Lecture 12. Center of mass Uniform circular motion

Transcription

1 Lecture 12 Center of mass Uniform circular motion

2 Today s Topics: Center of mass Uniform circular motion Centripetal acceleration and force Banked curves

3 Define the center of mass The center of mass is a point that represents the average location for the total mass of a system. For linear motion, we can consider all of the mass of the object to be concentrated at the center of mass.

4 Center of mass simplifies calculations involving conservation of momentum for combined linear and rotational motion, the CM of an object is the point that obeys the linear equations of motion we have developed DEMO: Shape

5 What if there are more than 2 y objects? Oxygen Oxygen nm 60.0 o Sulfur 60.0 o nm x A sulfur dioxide molecule consists of 2 oxygen atoms and a sulfur atom. The sulfur atom is twice as massive as the oxygen atom. Find the x and y coordinates of the center of mass of this molcule 1. Find the x and y components of all atoms with respect to the origin (sulfur atom) x-coordinate y-coordinate Oxygen atom (on the left) x 1 = (0.143 nm) sin 60.0 = nm Oxygen atom (on the right) x 2 = +(0.143 nm) sin 60.0 = nm y 1 = +(0.143 nm) cos 60.0 = nm y 2 = +(0.143 nm) cos 60.0 = nm Sulfur atom x 3 = 0 nm y 3 = 0 nm

6 Let m 1 and m 2 represent the masses of the oxygen atoms on the left and right, respectively. Likewise, let m 3 be the mass of the sulfur atom. The x-coordinate of the center of mass is x cm = m x + m x + m x = m 1( nm) + m nm m 1 + m 2 + m 3 m 1 + m 2 + m 3 Oxygen nm 60.0 o ( ) + m 3 ( 0 nm) y Sulfur 60.0 o Oxygen nm x Since m 1 = m 2 (both are oxygen atoms), the equation above yields x = cm 0 m The y-coordinate of the center of mass is y cm = m y + m y + m y = m 1 ( nm) + m nm m 1 + m 2 + m 3 m 1 + m 2 + m 3 ( ) + m 3 ( 0 nm) Substituting m 1 = m 2 = m and m 3 = 2m into the equation above yields y cm = m ( nm) + m ( nm) + 2m( 0 nm) m + m + 2m = nm

7 Continuous mass distributions Use symmetry! CM somewhere along this line The CM need not be inside the object! For a system composed of multiple parts with symmetry, first condense each part to its CM and then treat each CM as a pointlike particle 2 kg 1 kg 2 kg = CM 1 kg

8 Velocity of the center of mass Δx cm = m 1Δx 1 + m 2 Δx 2 m 1 + m 2 v cm = m 1v 1 + m 2 v 2 m 1 + m 2 = p 1 + p 2 m 1 + m 2 = p total m total In an isolated system, the total linear momentum does not change, therefore the velocity of the center of mass does not change.

9 A stationary 4-kg shell explodes into three pieces. Two of the fragments have a mass of 1 kg each and move along the paths shown with a speed of 10 m/s. The third fragment moves upward as shown. 2 kg What is the speed of the third fragment? 1 kg kg 10 m/s 10 m/s What is the speed of the center of mass of this system after the explosion? (a) zero m/s (b) 1 m/s (c) 3 m/s (d) 5 m/s (e)7 m/s

10 ACT: Velocity of the Center of Mass Two equal-mass particles (A and B) are located at some distance from each other. Particle A is held stationary while B is moves away at speed v. What happens to the center of mass of the two-particle system? a) it does not move b) it moves away from A with speed v c) it moves toward A with speed v d) it moves away from A with speed ½v e) it moves toward A with speed ½v Let s say that A is at the origin (x = 0) and B is at some position x. Then the center of mass is at x/2 because A and B have the same mass. If v = Δx/Δt tells us how fast the position of B is changing, then the position of the center of mass must be changing like Δ(x/2)/Δt, which is v/2.

