Physics 111: Lecture 3 Today’s Agenda Reference frames and relative motion Uniform Circular Motion 1

Inertial Reference Frames: Cart on track on track A Reference Frame is the place you measure from. It’s where you nail down your (x,y,z) axes! An Inertial Reference Frame (IRF) is one that is not accelerating. We will consider only IRFs in this course. Valid IRFs can have fixed velocities with respect to each other. More about this later when we discuss forces. For now, just remember that we can make measurements from different vantage points.

Relative Motion Consider a problem with two distinct IRFs: An airplane flying on a windy day. A pilot wants to fly from Champaign to Chicago. Having asked a friendly physics student, she knows that Chicago is 120 miles due north of Urbana. She takes off from Willard Airport at noon. Her plane has a compass and an air-speed indicator to help her navigate. The compass allows her to keep the nose of the plane pointing north. The air-speed indicator tells her that she is traveling at 120 miles per hour with respect to the air.

Relative Motion... The plane is moving north in the IRF attached to the air: Vp, a is the velocity of the plane w.r.t. the air. Air Vp,a

Relative Motion... But suppose the air is moving east in the IRF attached to the ground. Va,g is the velocity of the air w.r.t. the ground (i.e. wind). Air Vp,a Va,g

Relative Motion... What is the velocity of the plane in an IRF attached to the ground? Vp,g is the velocity of the plane w.r.t. the ground. Vp,g

Relative Motion... Tractor Vp,g = Vp,a + Va,g Is a vector equation relating the airplane’s velocity in different reference frames. Va,g Vp,a Vp,g

Lecture 3, Act 1 Relative Motion You are swimming across a 50m wide river in which the current moves at 1 m/s with respect to the shore. Your swimming speed is 2 m/s with respect to the water. You swim across in such a way that your path is a straight perpendicular line across the river. How many seconds does it take you to get across ? (a) (b) (c) 1 m/s 50 m 2 m/s

Lecture 3, Act 1 solution y Choose x axis along riverbank and y axis across river x The time taken to swim straight across is (distance across) / (vy ) Since you swim straight across, you must be tilted in the water so that your x component of velocity with respect to the water exactly cancels the velocity of the water in the x direction: 1 m/s 1m/s y 2 m/s m/s x

Lecture 3, Act 1 solution So the y component of your velocity with respect to the water is So the time to get across is m/s m/s 50 m y x

Uniform Circular Motion What does it mean? How do we describe it? What can we learn about it?

What is UCM? Puck on ice Motion in a circle with: Constant Radius R Constant Speed v = |v| y v (x,y) R x

How can we describe UCM? In general, one coordinate system is as good as any other: Cartesian: (x,y) [position] (vx ,vy) [velocity] Polar: (R,) [position] (vR ,) [velocity] In UCM: R is constant (hence vR = 0).  (angular velocity) is constant. Polar coordinates are a natural way to describe UCM! y v (x,y) R  x

Polar Coordinates: The arc length s (distance along the circumference) is related to the angle in a simple way: s = R, where  is the angular displacement. units of  are called radians. For one complete revolution: 2R = Rc c = 2 has period 2. 1 revolution = 2radians R v x y (x,y) s 

Polar Coordinates... R x y (x,y)  x = R cos  y = R sin  1 cos sin /2  3/2 2  -1

Polar Coordinates... v = R Tetherball In Cartesian coordinates, we say velocity dx/dt = v. x = vt In polar coordinates, angular velocity d/dt = .  = t  has units of radians/second. Displacement s = vt. but s = R = Rt, so: y v R s t x v = R

Period and Frequency  = 2 / T = 2f Recall that 1 revolution = 2 radians frequency (f) = revolutions / second (a) angular velocity () = radians / second (b) By combining (a) and (b)  = 2 f Realize that: period (T) = seconds / revolution So T = 1 / f = 2/ v R s   = 2 / T = 2f

Recap: x = R cos()= R cos(t) y = R sin()= R sin(t)  = arctan (y/x)  = t s = v t s = R = Rt v = R v (x,y) R s t

