vertical forces exerted by the hinge on the beam.

Remember to draw a free-body diagram first! (w)x = w cos(- 90o) = 0 (w)y = w sin(- 90o) = - 245 N (FV)x = FV cos(90o) = 0 (FV)y = FV sin(90o) = FV (FH)x = FH cos(0o) = FH

(FH)y = FH sin(0o) = 0 (T)x = T cos(145o) (T)y = T sin(145o) Fx = 0

FH + T cos(145o) = 0

Fy = 0

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

w = - (245 N)(1.70 m) = - 416.5 Nm T= + (T)(1.350 m) sin(35o) FH = 0  FV = 0 = 0 -416.5 Nm + (T)(1.350 m) sin(35o) = 0

T = 538 N

FH = 441 N FV = -63.6 N 9. Neptune is an average distance of 4.50 x 109 km from the Sun. Estimate the length of the Neptunian year given that the Earth is 1.50 x 108 km from the Sun on the average. TN2 = (42/GMS )RN3 TE2 = (42/GMS )RE3 TN2/TE2 = RN3/RE3 TN/TE = (RN3/RE3)1/2 TN/TE = ((4.50 x 109)3/(1.50 x 108)3)1/2 = 164 TN = 164 years What is the Neptune’s orbital speed?

V = 2R/T V = 2(4.50 x 1012m)/(164x365x24x3600 s) = 5467 m/s

10. A mass-spring system undergoes SHM in horizontal direction. If the mass is 0.250 kg, the spring constant is 12.0 N/m, and the amplitude is 15.0 cm, a. What would be the maximum speed of the mass? Vmax = {(k/m) A2}½ Vmax = {[(12.0 N/m)/(0.250 kg)](0.150 m)2}½ = 1.03 m/s

b. Where would it occur? @ x= 0 c. What would be the speed at a half-amplitude position? V = {(k/m)(A2 – x2)}½ V = {[(12.0 N/m)/(0.250 kg)][(0.150 m)2 - (0.0750 m)2 ]}½ = 0.900 m/s

d. How many oscillations per second does the system go through? f = (1/2 (k/m)½ = (1/2 {(12.0 N/m)/(0.250 kg)}½ = 1.10 Hz

e. What would be the magnitude of the mass acceleration at the half-amplitude position? a = F/m F=kx a = k x/m a = (12.0 N/m)(0.0750 m)/(0.250 kg) = 3.60 m/s2

TIME

OF COMPLETION_______________ NAME_____SOLUTION________________________





4.


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: Stop:

10:30 a.m. 11:45 a.m

PROBLEM

POINTS

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

CREDIT

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. An ice skater performs a pirouette (a fast spin) by pulling in his outstretched arms close to his body. What happens to his moment of inertia about the axis of rotation? a. It does not change. b. It increases. (4) c. It decreases. d. It changes, but it is impossible to tell which way. 2. A wheel with moment of inertia 3.00 kg∙m2 has a net torque of 3.50 N∙m applied to it. What angular acceleration does it experience? a. 0.857 rad/s2 b. 1.17 rad/s2 (4) c. 3.00 rad/s2 d. 3.50 rad/s2

3. A fan has blades 0.250 m long rotating at 20.0 rpm. What is the centripetal acceleration of a point on the outer tip of a blade? a. 1.10 m/s2. b. 0.87 m/s2. (4) c. 0.55 m/s2. d. 0.23 m/s2. 4. According to Kepler’s first law, Halley’s comet circles the sun in an elliptical path with the Sun at one focus of the ellipse. What is at the other focus of the ellipse? a. Nothing. b. The Earth. (4)

c. The comet itself passes through the other focus. d. The tail of the comet stays at the other ellipse. 5. The quantity “moment of inertia” is equivalent, in terms of fundamental quantities of mass, length, and time, to: a. ML2T-2. b. ML. (4) c. ML2. d. ML-1T-2.

6.

A 60.0-kg painter is on a uniform 25-kg scaffold supported from above by ropes. There is a 4.0-kg pail of paint to one side, as shown. What are the magnitudes of the tension forces exerted by the ropes on the scaffold?

