1. PHY 231 1 PHYSICS 231 Lecture 16: Centripetal acceleration Remco Zegers Walk-in hour: Tue 4-5 pm Helproom

2. PHY 231 2 Last lecture... Average angular velocity (rad/s) Instantaneous Angular velocity Average angular acceleration (rad/s 2 ) Instantaneous angular acceleration 2  rad  360 0 1 0  2  /360 rad 1 rad  360/2  deg

3. PHY 231 3 And... Angular and linear velocity are related: Angular and linear acceleration are related: demo: tangential velocity

4. PHY 231 4 Rotational motion Angular motion  (t)=  (0)+  (0)t+½  t 2  (t)=  (0)+  t

5. PHY 231 5 gears r 1 =0.3 r 2 =0.15 r 3 =0.7 m If  1 =3 rad/s, how fast Is the bike going? b) What if r 1 =0.1 m ? V 1 =  1 r 1 =3*0.3=0.9 m/s V 1 =V 2 (because of the chain)  2 =V 2 /r 2 =0.9/0.15=6 rad/s  3 =  2 (connected) V 3 =  3 r 3 =6*0.7=4.2 m/s b) V 1 =  1 r 1 =3*0.1=0.3 m/s V 1 =V 2 (because of the chain)  2 =V 2 /r 2 =0.3/0.15=2 rad/s  3 =  2 (connected) V 3 =  3 r 3 =2*0.7=1.4 m/s

6. PHY 231 6 quiz (extra credit) An object is moving in a circle with radius R=2 m. If the radius is doubled and the angular velocity remains the same, the linear velocity will: a)half b)remain the same c)double d)quadruple e)whatever

7. PHY 231 7 for old times sake… A man is standing on a platform resting on a frictionless surface. At the end of the platform a wall is placed and fixed to the platform.The man throws a ball against the wall and it bounces back over the man as shown in the figure. After the ball has bounced against the wall… a)the platform moves to the left b)the platform moves to the right c)the platform remains at its place

8. PHY 231 8 Driving a car through a bend Is there a force that pushes you away from the center of the circle? Newton’s first law: If no net force is acting on an object, it will continue with the same velocity (inertia of mass) Velocity is a vector (points to a direction) If no net force is acting on an object, it will not change its direction. A force is acting on the car (steering+friction) but you tend to go in the same direction as you were going! It is not a force that pushes you, but the lack of it! The side door will keep you from falling out: it exerts a force on you and you exert a force on the door (F 21 =-F 12 )

9. PHY 231 9 Centripetal acceleration The change in velocity is not the change in speed but in direction. Sin(  /2)=(  s/2)/r Sin(  /2)=(  v/2)/v  v=  s*(v/r)  t (  s/2)/r=(  v/2)/v v  s/  t a c =v 2 /r

10. PHY 231 10 Centripetal acceleration a c =v 2 /r directed to the center of the circular motion Also v=  r, so a c =  2 r This acceleration can be caused by various forces: gravity (objects attracted by earth) tension (object making circular motion on a rope) friction (car driving through a curve) etc This acceleration is NOT caused by a mysterious force

11. PHY 231 11 Forces that can cause centripetal acceleration. Object swinging on a rope. T T  F=ma T=ma c T=mv 2 /r=m  2 r An object with m=1 kg is swung with a rope of length 3 m around with angular velocity  =2 rad/s. What is the tension in the rope? T=m  2 r=1*2 2 *3=12 N

12. PHY 231 12 Lifting by swinging Swinging mass (m 1 ) with velocity v Hanging mass (m 2 ) r What is the relation between v and r that will keep m 2 stationary? T T F g =m 2 g m 1 : T=m 1 a m 2 : -T=m 2 g v out of paper a=-m 2 g/m 1 Also: a c =(-)v 2 /r v=linear/tangential velocity of m 1 v 2 /r= m 2 g/m 1 If m 1 slows down, r must go down so m 2 sinks. demo

13. PHY 231 13 Swinging mass (m 1 ) with velocity v Hanging mass (m 2 ) r T T F g =m 2 g v out of paper v 2 /r= m 2 g/m 1 question If the velocity of m 1 is increased to twice the original value, the radius of orbit a)decreases to ¼ of its original value b)decreases to ½ of its original value c)remains the same d)increases to 2x its original value e)increases to 4x its original value

14. PHY 231 14 A car going through a bend A car is passing through a bend with radius 100 m. The kinetic coefficient of friction of the tires on the road is 0.5. What is the maximum velocity the car can have without flying out of the bend?  F=ma  k n=ma c  k mg=mv 2 /r 0.5*9.81=v 2 /100 v=22 m/s

15. PHY 231 15 2001: A space odyssey A space ship rotates with a linear velocity of 50 m/s. What should the distance from the central axis to the crew’s cabin’s be so that the crew feels like they are on earth? (the floor of the cabins is the inside of the outer edge of the spaceship)  F=ma mg=mv 2 /r so r=v 2 /9.8 and thus r=255 m The rotating spaceship has an acceleration directed towards the center of the ship: the ‘lack’ of forces acting on the crew pushes them against the ship.

16. PHY 231 16 Conical motion  What is the centripetal acceleration if the mass is 1 kg and  =20 o ? mg T Tcos  Tsin   Vertical direction:  F=ma Tcos  -mg=0 So T=mg/cos  Horizontal direction:  F=ma c Tsin  =ma c mgsin  /cos  =mgtan  =ma c a c =gtan  =9.8*0.36=3.6 m/s 2 demo

17. PHY 231 17 A general strategy As usual, make a drawing of the problem, if not given. Draw all the forces that are acting on the object(s) under investigation. Decompose each of these into directions toward the center of the circular path and perpendicular to it. Realize that  F to center =ma c =mv 2 /r

18. PHY 231 18 question A point on the rim of a 0.25 m radius rotating wheel has a centripetal acceleration of 4.0 m/s 2. What is the linear speed of the wheel? (  F to center =ma c =mv 2 /r) a)1.0 m/s b)2.0 m/s c)3.2 m/s d)4.0 m/s e)8.0 m/s a c =v 2 /r 4=v 2 /0.25 v 2 =16 v=4 m/s