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Page 1/17 First Name: Last Name: Student #: Instructor: Dr. K. Foyle Class: Physics 1BB3, Modern Physics for the Life Sciences FIRST MIDTERM EXAMINATION DATE: Feb. 6th, 2014 TIME: 7pm PLACE: T29 101 & 105 DURATION OF EXAMINATION: 2 hours This examination paper includes 17 pages and 17 questions. You are responsible for ensuring that your copy of the paper is complete. Bring any discrepancy to the attention of your invigilator. Special Instructions: Answer multiple-choice questions (questions 1 to 16) on the provided examination answer sheet. Answer the written answer question (question 17) directly on this copy. Return both your examination answer sheet and this copy at the end of the examination, after verifying your name and student number are written on both. A calculator can be used if necessary. A formula sheet is provided on the back of this page.

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  • Page 1/17

    First Name: Last Name: Student #: Instructor: Dr. K. Foyle Class: Physics 1BB3, Modern Physics for the Life Sciences

    FIRST MIDTERM EXAMINATION DATE: Feb. 6th, 2014 TIME: 7pm PLACE: T29 101 & 105 DURATION OF EXAMINATION: 2 hours This examination paper includes 17 pages and 17 questions. You are responsible for ensuring that your copy of the paper is complete. Bring any discrepancy to the attention of your invigilator. Special Instructions:

    Answer multiple-choice questions (questions 1 to 16) on the provided examination answer sheet. Answer the written answer question (question 17) directly on this copy. Return both your examination answer sheet and this copy at the end of the examination, after verifying your name and student number are written on both. A calculator can be used if necessary. A formula sheet is provided on the back of this page.

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    Mark: Multiple-choice questions (#1-16): /85 Written answer questions (#17): /15 Total: /100 ---------DO NOT WRITE ABOVE THIS LINE------------

    FORMULA SHEET

    Moment of inertia of a solid disk or solid cylinder about its center: I=(1/2)MR2

    Moment of inertia of a hollow cylinder about its center: I=MR2 Moment of inertia of a sphere about an axis through its center: I=(2/5)MR2 Moment of inertia of a rod about one of its ends: I=(1/3)ML2 g=9.81 m.s-2

    a = v2

    R(t) =0 +t

    (t) = 0 +0t +12t

    2

    = F r sin

    I = miri2i

    Ekrotation =12 I

    2

    L = mv r sinL = I

  • Page 3/17

    Part 1: Multiple-choice questions (questions 1 to 16) Answer these questions on the provided examination answer sheet. (Make sure you write your name and student number on the answer sheet!). For questions with a numerical answer, if your answer does not match any of the provided answers, pick the one with the closest value. Questions 1 to 9 are completely independent questions. Questions 10 to 16 are part of a problem, but still mostly independent from one another.

    Question 1. You are driving at a constant speed of 39.0 m/s along a circular trajectory with radius 29.0 m. What is your acceleration? A- 52.4 m/s2 B- 1.34 m/s2 C- 21.6 m/s2 D- 0.7444 m/s2 E- 8.45 m/s2

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    Question 2. Which one of the following statements concerning the moment of inertia I is false?

    A- I may be expressed in units of kg.m2. B- I depends on the angular acceleration of the object as it

    rotates. C- I depends on the location of the rotation axis relative to

    the particles that make up the object. D- I depends on the orientation of the rotation axis relative

    to the particles that make up the object. E- Of the particles that make up an object, the particle with

    the smallest mass may contribute the greatest amount to I.

    Question 3. A high diver in midair pulls her legs inward toward her chest in order to rotate faster. Doing so changes which of these quantities: her angular momentum L , her rotational inertia, I , and her rotational kinetic energy torK ?

    A- L only B- I only C- torK only D- L and I only E- I and torK only

  • Page 5/17

    Question 4. A 2kg disk having a radius of 20cm is pivoted at a point on its circumference by a thin nail. The disk is pulled back so that its center of mass is at the same level as the nail. With what angular velocity must the disk be launched so that the center of mass of the disk will just reach a point vertically above the nail?

    !!CM!

    Nail!

    ini)al!

    nal!

    A- 0 rad/s B- 20.3 rad/s C- 15.4 rad/s D- 12.8 rad/s E- 8.1 rad/s

    Question 5. A 20 kg wooden beam 3.0 m long is tapered such that the center of mass is located nearer one end. Two women are carrying the beam horizontally, one woman at each end. If one is supporting 50% more weight than the other, how far from the lighter end of the beam is the center of mass located? A- 2.0 m B- 1.5 m C- 1.8 m D- 1.3 m E- None of the above

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    Question 6: 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 string? A- The tension is unchanged. B- The tension reduces to half its original value. C- The tension increases to twice its original value. D- The tension increases to four times its original value. E- The tension reduces to one-fourth of its original value. Question 7. If an object in circular motion increases its angular velocity from 250 rpm to 500 rpm in 5 seconds, the tangential acceleration of a point of the object at 6 cm from its center of rotation is approximately

    A- 0.3 m/s2 B- 0.8 m/s2 C- 1.5 m/s2 D- 3 m/s2 E- 6 m/s2

  • Page 7/17

    Question 8. Consider two masses, each of 2 kg at the ends of a light rod of length L with the axis of rotation through the center of rod. The rod is doubled in length and the mass on the left side is halved. Where can an additional mass of 2 kg be placed on the rod in order to keep the rod balanced on its center?

    A- At far left B- At the far right C- At L/4 from the left D- At L/2 from the left E- At L/3 from the right

    Question 9. A hockey stick, 2.00 m long and 1.2 kg, is rested up against a wall. The torque about the pivot point caused by the sticks weight is 5.0 Nm. What is the torque about the pivot point caused by the force of the wall on the top of the stick? (Assume that this stick is in static equilibrium)

    A- 1.0 Nm B- 3.0 Nm C- 5.0 Nm D- 10. Nm E- Need to know the angle between the hockey stick

    and the vertical.