(20) TLx = TL cos (90o) = 0 TLy = TL sin (90o) = TL TRx = TR cos (90o) = 0 TRy = TR sin (90o) = TR wpx = wp cos (-90o) = 0

wpy = wp sin (-90o) = - mp g = - (60.0 kg)(9.81 m/s2) = - 589 N wsx = ws cos (-90o) = 0 wsy = ws sin (-90o) = - ms g = - (25.0 kg)(9.81 m/s2) = - 245 N wpail x = wpail cos (-90o) = 0 wpail y = wpail sin (-90o) = - mpail g = - (4.00 kg)(9.81 m/s2) = - 39.2 N Fx = 0 => 0 = 0 Fy = 0 => TR + TL – 589 N – 245 N – 39.2 N = 0 TR + TL= 873 N  Assuming that the axis of rotation is at the point where the left cable is attached to the scaffold: TL = 0 TR = TR (4.00 m) wp = - (589 N) (2.00 m) = - 1178 N-m sp = - (245 N) (2.00 m) = - 490 N-m wp = - (39.2 N) (1.00 m) = - 39.2 N-m  = 0 TR (4.00 m) - 1178 N-m – 490 N-m – 39.2 N-m = 0 TR = 427 N TL= 873 N – 417 N = 446 N

7. A potter’s wheel is rotating around a vertical axis through its center at an angular velocity of 1.50 rad/s. The wheel can be considered a uniform disk of mass 5.00 kg and diameter 0.800 m. The potter then throws a 1.00-kg chunk of clay onto the rim of the rotating wheel. What is the angular velocity of the wheel after the clay sticks to it?

Ii = ½ mw R2 = ½ (5.00 kg)(0.400 m)2 = 0.400 kg m2 If = ½ mw R2 + mc R2 = ½ (5.00 kg)(0.400 m)2 + (1.00 kg)(0.400 m)2 = 0.560 kg m2 Ii i = If f f = (Ii /If) i = 1.07 rev/s 24 8. A satellite of mass 5500 kg orbits the Earth mass  6.010 kg  and has a period of 6200 s.

a. Find the magnitude of the Earth’s gravitational force on the satellite. Hint: find answer to b first and use it to calculate the gravitational force. F = G m1 m2/r2 F = G Ms ME/R2 = (6.673 x 10-11 Nm2/kg2) (5500 kg) (6.00 x 1024 kg )/(7.26 x 106 m)2 = 41,779 N

b. What is the orbital radius of the satellite? T2 = (42/GME )R3 T2 GME = 42 R3 R3 = T2 GME /(42) R = (T2 GME /(42))1/3 R = ((6200 s)2 (6.673 x 10-11 Nm2/kg2)(6.00 x 1024 kg) /(42))1/3 = 7.26 x 106 m 9. A teenager pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to an angular velocity of 1.75 rad/s in 10.0 s. Assume the merry-goround is a uniform disk of radius 2.50 m and has a mass of 760 kg, and two children (each with a mass of 25.0 kg) sit opposite each other on the edge. a. What angular acceleration is required to reach 1.75 rad/s velocity in 10.0 s?  = 0 +  t

 = ( - 0) /t  = (1.75 rad/s) / (10.0 s) = 0.175 rad/s2 c. What is the moment of inertia of the merry-go-round with two children on it? I = ½ M R2 + m R2 + m R2 = ½ (760 kg)(2.50 m)2 + 2 (25.0 kg)(2.50 m)2 = 2688 kg m2

b. Calculate the torque required to produce the acceleration, neglecting frictional torque.   = I   = (2688 kgm2)(0.175 rad/s2) = 470 N-m

Bonus (5 points): What is the magnitude of the tangential force required to produce this torque?  = F R F =  /R = (470 N-m)/(2.50 m) = 188 N

TIME

OF COMPLETION_______________ NAME______SOLUTION_______________________





4.


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: Stop:

6:00 p.m. 7:15 p.m

PROBLEM

POINTS

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

CREDIT

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. Two children are riding on a merry-go-round. Child A is at greater distance from the axis of rotation than child B. Which child has the larger angular displacement? a. Child A. b. Child B. (4) c. Both have the same angular displacement. d. Cannot be determined based on the given information.

