Inelastic Collisions. Experiment Number 8 Physics 109 Fall 2017

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1 Inelastic Collisions Experiment Number 8 Physics 109 Fall 2017

2 Midterm Exam Scores Number Score Range

3 Outline Ballistic Pendulum Physics of Rotation Angular Collision Clicker Questions

4 Ballistic Pendulum An excellent method for measuring the velocity of a bullet The pendulum has mass, M The bullet has mass, m Bullet strikes pendulum with velocity, v Pendulum then rises. Pendulum rotates thru an angle Radius to center of mass times cosine of angle then subtracted from the radius gives the height that the pendulum rises.

5 Ballistic Pendulum (cont.) This illustrates the geometry of the pendulum

6 Ballistic Pendulum (cont.) Note: Energy is not conserved in the collision, but is conserved after the collision. The kinetic energy of the ball-pendulum is converted to potential energy as the ballpendulum rises. From this we can calculate the ball velocity. Vb = ((mb + mp)/mb)sqrt(2gh)

7 Ballistic Pendulum (cont.) The height is just the radius (radius x cos(angle)) Remember the velocity you calculate is an approximation because we did not include the moment of inertia of the pendulum.

8 Translation and Rotation We have been dealing with changes in position which we call Translations There are also possibilities for angular changes and this is called Rotational Motion The rolling object experiment involves both as the objects roll down an inclined plane Please note that we are skipping Chapter 7 in the text which deals with vibrations we will return to this topic later.

9 Pure Rotation We must become used to expressing angles in terms of RADIANS. The angle in Radians, θ, is expressed as the arc length divided by the radius. θ l r θ = l/r Note that radians are dimensionless!!

10 A DVD Example Information stored/retrieved must be at a constant LINEAR velocity because data is fed at a constant rate, roughly 11 Megabits/second If each bit requires about.5 x 10-6 m of space, this would require about 550 cm/s of linear speed. At the inside of the disk, r = 3 cm and the circumference is 18 cm. So, 550/18 = rotations per second, or 1833 RPM At the outside of the disk, r = 6 cm and 876 RPM will give the required linear velocity

11 New Rotational Terms Angular displacement, θ = θ Angular velocity = ω = θ/ t units radians/sec. θ0 units-radians At constant angular velocity each rotation takes the same amount of time... this is the Period, T of the rotation. So for constant ω Angular acceleration = α = ω/ t = radians/sec2 = 2π/T

12 Kinematic Equations These are for rotational motion at constant acceleration... Translational Rotational Vx = vx0 + axt ω = ω0 + αt X = x0 + v0t + 1/2at2 θ = θ0 + ω0t + 1/2αt2 Vx2 = vx02 + 2ax x ω2 = ω02 + 2α θ

13 Rotational Analogs Translational Rotational Position, x Angular Position, θ Displacement, x Displacement, θ Velocity, vx Angular velocity, ω Acceleration, a Time, t Angular acceleration, α Time, t

14 A CD Example Given..rotation goes from 500 RPM to 200 RPM at an angular acceleration, α, = x radians/second Convert RPM to rad/s 500 = 52.4 ; 200 = 20.9 So, θ = (ω2 - ω02) / 2α = 1.63 x 105 rad = 25,900 revolutions Now compute the time = (ω ω0)/α = 4.45 x 103 seconds = minutes Corresponds to the capacity of an audio CD!

15 Rotational and Tangential Motion Tangential velocity and speed, given T the period of rotation Vt = distance/time = 2πr/T = 2πr/2π/ω = rω This is called the tangential speed. For the CD, outer radius, rw = 1.31 m/s and for the inner radius, rw = 1.25 m/s Centripetal Acceleration = vt2/r = (rw)2/r = rw2 This is valid for radian measure

16 Kinetic Energy and Rotational Inertia The kinetic energy, K, of a small mass at a radius 2 r, with a tangential velocity of vt = ½ mvt For a collection of such point masses in rotation: K = [ ] mi v ti = [ ] mi r i w 2 2 So we can write: K = 1 ( m r 2 ) w 2 i i 2 And we call the sum the Rotational Inertia The symbol is I, so K = ½ Iω 2

17 Rotational Inertia Be sure to look at the information given on page 179 of your text. The I values for a number of different shapes are given there. For purposes of our experiment we will need four different values... Hollow cylinder about its axis... I = MR 2 A disk or solid cylinder about its axis, I = ½ MR A hollow sphere about its diameter I = 2/3 MR2 A solid sphere about its diameter I = 2/5 MR2 2

18 Rotational Inertias

19 Shapes we will use

20 A shape you will see again The formula given here is for the rotational Inertia. There is another related property called the Moment of Inertia, or first moment. It does not contain the mass term, but does contain the area. The formula becomes : I = 1/12 ba3

21 And where will you see it? The bending of a beam involves knowledge of the bending moment M and the moment of inertia of the beam cross section, I. The stress in the outer portion of the beam is given by σ = Mc/EI where c is the beam half height and E is the elastic modulus of the material. L/2 P c h P/2 M= P(L/2), I = 1/12bh P/2 3 b

22 An Aside: Why do you think that a popular beam cross section is called an I-Beam? Because most of the material is far from the center line, where it is most resistant to bending.

23 Rotational Collision Measure masses and radii of disk and hoop We will apply the conservation of angular momentum Spin the disk and measure the revolution rate Drop the hoop and measure the new revolution rate Do this for three trials

24 Rotational Collision (cont.) Conservation of angular momentum: Idωi = (Id + Ih)ωf Remember to multiply by 2π to get angular velocity from revolutions per second.

25 The moment of inertia of the disk is? A. I = mr^2 B. I = ½ mr^2 C. I = 2 mr^2 D. None of the above

26 The moment of inertia of the thick walled cylinder? A. I = m(router2 rinner2) B. I = m(router2 + rinner2) 2 outer r 2 inner C. I = ½ m(r ) D. I = ½ m(router2 + rinner2)

27 What is lost in these collisions? A. Temperature B. Heat C. Kinetic Energy D. None of the above

28 What would improve the accuracy of the ballistic pendulum? A. Position the angle recorder at zero B. Measure the moment of inertia of the pendulum C. Position the angle recorder one degree below the actual measurement. D. All of the above. E. B and C.

29 A CD maintains a constant linear rate as it rotates. How does it accomplish this? A. The head moves out faster at the rim. B. It changes RPM as the distance from the center changes. C. Neither of the above. D. Both of the above. l

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