• If your clicker answers are not being recorded on Blackboard, let us know. • There is a set of prac;ce problems on vectors in MasteringPhysics. There are also some more in online HW1 (s;ll accessible if you haven’t done them). • Please make sure your Phys 221 Student ID in MP is your ISU netID, not your ISU email or ISU student ID #. (This is diﬀerent than your MP login ID.) You can change it in your Proﬁle. Example: Maximum height A person throws a ball from a window 10 m above the ground, with an ini9al velocity of 10 m/s and at a 30-‐degree angle with the horizontal. Air resistance can be neglected. Calculate the maximum height the ball will reach above the ground. v0 = 10 m/s ymax ⇔ v y = 0 y y0 = 10 m θ = 300 Time to get to the top: 0 = v0 sinθ − gt → t = ymax ? v0 sinθ g x 1 ymax = y0 + v0 sinθ t − gt 2 2 2 What determines the maximum height of a projec9le? Answer: the ini9al velocity v 0y in the ver9cal direc9on v sinθ 1 ⎛ v0 sinθ ⎞ = y0 + v0 sinθ 0 − g⎜ g 2 ⎝ g ⎟⎠ 2 (10 m/s ) sin2 (30° ) v02 sin2 θ = y0 + = 10 m + = 11.3 m 2 2g 2 ( 9.8 m/s ) ( ) How does the range R depend on θ ? y v0 yﬁnal = 0 θ 1 y = y0 + v0 y t + ay t 2 2 1 0 = 0 + v0 y t − gt 2 2 x = x0 + v 0 x t x ⎧t = 0 (start!) Transfer ;me (t) from ver;cal mo;on ⎪ ⎨ 2v0 y 2v0 sinθ to horizontal mo;on ⎪t = g = g ⎩ R = x − x0 = v 0 x t 2v0 sinθ 2v02 sinθ cosθ v 02 sin2θ = v0 cosθ = = g g g R 450 900 Angle (θ) Maximum range is for θ = 450 Lecture -‐ 06 Circular mo9on Review: accelera;on in 2D (and 3D) ! ! ! dv d(vv) ˆ dv dvˆ a= = = vˆ + v v = vvˆ dt dt dt dt ! ! ! a = a|| + a⊥ a|| or tangen9al accelera9on: Parallel/an9parallel to velocity Changes the speed. aperp or normal or centripetal/ radial accelera9on: • Perpendicular to velocity • Inwards (towards the center of curvature) • Does not change the speed but shiVs the direc9on of the mo9on. Circular mo;on Circular mo9on is the mo9on in a circle with constant radius. Polar coordinates Polar coordinates (r, θ ) are more convenient than Cartesian coordinates to describe circular mo;on: r = R, only θ = θ (t) Arch: s = R θ this equa;on only works if θ is in radians Deﬁni;on: y • 1 radian = angle so that s=R • 1 revolu;on = 2π radians Rela9on to Cartesian coordinates: x = r cos θ y = r sin θ R θ s x Velocity dx dy Cartesian coordinates: v x = ; vy = dt dt Polar coordinates: dr dθ vr = ; ω = dt dt Radial velocity Angular velocity For circular mo9on: vr = 0 ds dθ s = Rθ ⎯⎯ → =R ⎯⎯ → v = Rω dt dt (where ω is in radians/unit 9me) EXAMPLE: Two balls Two balls connected by thin rod as shown, at distances R and 2R from the center, move in circles. Same angular speed ω for both (same angle in any Δt) Diﬀerent (linear) speeds (ball 2 travels twice the distance in any Δt) v1 = Rω v2 = 2Rω Uniform circular mo;on (UCM) ω is constant. This is an example of periodic mo;on. Period T : Time it takes to go back to the same situa9on (same posi9on, same velocity). Frequency f : Number of revolu9ons per unit 9me. Units: Hz (turns per second), rpm (rev per min) 2π T= ω 1 f= T ω = 2π f Example: Ferris Wheel A Ferris Wheel of radius 8.0 m rotates at a constant rate of 1.5 rpm. Find: a. The period b. The linear speed of a cabin. ACT: Ferris wheel The ferris wheel in the ﬁgure rotates counterclockwise at a uniform rate. What is the direc;on of the average accelera;on of a gondola as it goes from the top to the bo`om of its trajectory? A. Down B. C. The accelera9on is 0 because the mo9on is uniform. Radial or centripetal accelera;on During uniform circular mo9on, speed is constant, but velocity is not!!! The direc9on keeps changing! a =ûa || + a⊥ (constant speed) Perpendicular/normal/radial/ centripetal accelera9on • Points to the center of the circle UCM in Cartesian components Angle covered in 9me interval ∆t: θ= θ0 + ω ∆t. If we choose θ0 = 0 at t = 0, it’s θ = ω t y x ωt y t x The Cartesian coordinates are sine-‐func9ons: ( ) y = Rsin(ω t ) x = Rcos ω t Magnitude of Radial Accelera;on For UCM: x = R cos (ωt ) y = R sin (ωt ) ax = −Rω 2 cos (ωt ) v x = −Rω sin (ωt ) d dt v y = Rω cos (ωt ) |r |= R d dt | a |= Rω 2 ! ! a = −ω 2 r = −ω 2 R rˆ |v |= Rω ay = −Rω 2 sin (ωt ) | a |= Rω 2 a = −ω 2 r = −ω 2 R rˆ | a |= Rω 2 a = −ω 2 r = −ω 2 R rˆ |a|= Rω In UCM, all the accelera9on is centripetal. Thus, 2 v ar = Rω 2 = R Radial accelera9on 2 Accelera;on simulator In the movie “The Right Stuﬀ”, a seat at the end of a long arm that rotates very fast is used to prepare astronauts for high accelera9ons. If R = 5 m, what is the speed needed to have a = 5g? ω R ACT: Angular speed A horizontal rod of length 1 m turns with constant speed about a ver9cal axis. The blue 9p of the rod makes one turn each second. Find the angular speed of the red spot painted in the middle of the rod. A. π rad/s B. 2π rad/s C. 4π rad/s The linear speed is not: vblue tip = R ω = (1 m)(2π rad/s) = 6.28 m/s vred spot = R ω = (0.5 m)(2π rad/s) = 3.14 m/s 6.28 m 3.14 m 1 m It makes sense: The blue 9p is covering twice the distance covered by the red spot in the same 9me. v2 Does this make sense? |ar |= R 1 As you decrease the radius, the rate at |a|∝ which the velocity shiVs, hence the R accelera9on, grows. As you increase the velocity, the |a|∝ v 2 accelera9on grows in two ways: -‐ The rate at which the velocity is shiVing grows. -‐ The amount of velocity which needs to be shiVed around also grows. R Non-‐uniform circular mo;on Slowing down R ar v Speeding up v a a at at R r a Radial accelera9on (towards the center) → changes in direc9on Tangen9al accelera9on (tangen9al to trajectory) → changes in speed Angular accelera;on If linear speed v is changing, the angular speed ω is also changing Angular accelera9on dω d 2θ α= = 2 dt dt d dt v = Rω ⎯⎯ → at = Rα ! ! dv dv because'at = '''''''(not'to'be'confused'with'a = ''!!!) dt dt Equa;ons of mo;on when α= constant The deriva9on is formally iden9cal to what we did in 1D: dθ ω= dt dω α= dt ω = ω0 + α t 1 2 θ = θ 0 + ω0 t + α t 2 Δθ ω + ω 0 ω= = Δt 2 ω 2 − ω02 = 2αΔθ Example: Wheel At t = 0, a wheel has an angular velocity of 20 rad/s. It speeds up at a uniform rate of 4.0 rad/s2 un9l a circuit breaker trips at t = 5.0 s. From then on, it turns through 1600 rad as it coasts to a stop with constant angular accelera9on. What was its accelera9on as it slowed down? Example Test Ques;on(ETQ)

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