2. What is the quantity used to measure an object’s resistance to changes in rotational motion?

a. Mass. b. Moment of inertia. (4) c. Angular velocity. d. Angular acceleration.

3. Two forces are applied to a doorknob, perpendicular to the door. The first force is twice as large as the second force. The ratio of the torque of the first force to the torque of the second is: a. 1/4. b. 1/2. (4) c. 2. d. 4.

4. When you ride a bicycle, what is the direction of the angular velocity of the wheels? a. To your left. b. To your right. (4) c. Forward. d. Backward.

5. You sit on a freely rotating stool (no friction). When you extend your arms, a. Your moment of inertia decreases and your angular velocity increases. b. Your moment of inertia increases and your angular velocity increases. (4)

c. Your moment of inertia decreases and your angular velocity decreases. d. Your moment of inertia increases and your angular velocity decreases.

6. Two identical spheres, each of mass M and radius R just touch each other. What is the magnitude of the gravitational force that they exert on each other? a. (GM2)/(R2). b. (GM2)/(2R2). (4) c. (GM2)/(4R2). d. (2GM2)/(R2).

7. Shekema and Rodney are carrying a 40.0 kg block on a 2.00 m board as shown below. The mass of the board is 5.00 kg. Shekema insists that she is very strong and could have carried the block all by herself, so they place the block 0.500 m away from her and 1.50 m away from Rodney. Find the forces (in Newtons) exerted by each to carry the block.

NRx = NR cos (90o) = 0

NRy = NR sin (90o) = NR NSx = NS cos (90o) = 0 NSy = NS sin (90o) = NS Wboard x = wboard cos (-90o) = 0 Wboard y = wboard sin (-90o) = - mboard g = - (5.00 kg)(9.81 m/s2) = - 49.1 N Wblock x = wblock cos (-90o) = 0 Wblock y = wblock sin (-90o) = - mblock g = - (40.0 kg)(9.81 m/s2) = - 392 N Fx = 0 => 0 = 0 Fy = 0 => NR + NS – 589 N – 49.1 N – 392 N = 0 NR + NS= 442 N Assuming that the axis of rotation is at the point where Rodney is holding the board: NR = 0 NS = NS (2.00 m) w board = - (49.1 N) (1.00 m) = - 49.1 N-m w block = - (392 N) (1.50 m) = - 588 N-m  = 0 NS (2.00 m) – 49.1 N-m – 588 N-m = 0 NS = 319 N NR= 442 N – 319 N = 123 N 8. A compact disc rotates from rest to 500 rev/min in 5.50 s. a. What is its angular acceleration, assuming that it is constant? 0 =    = (500 rev/min)(1 min/60.0 s)(2 rad/1 rev) = 52.4 rad/s 

 = 0 +  t  = ( - 0) /t  = (52.4 rad/s) / (5.50 s) = 9.52 rad/s2

b. How many revolutions does it make in 5.50 s?  =0 +0 t +t2 = ½ (52.4 rad/s2)(5.50 s)2 = 793 rad = 126 rev

c. At the moment the disk reaches the angular velocity of 500 rev/min, what is the tangential velocity of a point on the rim of the disc 6.00 cm from the center of rotation ? Vt =  R = (52.3 rad/s)(0.0600 m) = 3.14 m/s 9. Penny goes bowling with her friends. She notices that a bowling ball of radius 11.0 cm and mass m = 7.20 kg is rolling without slipping on a horizontal ball return at 2.00 m/s. It then rolls without slipping up a hill to a height h before momentarily coming to rest. Find h. Ei = Ef m g yi + ½ m Vi 2 + ½ I i2 = m g yf + ½ m Vf 2 + ½ I f2 yi = 0 Vi = 2.00 m/s i = Vi /R yf = h Vf = 0 f = 0 ½ m Vi 2 + ½ I i2 = m g yf m Vi 2 + I i2 = 2 m g yf

m Vi 2 + (2/5 m R2)(Vi /R)2 = 2 m g yf Vi 2 + (2/5) Vi2 = 2 g yf (7/10) Vi2 = g yf yf = 0.285 m 10. Amanda, David, and Rotasha play with two children who want a ride on a merry-go-round. The merry-go-round is a solid disc that rotates freely (without friction) around its center and has a moment of inertia of 500 kg-m2 relative to this axis. The children have the same masses of 25.0 kg and sit at distances of 1.00 m and 2.00 m away from the center of merry-go-round. The radius of marry-go-round is 3.00 m.

