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VAAL UNIVERSITY OF TECHNOLOGY

FACULTY OF APPLIED AND COMPUTER SCIENCES

Assessment 2: LU 5-9 Physics II SUBJECT DATE TIME TOTAL MARKS FULL MARKS

: Physics II (APPHT2A) : 23 September 2016 : 14H00-16H00 : 106 : 100

EXAMINER MODERATOR

: Mr. G.S. NKOSI : Prof. W.J. BEKKER

   

Non-programmable Calculators may be used. Read the exam rules and regulations. Number the answers correctly. The question paper consists of 8 typed pages including the front page, and the information sheet.

DO NOT TURN THIS PAGE BEFORE PERMISSION IS GRANTED

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QUESTION 1

[30]

Write the correct answer for each of the following in your answer book. (Example 1.1. A) 1.1. A football is kicked at an angle ď ąď€ with respect to the horizontal. Which one of the following statements best describes the acceleration of the football during this event if air resistance is neglected? A. The acceleration is zero m/s2 at all times. B. The acceleration is zero m/s2 when the football has reached the highest point. C. The acceleration is positive as the football rises, and it is negative as the football falls. D. The acceleration is 9.8 m/s2 at all times. E. None of the above. 1.2. A race car is travelling at constant speed around a circular track. What happens to the centripetal acceleration of the car if the speed is doubled? A. The centripetal acceleration remains the same. B. The centripetal acceleration is doubled. C. The centripetal acceleration increases by a factor of 4. D. The centripetal acceleration is decreased by a factor of one-half. E. The centripetal acceleration is decreased by a factor of one-fourth. 1.3.

A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the tension in the string is FT. The boy then speeds up the stone, keeping the radius of the circle unchanged, so that the string makes two complete revolutions every second. What happens to the tension in the sting? A. The tension is unchanged. B. The tension increases to four times its original value. C. The tension reduces to one-fourth of its original value. D. The tension reduces to half of its original value. E. The tension increases to twice its original value.

1.4. A rigid body rotates about a fixed axis with a constant angular acceleration. Which one of the following statements is true concerning the tangential acceleration of any point on the body? A. The tangential acceleration depends on the change in the angular velocity. B. The tangential acceleration is zero m/s2. C. The tangential acceleration depends on the angular velocity. D. The tangential acceleration is equal to the centripetal acceleration. E. The tangential acceleration is constant in both magnitude and direction.

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1.5. Which statement concerning a wheel undergoing rolling motion is true? A. The angular acceleration of the wheel must be zero m/s2. B. There is no slipping at the point the wheel touches the surface on which it is rolling. C. The tangential velocity is the same for all points on the rim of the wheel. D. The linear velocity for all points on the rim of the wheel is non-zero. E. The tangential velocity is the same for all points on the wheel. 1.6. The figure below shows the top view of a door that is 2 m wide. Two forces are applied to 10.0 N 60.0° 1.0 m

60.0° 10.0 N

the door as indicated. What is the magnitude of the net torque on the door with respect to the hinge? A. 0 N.m. B. 5.0 N.m. C. 8.7 N.m. D. 10.0 N.m. E. 26.0 N.m.

1.7. The figure below shows A string that is wrapped around a pulley which has a radius of F

0.05 m and a moment of inertia of 0.2 kg ďƒ— m2. If the string is pulled with a force F, the resulting angular acceleration of the pulley is 2 rad/s2. The magnitude of the force F is: A. 0.5 N. B. 4 N. C. 8 N. D. 16 N. E. 40 N. 1.8. A 2.0-kg hollow cylinder or hoop rolls without slipping on a horizontal surface so that its center proceeds to the right with a constant linear speed of 6.0 m/s. [IS = 1/2MR2] 6 m/s 2.0 kg

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What is the total kinetic energy of the hoop? A. 36 J. B. 54 J. C. 72 J. D. 96 J. E. 140 J.

1.9. A simple pendulum on earth has a period of 6.0 s. What is the approximate period of this pendulum on the moon where the acceleration due to gravity is roughly 1/6 that of earth? A. 1.0 s. B. 2.4 s. C. 6.0 s. D. 15.0 s. E. 36.0 s. 1.10. A cable stretches by an amount d when it supports a crate of mass M. The cable is then cut in half. If the same crate is supported by either half of the cable, by how much will the cable stretch? A. d. B. d/2. C. d/4. D. 2d. E. 4d.

QUESTION 2

[76]

2.1. Define, state or explain each of the following. (a) Angular velocity. (b) Lever arm. (c) Principle of conservation of angular momentum. (d) Newton’s second law for rotational motion about a fixed axis. (e) Hooke’s law for stress and strain.