    Pivot Point

  • Page 8/17

    Read the following carefully before answering questions 10 to 16. Those questions are all independent, except for question 15 (you will need to answer several previous questions correctly to answer question 15). A yoyo is a toy consisting of a flattened spool around which a string can be wrapped. If a yoyo is released without initial velocity, with the end of the string kept in a fixed position, the downward speed of the yoyo starts by increasing and but then it ends up decreasing. In the following series of questions (10 to 16), we try to understand why. We will consider a yoyo made of two identical solid disks (each with mass M1 = 0.050 kg and radius R1 = 0.075 m), joined by a concentric solid cylindrical shaft (mass M2 = 0.0050 kg, radius R2 = 0.010 m), as shown in Fig. 1. A string with negligible mass is wrapped around the yoyo, a shown in Fig. 2 and 3. Initially, the length of string wrapped around the yoyo is L = 0.85 m. The string is attached to the yoyo at one end (so that the wrapped up part of the string cannot slip around the shaft) and its other end is kept in a fixed position as the yoyo falls (Fig. 3).

  • Page 9/17

    Figure 1 Question 10: The moment of inertia of the yoyo with respect to a horizontal axis passing through its center of mass (this axis is indicated by a dashed line in Fig. 1 above) is: A- I = 2.5 10-7 kg.m2 B- I = 1.4 10-4 kg.m2 C- I = 2.8 10-4 kg.m2 D- I = 5.3 10-4 kg.m2 E- I = 5.6 10-4 kg.m2

  • Page 10/17

    Figure 2

    Question 11: Which of the following statements correctly applies to the yoyo during its fall (see Fig. 2 above for the orientation of the wrapped up string with respect to the yoyo, and consider that a counterclockwise rotation is positive, as indicated in the figure): A- The net force exerted on the yoyo is zero. B- The net torque exerted on the yoyo is zero. C- The net torque exerted on the yoyo is negative. D- The net torque exerted on the yoyo is positive. E- The angular momentum of the yoyo is conserved.

  • Page 11/17

    Question 12: During its fall, the mechanical energy of the yoyo is conserved. Why? A- Because the net torque exerted on the yoyo is zero. B- Because the net force exerted on the yoyo is zero. C- Because all the forces exerted on the yoyo are conservative. D- Because, although at least one force applied to the yoyo is not conservative, it does no work. E- Because none of the forces exerted on the yoyo does any work.

  • Page 12/17

    Figure 3 Question 13: If the zero for gravitational potential energy is chosen at z=0 (that is when the yoyo is at its lowest point, as indicated in Fig. 3 above), what is the value of the mechanical energy of the yoyo during its fall? A- 0 J B- 0.46 J C- 0.88 J D- 1.8 J E- One would need more information to answer this question.

  • Page 13/17

    Question 14: If the thickness of the layer of wound-up string is d (see Fig. 2 on page 6), what is the relationship between the angular velocity of the yoyo () and its downward speed (v)? (If you find it hard to answer this question, try considering how much string is unwound during one rotation of the yoyo, and how that length of string is related to the downward speed and angular velocity of the yoyo). A- v=R1. B- v=(R1+d). C- v=R2. D- v=(R2+d). E- v=(R1+R2).

    Question 15: When the yoyo is halfway through its fall, at z=L/2, the thickness of the layer of wound-up string is d = 0.030 m, but at z=0 it has decreased to d = 0.000 m. Making use of this information, calculate the downward speed of the yoyo at z=L/2 and z=0 (these speeds will be noted v1 and v2, according to the notation in Fig. 3). A- v1 = 3.1 m/s and v2= 1.8 m/s B- v1= 2.5 m/s and v2= 0.78 m/s C- v1= 2.5 m/s and v2= 0.62 m/s D- v1= 1.8 m/s and v2=0.78 m/s E- v1= 1.8 m/s and v2=0.62 m/s

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    Question 16: What happens to the linear and rotational components of the kinetic energy of the yoyo as it falls from z=L/2 to z=0? A- Both the linear kinetic energy and rotational kinetic energy of the yoyo decrease. B- Both the linear kinetic energy and rotational kinetic energy of the yoyo increase, but the linear kinetic energy increases more than the rotational kinetic energy. C- Both the linear kinetic energy and rotational kinetic energy of the yoyo increase, but the rotational kinetic energy increases more than the linear kinetic energy. D- The linear kinetic energy of the yoyo increases while its rotational kinetic energy decreases. E- The linear kinetic energy of the yoyo decreases while its rotational kinetic energy increases. --------------End of Part 1: Multiple-Choice Questions----------

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    Part 2: Written answer question (question 17) Answer this question directly on your examination copy, explaining clearly and concisely how you arrived to the answer. Question 17. A hollow, vertical cylindrical drum of radius R = 2.45 m is rotated with angular velocity about its central axis, as shown in the figure below. A smaller cylinder (with mass M = 80.0 kg) rotates around on its inner wall. The smaller cylinder is NOT in anyway attached to the cylindrical drum.

    a. Draw a free-body diagram of the mass, labeling all the forces acting on it, and write a list of what the labels mean. b. What angular velocity (in radians/second) would be required to make the normal force pressing against the mass be twice the magnitude of its own weight? c. At the value of obtained in (b), what is the minimum coefficient of static friction, S, necessary to keep the mass suspended on the inner wall of the cylinder as it rotates?

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    ----------End of Part 2: Written Answer Question---------- -------------------END OF EXAMINATION----------------