Bill the Cat, tied to a rope, is twirled around in a vertical circle. Draw the free-body diagram for Bill in the positions shown. Then sum the X and

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2 Assignment (a) No assigned WH. (b)read motion in the presence of resistive forces (finish the chapter). Go over problems covered in classes. (c)read: System and Environments, Work done by a constant force, the scalar product of two vectors. Work done by a varying force. (d) Exam on Thursday

3 Bill the Cat, tied to a rope, is twirled around in a vertical circle. Draw the free-body diagram for Bill in the positions shown. Then sum the X and Y forces.

4 ΣF y = ma c F T + mg = mv²/r F T = mv²/r - mg F T = m ((v²/r) - g) a c mg F T ΣF y = ma c F T - mg = mv²/r F T = mv²/r + mg F T = m ((v²/r) + g) a c F T mg

5 Minimum velocity needed for an object to continue moving in a vertical circle. Any less velocity and the object will fall. At this point, F T = 0, so ΣF y = ma c F T + mg = mv²/r 0 + mg = mv²/r g = v 2 /r rg = v 2 or, v c = rg

6 Example 3: The Effect of Speed on Centripetal Force The model airplane has a mass of 0.90 kg and moves at a constant speed on a circle that is parallel to the ground. The path of the airplane and its guideline lie in the same horizontal plane, because the weight of the plane is balanced by the lift generated by its wings. Find the tension T in the guideline(length=17m) for speeds of 19 and 38m/s. 6

7 Equation 5.3 gives the tension directly: F C =T=mv 2 /r Speed =19m/s Speed =38m/s 7

8 Conceptual Example 6: A Trapeze Act In a circus, a man hangs upside down from a trapeze, legs bent over the bar and arms downward, holding his partner. Is it harder for the man to hold his partner when (a) the partner hangs straight down and is stationary or (b)when the partner is swinging through the straight-down position? 8

9 Reasoning and Solution: When the man and his partner are stationary, the man s arms must support his partner s weight. When the two are swinging, however, the man s arms must do an additional job. Then the partner is moving on a circular arc and has a centripetal acceleration. The man s arms must exert and additional pull so that there will be sufficient centripetal force to produce this acceleration. Because of the additional pull, it is harder for the man to hold his partner while swinging than while stationary. 9

10 Level Curves (your turn) (what is the maximum speed of the car) A 1500 kg car moving on a flat, horizontal road negotiates a curve as shown. If the radius of the curve is 35.0 m and the coefficient of static friction between the tires and dry pavement is 0.523, find the maximum speed the car can have and still make the turn successfully. v rg October 22, 2013

11 Level Curves (calculations in the book) The force of static friction directed toward the center of the curve keeps the car moving in a circular path. f N v s,max max F y 2 v sn m r N mg 0 mg snr m (0.523)(9.8m / s max smgr m 2 )(35.0m) v rg gr s 13.4m / s October 22, 2013

12 Example 4: Centripetal Force and Safe Driving Compare the maximum speeds at which a car can safely negotiate an unbanked turn (r= 50.0m) S dry 0.9 S --dry = 0.9 S --icy = 0.1 F S N f s max N s F S mg 12

13 Check your understanding A car is traveling in uniform circular motion on a section of road whose radius is r. The road is slippery, and the car is just on the verge of sliding. If the car s speed was doubled, what would have to be the smallest radius in order that the car does not slide? 1. 3R 2. 2R 3. R 4. 4R 0% 0% 0% 0%

14 Check your understanding A car is traveling in uniform circular motion on a section of road whose radius is r. The road is slippery, and the car is just on the verge of sliding. What would be your answer to part (a) if the car were replaced by one that weighted twice as much? 1. 3R 2. 2R 3. R 4. 4R 0% 0% 0% 0%

15 Banked Curves 15

16 A car is going around a friction-free banked curve. The radius of the curve is r. F N sin F that points toward the center C C F N sin mv F N cos and, since the car does not accelerate in the vertical direction, this component must balance the weight mg of the car. F cos mg N r 2 16

17 F sin mv / N F cos mg N 2 r tan v 2 rg At a speed that is too small for a given, a car would slide down a frictionless banked curve: at a speed that is too large, a car would slide off the top. 17

18 Example 5: Banked Curves Roads designed for high-speed travel have banked curves to give the normal force a component toward the center of the curve. That lets cars turn without relying on friction between tires and road. At what angle should a road with 200-m curvature radius be banked for travel at 90 km/h (25 m/s)? Net/Mechanics/Circular%20Motion/banked _no_friction.htm October 22, 2013

19 Vertical Circular Motion Usually, the speed varies in this stunt. non-uniform 19

20 (1) 2 mv1 F N 1 mg r (3) 2 mv3 F N 3 mg r =F C1 =F C3 (2) mv r 2 F 2 N 2 (4) mv r 2 F 4 N 4 =F C2 =F C4 The magnitude of the normal force changes, because the speed changes and the weight does not have the same effect at every point. 20

21 The weight is tangent to the circle at points 2 and 4 and has no component pointing toward the center. If the speed at each of the four places is known, along with the mass and radius, the normal forces can be determined. They must have at least a minimum speed at the top of the circle to remain on the track. v 3 is a minimum when F N3 is zero. v 3 Weight mg provides all the centripetal force. The rider experiences an apparent weightlessness. rg 21

22 Example 6: Roller Coaster A roller-coaster car has a mass of 500 kg when fully loaded with passengers. (a) If the vehicle has a speed of 20.0 m/s at point A, what is the force exerted by the track on the car at this point? (b) What is the maximum speed the vehicle can have at point B and still remain on the track? October 22, 2013

23 step 1: draw free-body diagram step 2: choose a coordinate system: centripetal acceleration? step 3: Newton s 2nd law N mg October 22, 2013

24 What is the maximum/minimum speed at point B? F F ma net, y y y N mg October 22, 2013