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PHYSICS 231 Lecture 18: Centripetal acceleration

PHYSICS 231 Lecture 18: Centripetal acceleration. Remco Zegers Walk-in hour: Monday 9:15-10:15 am Helproom. Last week. Average angular velocity (rad/s). Instantaneous Angular velocity. Average angular acceleration (rad/s 2 ). 2  rad  360 0 1 0  2/360 rad 1 rad 360/2 deg.

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PHYSICS 231 Lecture 18: Centripetal acceleration

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  1. PHYSICS 231Lecture 18: Centripetal acceleration Remco Zegers Walk-in hour: Monday 9:15-10:15 am Helproom PHY 231

  2. Last week... Average angular velocity (rad/s) Instantaneous Angular velocity Average angular acceleration (rad/s2) 2 rad  3600 10 2/360 rad 1 rad 360/2 deg Instantaneous angular acceleration PHY 231

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

  4. Rotational motion Angular motion (t)= (0)+(0)t+½t2 (t)= (0)+t PHY 231

  5. 3=v3/r3=3.5/0.7=5 rad/s 2=3 v2=2r2=5*0.15=0.75 m/s v1=v2 (chain) r1=v1/1=0.75/3=0.25 m quiz extra credit If the bike is going at 3.5 m/s and 1=3 rad/s, what is the radius of r1? r3=0.7 m r1=??? r2=0.15 • 0.15 m • 0.20 m • 0.25 m • 0.3 m • 0.35 m 3.5 m/s PHY 231

  6. 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 (F21=-F12) PHY 231

  7. Sin(/2)=(s/2)/r Sin(/2)=(v/2)/v (s/2)/r=(v/2)/v v=s*(v/r) t t vs/t ac=v2/r Centripetal acceleration The change in velocity is not the change in speed but in direction. PHY 231

  8. Centripetal acceleration ac=v2/r directed to the center of the circular motion Also v=r, so ac=2r • 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 PHY 231

  9. A race car accelerating on a track. atangential atotal= (ac2+atangemtial2) ac PHY 231

  10. Forces that can cause centripetal acceleration. Object swinging on a rope. T F=ma T=mac T=mv2/r=m2r T 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2r=1*22*3=12 N PHY 231

  11. T m1: T=m1a m2: -T=m2g T Also: ac=(-)v2/r v=linear/tangential velocity of m1 v2/r= m2g/m1 Fg=m2g If m1 slows down, r must go down so m2 sinks. Lifting by swinging Swinging mass (m1) with velocity v What is the relation between v and r that will keep m2 stationary? v out of paper r a=-m2g/m1 Hanging mass (m2) demo PHY 231

  12. 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 kn=mac kmg=mv2/r 0.5*9.81=v2/100 v=22 m/s PHY 231

  13. 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. 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=mv2/r so r=v2/9.8 and thus r=255 m PHY 231

  14. T Tcos  Vertical direction: F=ma Tcos-mg=0 So T=mg/cos Tsin mg Conical motion What is the centripetal acceleration if the mass is 1 kg and =20o?  Horizontal direction: F=mac Tsin=mac mgsin/cos=mgtan=mac ac=gtan=9.8*0.36=3.6 m/s2 demo PHY 231

  15. 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 Fto center=mac=mv2/r PHY 231

  16. ac=v2/r 4=v2/0.25 v2=16 v=4 m/s question • A point on the rim of a 0.25 m radius rotating wheel has • a centripetal acceleration of 4.0 m/s2. What is the angular • speed of the wheel? (Fto center=mac=mv2/r) • 1.0 rad/s • 2.0 rad/s • 3.2 rad/s • 4.0 rad/s • 8.0 rad/s PHY 231

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