a. What is the moment of inertia of the merry-go-round with the children on it relative to the center of rotation? I = 500 kg-m2 +(25.0 kg)(1.00 m)2 + (25.0 kg)(2.00 m)2 = 625 kg-m2

b. Amanda, David, and Rotasha pull the merry-go-round with tangential forces as shown in the figure. (The magnitudes are: A = 300 N, D = 150 N, R = 200 N) What is the net torque that they exert on the merry-go-round?  = (300 N)(3.00 m) + (200 N)(3.00 m) – (150 N)(3.00 m) = 1050 N-m

c. What is the angular acceleration of the merry-go-round? /I = (1050 N-m)/(625 kg-m2) = 1.68 rad/s2

TIME

OF COMPLETION_______________ NAME___SOLUTION__________________________





4.


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: Stop:

6:00 p.m. 7:15 p.m

PROBLEM

POINTS

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

CREDIT

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. A particle is traveling in a horizontal circle at a constant speed. One can conclude that __________ is constant. (A) Velocity. (B) Acceleration. (C) Net force. (D) All of the above. (E) None of the above.

2. The dimension of torque is the same as (A) Impulse. (B) Momentum. (C) Force. (D) Energy. (E) None of the above.

3. Two forces are applied to a doorknob, perpendicular to the door. The first force is twice as large as the second force. The ratio of the torque of the first force to the torque of the second is: (A) 1/4. (B) 1/2. (C) 2. (D) 4. (E) 8.

4. Two children are riding on a merry-go-round. Child A is at greater distance from the axis of rotation than child B. Which child has the larger linear displacement? (A) Child A. (B) Child B. (C) Both have the same zero linear displacement. (D) Both have the same non-zero linear displacement. (E) Cannot be determined based on the given information.

5. Two children are riding on a merry-go-round. Child A is at greater distance from the axis of rotation than child B. It is true statement that (A) Child A has greater angular velocity than child B. (B) Child B has greater angular velocity than child B. . (C) Both have the same zero angular velocity. (D) Both have the same non-zero angular velocity. (E) Cannot be determined based on the given information.

6. You sit on a rotating stool which rotates clockwise (seen from above). What is the direction of your angular acceleration if the stool is slowing down? (A) Vertically upward. (B) Vertically downward. (C) Horizontally and to your left. (D) Horizontally and to your right. (E) Your angular acceleration has no direction.

7. A wheel starts to rotate around its center with a constant angular acceleration of 2.60 rad/s2. The wheel starts from rest and rotates for 6.00 s. At the end of that time a. What is its angular velocity? t 2.60 rad/s2s) = 15.6 rad/s b. Through what angle has the wheel turned? ) ) = ) ) = (15.6 rad/s)2/(2 x 2.60 rad/s2) = 46.8 rad

c. How many revolutions has it made? 46.8 rad/(2 rad/rev) = 7.45 rev d. What is the magnitude of tangential velocity and tangential acceleration of a point 0.300 m from the axis of rotation? Vt =  r = (15.6 rad/s)(0.300 m) = 4.68 m/s at =  r = (2.60 rad/s2)(0.300 m) = 0.780 m/s2

8. To determine the location of his center of mass, a physics student lies on a lightweight plank supported by two scales 2.50 m apart. The left scale reads 290 N, and the right scale reads 122 N. (Hint: the readings correspond to the normal forces exerted on the plank.)

a. Draw a free-body diagram.

b. Find the student’s mass. NLx = NL cos (90.0o) = 0 NLy = NL sin (90.0o) = NL = 290 N NRx = NR cos (90.0o) = 0 NRy = NR sin (90.0o) = NR = 122 N wx = m g cos (-90.0o) = 0

wy = m g sin (-90.0o) = - m g Fx = 0 Fy = 0 290 N + 122 N - mg = 0 mg = 412 N m = 42.0 kg c. Find the distance from the student’s head to his center of mass. NL = 0 NR = (122 N) (2.50 m) = 305 N-m w = -w (x) = -(412 N) (x)  = 0 305 N-m – (412 N) (x) = 0 x = 0.740 m