[2] [2] [2] [2] [2]

2.2. A diver springs upward from a board that is three meters above the water. At the instant she contacts the water her speed is 8.90 m/s and her body makes an angle of 75.0° with respect to the horizontal surface of the water. Determine her initial velocity, both magnitude and direction. [5] 2.3. A car travels at a constant speed around a circular track whose radius is 2.6 km. The car goes once around the track in 360 s. What is the magnitude of the centripetal acceleration of the car? [4] 4


2.4. A “swing� ride at a carnival consists of chairs that are swung in a circle by 15.0-m cables attached to a vertical rotating pole, as the drawing shows. Suppose the total mass of a chair and its occupant is 179 kg.

(a) Determine the tension in the cable attached to the chair. (b) Find the speed of the chair.

[3] [4]

2.5. The drawing shows a device that can be used to measure the speed of a bullet. The device consists of two rotating disks, separated by a distance of d = 0.850 m, and rotating with an angular speed of 95.0 rad/s. The bullet first passes through the left disk and then through the right disk. It is found that the angular displacement between the two bullet holes is θ = 0.240 rad. From these data, determine the speed of the bullet. [4]

2.6. A motorcycle, which has an initial linear speed of 6.6 m/s, decelerates to a speed of 2.1 m/s in 5.0 s. Each wheel has a radius of 0.65 m and is rotating in a counterclockwise (positive) direction. What is (a) the constant angular acceleration (in rad/s2)? (b) the angular displacement (in rad) of each wheel?

[3] [3]

2.7. The sign over the page is 4.5m long, weighs 1350N, and is made of uniform material. A weight of 225N hangs 1m from the end. Determine the tension in each support cable. [7]

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2.8. A 2.0-kg solid cylinder of radius 0.5 m rotates at a rate of 40 rad/s about its cylindrical axis. What power is required to bring the cylinder to rest in 10 s? [8] 2.9. A thin walled hollow cylinder (mass = mh and radius= rh ) and a solid cylinder (mass = mS and radius = rS ) start from rest at the top of the incline. Both cylinders start at the same initial vertical height h0 and roll down the incline without slipping. [IS = 1/2MR2: IH = MR2]

Determine the ratio

S

h of the translational speed of the solid cylinder  S to that of the  h

hollow cylinder at the bottom of the incline.

\

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[10]


2.10. The figure below shows an artificial satellite placed into an elliptical orbit about the earth.

Telemetry data indicate that the points of closest approach (called perigee) is rp = 8.37  106m from the center of the earth, and its point of greatest distance, rA = 25  106m from the center of the earth is (called the apogee). The speed of the satellite at perigee is  p  8540m / s . Determine the speed  A at the apogee.

[5]

2.11. A spring with constant k = 30.0 N/m is at the base of a frictionless 30.0 inclined plane. A 0.50kg block is pressed against the spring, compressing it 0.25m from its equilibrium position.

The block is then released. If the block is not attached to the spring, how far along the incline is will it travel before it stops? [5] 2.12. A copper cylinder and a brass cylinder are stacked end to end as shown in the figure below. Each cylinder has a radius of 0.25cm. A compressive force of F = 6500N is applied to the

right end of the brass cylinder. Determine the amount by which the length of the stack decreases. [NB, The Young’s modulus for brass and copper are, 9.0  1010N/m2, 1.1  1011 N/m2, respectively.] [5]

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EXAMINATION FORMULAE SHEET AND CONSTANTS

INFORMATION SHEET Acceleration due to gravity Density of water Boltzmann’s constant Universal gas constant Permeability of free space

g = 9.8 m.s-2 water = 1 000 kg.m-3 k = 1.38 × 10-23 J.K-1 R = 8.314 J.K-1.mol-1 0 = 4 × 10-7 T.m-1.A-1

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VAAL UNIVERSITY OF TECHNOLOGY Memorandum Assessment 2: LU 5-9 Physics II SUBJECT DATE TIME TOTAL MARKS FULL MARKS Question 1 1.1 D

[30]

1.2 C

Question 2

: Physics II (APPHT2A) : 30 April 2010 : 2 HOURS : 106 : 100

1.3 B

1.4 A

1.5 B

1.6 C

1.7 C

1.8 C

1.9 D

1.10 B

[80]

2.1. (a) ω = θ/t, where θ is the angular displacement, t is time

[2]

(b) Lever arm is shortest (or perpendicular) distance between the axis of rotation and the work line of force. [2] (c) The total angular momentum of a system remains constant (conserved) if the net average external torque acting on the system is zero. [2] (d) The angular acceleration of a body is directly proportional to the net torque applied to it and inversely proportional to the moment of inertia. τ =Iα [2] (e) Stress is directly proportional to strain

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[2]


Physics II

MEMORANDUM

Assessment2 - September 2016

2.2 a.