Aside: Polar Unit Vectors We are familiar with the Cartesian unit vectors: i j k Now introduce “polar unit-vectors” r and  : r points in radial direction  points in tangential direction ^ ^ ^ ^  ^ r ^ (counter clockwise) y R  j x i

Acceleration in UCM: Even though the speed is constant, velocity is not constant since the direction is changing: must be some acceleration! Consider average acceleration in time t aav = v / t v1 v2 v v2 R v1 t

Acceleration in UCM: Even though the speed is constant, velocity is not constant since the direction is changing. Consider average acceleration in time t aav = v / t v R seems like v (hence v/t ) points at the origin!

Acceleration in UCM: Even though the speed is constant, velocity is not constant since the direction is changing. As we shrink t, v / t dv / dt = a a = dv / dt R We see that a points in the - R direction.

Acceleration in UCM: This is called Centripetal Acceleration. Now let’s calculate the magnitude: Similar triangles: v v1 v2 But R = vt for small t v2 R v1 So: R

Centripetal Acceleration UCM results in acceleration: Magnitude: a = v2 / R Direction: - r (toward center of circle) ^ R a 

Derivation: We know that and v = R Substituting for v we find that:  a = 2R

Lecture 3, Act 2 Uniform Circular Motion A fighter pilot flying in a circular turn will pass out if the centripetal acceleration he experiences is more than about 9 times the acceleration of gravity g. If his F18 is moving with a speed of 300 m/s, what is the approximate diameter of the tightest turn this pilot can make and survive to tell about it ? (a) 500 m (b) 1000 m (c) 2000 m

Lecture 3, Act 2 Solution 2km

Example: Propeller Tip The propeller on a stunt plane spins with frequency f = 3500 rpm. The length of each propeller blade is L = 80cm. What centripetal acceleration does a point at the tip of a propeller blade feel? what is a here? f L

Example: First calculate the angular velocity of the propeller: so 3500 rpm means  = 367 s-1 Now calculate the acceleration. a = 2R = (367s-1)2 x (0.8m) = 1.1 x 105 m/s2 = 11,000 g direction of a points at the propeller hub (-r ). ^

Example: Newton & the Moon What is the acceleration of the Moon due to its motion around the Earth? What we know (Newton knew this also): T = 27.3 days = 2.36 x 106 s (period ~ 1 month) R = 3.84 x 108 m (distance to moon) RE = 6.35 x 106 m (radius of earth) R RE

Moon... Calculate angular velocity: So  = 2.66 x 10-6 s-1. Now calculate the acceleration. a = 2R = 0.00272 m/s2 = 0.000278 g direction of a points at the center of the Earth (-r ). ^

Moon... So we find that amoon / g = 0.000278 Newton noticed that RE2 / R2 = 0.000273 This inspired him to propose that FMm  1 / R2 (more on gravity later) R RE amoon g

Lecture 3, Act 3 Centripetal Acceleration The Space Shuttle is in Low Earth Orbit (LEO) about 300 km above the surface. The period of the orbit is about 91 min. What is the acceleration of an astronaut in the Shuttle in the reference frame of the Earth? (The radius of the Earth is 6.4 x 106 m.) (a) 0 m/s2 (b) 8.9 m/s2 (c) 9.8 m/s2

Lecture 3, Act 3 Centripetal Acceleration First calculate the angular frequency : Realize that: RO RO = RE + 300 km = 6.4 x 106 m + 0.3 x 106 m = 6.7 x 106 m 300 km RE

Lecture 3, Act 3 Centripetal Acceleration Now calculate the acceleration: a = 2R a = (0.00115 s-1)2 x 6.7 x 106 m a = 8.9 m/s2

Recap for today: Reference frames and relative motion. (Text: 2-1, 3-3, & 4-1) Uniform Circular Motion (Text: 5-2, also 9-1) Look at Textbook problems Chapter 3: # 47, 49, 97, 105 29