9. A merry-go-round of radius 2.00 m is rotating about its center making one revolution every 5.00 seconds. A child of mass 25.0 kg stands at the rim. a. What is the magnitude of the angular velocity of merry-go-round in rad/s?  = (2  rad)/(5.00 s) = 1.26 rad/s

b. What is the magnitude of centripetal acceleration of the child? acp =  r = (1.26 rad/s)2(2.00 m) = 3.18 m/s2

c. The merry-go-round is very slippery (frictionless) and the child stays on it by holding onto a rope attached to the pivot at the center of merry-go-round. What has to be the magnitude of tension in the rope for the child to stay on merry-go-round? T = m acp = (25.0 kg)(3.19 m/s2) = 79.4 N

d. The rope suddenly breaks. What is the motion of the child immediately after it? The child continues to move in the direction which was tangential to the circle the moment the rope broke, until some other force changes his velocity.

10. A square metal plate 0.180 m on each side is pivoted about an axis through point O at its center and perpendicular to the plate. Calculate the net torque about this axis due to the three forces shown in the figure if the magnitudes of the forces are F1 = 18.0 N, F2 = 26.0 N, and F3 = 14.0 N. The plate and all forces are in the plane of the page.

r1 = r2 = r3 = ((0.0900 m)2 + (0.0900 m)2)1/2 /2 = 0.127 m 1 = F1 r1 sin(1) = (18.0 N)(0.127 m) sin (45.0o) = 1.62 Nm 2 = - F2 r2 sin(2) = - (26.0 N)(0.127 m) sin (45.0o) = -2.33 Nm 3 = F3 r3 sin(3) = (14.0 N)(0.127 m) sin (90.0o) = 1.78 Nm

 = 1.62 Nm -2.33 Nm + 1.78 Nm = 1.07 Nm

TIME

OF COMPLETION_______________ NAME_____SOLUTION________________________


Section 1 August 1, 2005

1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on eight (8) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3.


4.


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: Stop:

11:30 a.m. 12:50 p.m

PROBLEM

POINTS

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

CREDIT

PERCENTAGE

CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 2. Which one of the following quantities is zero when an object in simple harmonic motion is at its maximum displacement? e. Potential energy. (5) f. Acceleration. g. Kinetic energy. h. Frequency.

2. Two equal forces are applied to a door. The first force is applied at the midpoint of the door; the second force is applied at the doorknob. Both forces are applied perpendicular to the door. Which force exerts the greater torque? a. The first at the midpoint. (5) b. The second at the doorknob. c. Both exert equal non-zero torques. d. Both exert zero torques. 3. The quantity “angular momentum” (in terms of the fundamental quantities of mass, length, and time) is equivalent to: a. MLT -2. b. ML2T -1. (5) c. ML2T -3. d. ML3T. 4. Two moons orbit a planet in nearly circular orbits. Moon A has orbital radius r, and moon 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. 20 days. b. 80 days. (5) c. 160 days. d. 320 days.

5. The moment of inertia of a spinning body about its spin axis depends on its: a.

Angular speed, shape, and mass.

b. Angular acceleration, mass, and axis position. (5) c. Mass, shape, axis position, and size. d. Mass, size, shape, and speed.

6. A boy and a girl are riding on a merry-go-round 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 linear speed? a. The boy. b. The girl. (5) c. Both have the same non-zero translational velocity. d. Both have zero translational velocity.

