Physics 2210 Fall smartphysics 13 Collisions, Impulse, and Reference Frames 14 Rotational kinematics (?) 10/30/2015

Transcription

1 Physics 2210 Fall 2015 smartphysics 13 Collisions, Impulse, and Reference Frames 14 Rotational kinematics (?) 10/30/2015

2 Exam 3: smartphysics units Midterm Exam 3: Day: Fri ov. 06, 2015 Time: regular class time Section 01 12:55 1:45 pm Section 10 02:00 02:50 pm Location: FMAB Practice problems will be posted by Sunday

3 Poll Who played the drummer of the Queen-Haters? A. Johnny Depp B. Leonardo DiCaprio C. John Candy D. Keifer Sutherland

4 Unit 13 Relative Velocities are independent of reference frames Relative Velocities are independent of reference frames v 2i v 1i = v 2i + V CC v 1i + V CC = v 2i v ii v 2f v 1f = v 2f + V CC v 1i + V CC = v 2i v ii

5 Poll Revisited Before: v 2 v 1 = V After: v 2 v 1 = 2V 3 ot the same RELATIVE SPEED A block slides to the right with speed V on a frictionless floor and collides with a bigger block which is initially at rest. After the collision the speed of both blocks is V/3 in opposite directions. Was the collision elastic? A. Yes B. o

6 Impulse and Propulsion Definition: An impulse is the change in (vector) momentum of an object, p, resulting from a the action of a force F on the body over a time interval t = t f t i. ewton s 2 nd Law says: dp dd = F So by the Fundamental Theorem of Calculus the impulse is given by p = F t dd t i where the force can vary with time and is given by the function F t If we have two bodies interacting with no external forces then dp 1 dd = F dp 2 21, dd = F 12 But by ewton s 3rd Law F 12 = F 21 and so dp 2 dd = dp 1 p dd 2 = p 1 P p 1 + p 1 = 0 t f Remember this is a vector equation Remember this is a vector equation

7 Unit 13 ote from instructor: Remember integrating the net force in distance gives K, the change in kinetic energy K r 2 K = F dr = W, W j F j dr r 1 Integrating the net force in time gives p, the change in momentum p, p is also known as the net/total impulse I I 1 + I 2 + I I t 2 p = F dd = I, I j F j dd t 1 r 1 r 2 t 2 t 1

8 Poll A constant force acts for a time Δt on a block that is initially at rest on a frictionless surface, resulting in a final velocity V. Suppose the experiment is repeated on a block with twice the mass using a force that s half as big. For how long would the force have to act to result in the same final velocity? A. Four times as long. B. Twice as long. C. The same length. D. Half as long. E. One quarter as long.

9 Poll Identical balls are dropped from the same initial height and bounce back to half the initial height. In Case 1 the ball bounces off a cement floor and in case 2 the ball bounces off a piece of stretchy rubber. In which case is the average force acting on the ball during the collision the biggest? A. Case 1 B. Case 2

10 Example 13.1 An automatic rifle fires bullets of mass 55.0 grams each, at a maximum burst rate of 800 rounds/min. The muzzle velocity (actually this is a speed) of the bullets exiting the barrel is 975 m/s. (a) Each bullet takes 1.5 ms to travel down the barrel. Find the peak force of recoil of the rifle during the firing of one bullet. (b) Find the average (over many rounds) force of recoil of the rifle when firing at the maximum rate. m = 55 g F rrrrrl v ee = 975 m/s 800 bullets/min Magnitude of impulse imparted on rifle per bullet: p p = mv ee During the firing of one bullet over time τ = 1.5 ms = s, then the peak recoil force has magnitude F pppp = p τ = mv ee τ = kg 975 m s s =

11 Unit 13

12 Momentum and Energy of a System of Particles For a group of particles labeled i = 1,2,,, each with mass m i, (timedependent) position r i, and (time-dependent) velocity v i : Total momentum is given by: P = m i v i = m i v i + V CC Where v i v i V CC is, again, as indicated by the prime, the velocity of the ith particle in the CM frame. It follows then P = m i v i + m i V CC The first term (total momentum of group in CM frame) can be shown to vanish: m i v i = d dd m ir i = d dd m ir i = d dd m ir i d dd m ir CC = d dd MR CC d dd MR CC = 0 P = m i V CC = MV CC m i R CC

13 Momentum and Energy of a System of Particles The total kinetic energy of the group is given by (using the dot product form): K = 1 2 m 2 i v i = 1 2 m iv i v i = 1 2 m i v i + V CC v i + V CC = 1 2 m iv i v i m iv i V CC m iv CC V CC The second term can be shown to vanish! m iv i V CC = m i v i V CC Where the factor in the parenthesis is the total momentum in CM frame, which is always zero (from previous page)!!! And so K = 1 2 m i vv i MV CC 2 = K + K CC The total kinetic energy of a system of particles is the sum of the total kinetic energy about the center-of-mass (i.e. the 1 st term is the total kinetic energy measured in the CM frame) and the kinetic energy OF THE CETER OF MASS ote: (1) unlike the total momentum in the CM frame, the total kinetic energy in the CM frame is OT zero. (2) The kinetic energy of the center-of-mass is COSERVED if the net external force (from outside the group) is zero.

14 Rigid Body Motion

15 Digitized x and y positions of both ends over 16 frames

16 t (s) x1 (pix) y1 (pix) x2 (pix) y2 (pix) xcm (pix) ycm (pix) xcmf ycmf x1' y1' x2' y2' Center of mass: Constant velocity (almost zero) motion in the x-direction Constant (negative) acceleration motion in the y-direction

17 t (s) x1 (pix) y1 (pix) x2 (pix) y2 (pix) xcm (pix) ycm (pix) xcmf ycmf x1' y1' x2' y2' Coordinates in CM frame: When x is near zero then y is near max/min And when y is near zero then x is near max/min: characteristic of ROTATIO!

18 Rotational Motion Most manufactured objects consist of rigid bodies: where the particles (i.e. atoms) that make up the object are constrained to maintain a designed shape. The motion of such objects can be studied by separating A. the motion of the center-of-mass (like a point particle located at the CM with mass M of the body) B. Rotations of the body about the center-of-mass We will start by studying the rotational motion of a rigid body about a fixed axis: found everywhere in the most simple machines: wheels Pulleys Motors Propellers

19 Rotation About a fixed axis It is typically convenient to choose the rotation axis to be the z-axis. Here we have chosen z to point out of this page. The symbol indicates an arrow pointing out of the page. (And this symbol indicates an arrow into the page) The angular position of the body can be determined by the position r of an ARBITRARY reference point (which we will mark with a peg ) z y θ r x In the position shown the reference point lies on the x-axis, which we generally choose to be the ZERO (i.e. ORIGI) of rotation about the z-axis. And we generally pick the counter-clockwise (CCW) direction to be the positive rotational sense The angular position is then the (azimuthal) angle θ of the position vector r of the reference point measured CCW from the +x axis. smartphysics uses θ for the angular position and so we will too

20 θ = + 11π 18 rad y θ = + 17π 18 rad θ = + 5π 18 rad z x θ = π 18 rad θ = + 23π 18 rad θ = + 29π 18 rad As the object rotates about the axel (rotation axis), the reference point (remember this is an arbitrary fixed point in the body) traces out a circle