11 Uniform Circular Motion Uniform circular motion is the motion of an object traveling at a constant speed on a circular path. Let T be the time it takes for the object to travel once around the circle. T is also known as the period of the motion

12 Example A ball moves with a constant speed of 4 m/s around a circle of radius 0.25 m. What is the period of the motion?

13 Circular Motion In uniform circular motion, the speed is constant, but the direction of the velocity vector is not constant. So even if the speed is constant, there has to be an acceleration. centripetal acceleration What is the direction and magnitude of this centripetal acceleration?

14 Centripetal acceleration D v! v! 0! a = Δ! v c Δt Δ! v =! v F! v 0 - v! 0 v! F The centripetal acceleration in uniform circular motion points to the center of the circle.

15 Example 1 A racecar is traveling 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 increases by a factor of 2. (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

16 Centripetal Force! å F = a! m The centripetal force is the net force required to keep an object moving on a circular path. The direction of the centripetal force always points toward the center of the circle and continually changes direction as the object moves.! F c = ma! c = -m v r 2 rˆ

17 What s the origin of the Centripetal Force? A 0.25-kg ball attached to a string is rotating in a horizontal circle of radius 0.5 m. If the ball revolves twice every second, what is the tension in the string? Here I am calculating the magnitude of F C F = 2pr v = = T F T T F c = m = v r 2 v mac = m r 2p (0.5 m) 0.5 s 2 = (0.25 kg) = 6.28 m/s ( 6.28 m/s) 0.5 m 2 = 20 N In this case, it s the tension in the string

18 ACT: Tetherball In the game of tetherball, the struck ball whirls around a pole. In what direction does the net force on the ball point? a) toward the top of the pole b) toward the ground c) along the horizontal component of the tension force d) along the vertical component of the tension force e) tangential to the circle The vertical component of the tension balances the weight. The horizontal component of tension provides the centripetal force that W F T F T points toward the center of the circle. W

19 ACT: Ball and String Two equal-mass rocks tied to strings are whirled in horizontal circles. The radius of circle 2 is twice that of circle 1. If the period of motion is the same for both rocks, what is the tension in cord 2 compared to cord 1? a) T 2 = ¼T 1 b) T 2 = ½T 1 c) T 2 = T 1 d) T 2 = 2T 1 e) T 2 = 4T 1 The centripetal force in this case is given by the tension, so F T = mv 2 /r. For the same period, we find that v 2 = 2v 1 (v 22 = 4 v 12 ). However, for the denominator, we see that r 2 = 2r 1 which gives us the relation F T,2 = 2F T,1.

20 How about for a car going around a turn? On an unbanked curve, the static frictional force provides the centripetal force. Looking from behind F N f S mg

21 Or for a car going around a banked turn? On a frictionless banked curve, the centripetal force is the horizontal component of the normal force. The vertical component of the normal force balances the car s weight.

22 F c = F sin q = N m v 2 r F cosq = mg N

23 ACT: Barrel of Fun A rider in a barrel of fun finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? a b c d e The normal force of the wall on the rider provides the centripetal force needed to keep her going around in a circle. The downward force of gravity is balanced by the upward frictional force on her, so she does not slip vertically. Follow-up: What happens if the rotation of the ride slows down?

24 An Amusement Park Ride A small child rides in a barrel of fun ride. The floor drops downward and the child remains pinned against the wall. If the radius of the device is 2.15 m and the coefficient of static friction between the child and the wall is 0.400, with what minimum speed is the child moving if he is to remain pinned against the wall? F N mv 2 F N = r f S f max S mg = 0 mg Notice v does not depend on the mass of the child.

25 ACT: Going in Circles A skier goes over a small round hill with radius R. Because she is in circular motion, there has to be a centripetal force. At the top of the hill, what is F c of the skier equal to? a) F c = N + mg b) F c = mg N c) F c = N mg f K d) F c = N e) F c = mg F c points toward the center of the circle (i.e., downward in this case). The weight vector points down and the normal force (exerted by the hill) points up. The magnitude of the net force, therefore, is F c = mg N. v mg N R