vx2  v02x  7.70 m/s    0 m/s    2.47 m/s 2 2x 2  12.0 m  2

ax 

2

[4]

b. aparallel = ax cos 25.0  2.47 m/s2 cos 25.0  2.24 m/s2

2.3 vx = v cos 75.0° = (8.90 m/s) cos 75.0° = 2.30 m/s 2

2

2

2

2

v0y = vy + 2gy = (8.90 m/s) sin 75.0° + 2(9.80 m/s )(–3.00 m) v0y = 3.89 m/s

[5]

v0   2.30 m/s    3.89 m/s   4.52 m/s 2

0 = tan

1

2

(voy /vox) = 59.4

2.4 ac  v 2 / r and v   2 r  / T

 2 r  2 3   v 4 2 r 4  2.6 10 m  T  ac     2   0.79 m/s2 2 r r T  360 s  2

2

[4]

2.5 a. T cos 60.0  mg  0

 Fy

179 kg   9.80 m/s mg  cos 60.0 cos 60.0  v2  b. T sin 60.0  mac  m    r   Fx T

2

  3510 N

13.0 m 3510 N  sin 60.0  14.9 m/s r T sin 60.0  m 179 kg  2.6   t  0.240rad t   2.53  103 s  95.0rad/s x d 0.850m v    336 m/s t t 2.53  103 s v

v v0  v  v0 r r  2.7 a.   t t rt v  v0 2.1 m/s  6.6 m/s    1.4 rad/s2  0.65 m  5.0 s  rt

[3]

[4]

[4]

  0

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[3]


Physics II

MEMORANDUM

Assessment2 - September 2016

 v0  1 2 t   t  r  2

  0t  12  t 2   b.

[3]

2  6.6 m/s   2 1   5.0 s   2  1.4 rad/s   5.0 s   33 rad  0.65 m 

2.8

  0  0.75T

2

 1.25T1  1.2WM  0  0.75T2  1.25T1  1.25 * 225 N

 0.75T2  1.25T1  281.25 N

F

y

 0  T2  T1  WR  WM  0  T2  T1  WR  WM

T2  T1  1575N --

 T2  T1  1350 N  225N

-------------------------------------------------------------(2)

Multiply (2) by 1.25

1.25T2  1.25T1  1968.75 N --------------------------------------------------------------------------(3) (1) + (3) 2 2.0T2  2250 N  T2  1225 N  T1  450 N

2.9

WR    WR  I   f  0 _  t    

[7]

0 t

  

40rad / s 10.0s

  4.0rad / s 2

Also

  0t  t 2    0t  t 2    40rad / s 10s)  

1 1 1 2  40rad / s 2 10s  2 2 2   200rad 1 2 1  WR   mr 2   WR  2.0kg 0.5 4.0rad / s 2 200rad  WR  200 J 2 2   WR 200 J P  P  P  20W t 10s

[8]

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Physics II

MEMORANDUM

Assessment2 - September 2016

2.10 1 1 1 1 1 1 E0  E f  m 02  I02  mgh0  m 2f  I 2f  mgh f  mgh0  m 2f  I 2f 2 2 2 2 2 2 2    2 mgh 1 1 f 0  mgh0  m 2f  I     f  I 2 2  r  m 2 r (1) Hollow cylinder 2mh gh0 2mh gh0  f  gh0 h   f  2 mr 2mh mh  h 2 h rh (2) Solid cylinder 2ms gh0 4 2ms gh0  s   s  gh0  s  1.15 gh0 s  3 1 3 ms ms rs2 2 ms  2 2 rs

Ratio

 s 1.15 gh0  s   1.2  h h gh0

[10]

2.11

    LA  LP  I A A  I PP  mrA2  A  mrP2 P  rA2  A   rP2  P   rA A  rP P r  A  rP  r   8.37  106 m  8450m / s  A  2820m / s   A   P P   A   6 [5]  25.1 10 m   rA  2.12

1 1 1 1 1 E0  E f  m 02  kx02  m 2f  mgy f  kx02  mgy f  kx02  mgx f sin 30 0 2 2 2 2 2 1 2 1 2 30 N / .m0.25 k x0 0.9357 J 2  xf   x f  0.383m  xf  2  x  f 0 2 0 2.45 N 0.5k g9.8m / s sin 30 mg sin 30

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Physics II

MEMORANDUM

Assessment2 - September 2016

2.13 Y

F stress  A  F  Y  L  A Y  Y  L  L strain   0 L0

For copper: FL0 6500 N 3.0 10 2 m 5 Lcu   Lcu  Ycu A 1.11011 N .m2  0.25 10 2 m 2  Lcu  9.02 10 m For brass: Lbrass 

FL0 6500 N  5.0 10 2 m  Lbrass   Lbrass  1.8 10 4 m Ybrass A 9.0 1010 N .m  2  0.25 10  2 m 2



 Lstack  Lcu  Lbrass  2.72 10 4 m [5]

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2016 sem 2 assessment 2  
2016 sem 2 assessment 2  
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