7. A 64.0-kg person stands on a weightless 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.

PHYS 1111 Exam 3, Version 1 Fall 2005

47

a. How far from that pillar is the person standing? N1x = N1 cos (-90o) = 0 N1y = N1 sin (-90o) = - N1 = -828 N N2x = N2 cos (90o) = 0 N2y = N2 sin (90o) = N2 wx = w cos (-90o) = 0 wy = w sin (-90o) = - w = - m g = - (64.0 kg)(9.80 m/s2) = - 627 N Fx = 0 => 0 = 0 Fy = 0 => (-828 N) + N2 – (627 N) = 0 N2 = 1455 N N2 = N2 (1.10 m) = (1455 N)(1.10 m) = 1601 Nm N1 = 0 w = - (627 N) (x)  = 0 - (627 N) x + 1601 N m = 0 x = 2.55 m


48

b. Find the force exerted by the second pillar. N2 = 1455 N 8.

A potter’s wheel is rotating around a vertical axis through its center at a frequency of 1.5 rev s . The wheel can 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 radius 8.00 cm, onto the center of the rotating wheel. a. What is the frequency of the wheel after the clay sticks to it? Ii = ½ mw R2 = ½ (5.00 kg)(0.200 m)2 = 0.100 kg m2 If = ½ mw R2 + ½ mc R2 = ½ (5.00 kg)(0.200 m)2 + ½ (3.10 kg)(0.0800 m)2 = 0.10992 kg m2 Ii i = If f f = (Ii /If) i = 1.36 rev/s b. (Bonus) If the wheel rotates in clockwise direction as viewed from above, what is the direction of the wheel’s angular momentum?

9. A 0.500-kg mass resting on a horizontal frictionless surface is connected to a fixed spring with


49

spring constant of 80.0 N/m. The mass is displaced 16.0 cm from its equilibrium position and released. a. What is the amplitude of oscillations? A = 0.160 m b. What is the maximum speed of the mass? Vmax = (k/m)1/2 A = ((80.0 N/m)/(0.500 kg)) 1/2 (0.160 m) = 2.02 m/s

c. What is the total mechanical energy of the system? TE = ½ m Vmax2 = ½ (0.500 kg) (2.02 m/s)2 = 1.02 J d. What is the frequency of oscillations? f = (1/(2)(k/m)1/2 = 2.01 Hz e. What is the period of oscillations? T = 1/f = 0.497 s 

10. A uniform disk of mass 1.00 kg and radius 10.0 m has four symmetrically placed 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.

d. What is the moment of inertia of the disk relative to its center? I = ½ mdisk R2 + 4 m r2 = ½ (1.00 kg)(10.0 m)2 + 4 (0.250 kg)(5.00 m)2 = 75.0 kg m2 PHYS 1111 Exam 3, Version 1 Fall 2005

50

e. Two tangential forces are applied to the disk as shown in the figure. What is the net torque acting on the disk?

1 = F1 d1 = (40.0N)(10.0m) = 400 Nm 2 = - F2 d2 = - (20.0N)(5.00m) = -100 Nm  = 400 Nm – 100 Nm = 300 Nm f. What is the angular acceleration of the disk?  = I   =  / I = (300 Nm)/ (75.0 kgm2) = 4.00 rad/s2

g. 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 = ½ I 2 = ½ I ( t)2 = ½ (75.0 kgm2)((4.00 rad/s2)(10.0 s))2 = 60,000 J


51


52

          


53

     

9. After you pick up a spare, your bowling ball rolls without slipping back toward 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 = m g yi + ½ mVi2 + ½ Ii2 = 0 + ½ mVi2 + ½ Ii2 = ½ mVi2 + ½ (2/5mR2)i2 = ½ mVi2 + ½ (2/5)mVi2 Ef = m g yf + ½ mVf2 + ½ If2 = m g yf + ½ mVf2 + ½ (2/5mR2)f2 = m g yf + ½ mVf2 + ½ (2/5)mVf2 Ei = Ei Vi2 + (2/5)Vi2 = 2 g yf + Vf2 + (2/5)Vf2 (7/5)Vi2 = 2 g yf + (7/5)Vf2 Vi2 = (5/7)(2 g yf)+ Vf2 Vf2 = - (5/7)(2 g yf)+ Vi2 Vf = (- (5/7)(2 g yf)+ Vi2)

1/2

Vf = (- (5/7)(2 (9.81 m/s2) (0.530 m))+ (2.85 m/s)2) Vf = 0.834 m/s


54

1/2

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


55

NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES

Recommend Documents