Lecture5 - Applica'ons of Newton’s Laws of Mo'on...

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Unformatted text preview: Applica'ons of Newton’s Laws of Mo'on We will con'nue to use Newton’s laws of mo'on to solve problems. A lot of chalk board stuffs. Take notes. Then we will go to Chapter 6. Let’s start with the problem, we couldn’t finish in the last class. A shu&le of mass M slides on a fric3onless track. A rope is a&ached to the shu&le, which passes over a pulley and is a&ached to a block of mass m, hanging from the rope. Assuming the rope and the pulley are massless, what is the accelera3on of the shu&le ? M y (1) M a = T (2) T - m g = - m a a= m g M +m m x y M y FN T x m T x FgM Fgm In the situa3on shown in the adjacent diagram, a) what are the magnitudes of T and FN ? b) if I cut the rope, what’s the accelera3on of the block (assume no fric3on)? Note special choice of coordinate system & direc'on of FN. a) Block is in equilibrium: Free ­body diagrams: b) If I cut the rope, the y component of the weight will s'll be balanced by the normal force. But the tension will be removed. − mg sin θ = ma ⇒ a = − g sin θ Compare with Example 5.10 A spring balance at rest shows that a body weighs 19.6 N. How much weight does the spring balance show if it is weighed in an elevator 1.  going up with constant velocity of 5 m/s? 2.  accelera3ng upward with an accelera3on of 2 m/s2? 3.  accelera3ng downward with an accelera3on of 2 m/s2? Textbook Example 5.9 Determine the the accelera3on of the two objects and the tension in the lightweight cord. (Atwood machine) Textbook, Ch. 5 Problem 40 Calculate θ and n. n F θ f mg y n f θ mg F x F = 35.0 N f = 20.0 N m = 20.0 kg Textbook, Ch. 5 Problem 37 vi = 20 m/s2, Δx = 45 m μs = 0.650, μk = 0.550 m = 3.80 kg a)  Will the book start to slide over the seat? b)  What force does the seat exert on the book in this process? f mg y n a x Example 5.13 A block of mass m1 on a rough, horizontal surface is connected to a ball of mass m2 by a lightweight cord over a lightweight, fric'onless pulley, as shown in Figure 5.20a. A force of magnitude F at an angle θ with the horizontal is applied to the block as shown. The coefficient of kine'c fric'on between the block and surface is μk . Determine the magnitude of the accelera'on of the two objects. Chapter 6 Circular mo'on and other applica'ons of Newton’s Laws The average skater achieves speeds of about 10 ­15 mph, but when approaching high risk maneuvers speeds of about 30 mph are necessary. How is it that Tony Hawk is able to stay on his skateboard and not fall onto his neck while doing the “loopty loo” shown in this picture? Uniform circular mo'on We already know that in uniform ( = constant speed) circular mo'on, the accelera'on is directed toward the center and its magnitude is given by v2 a= Here v is the constant speed and r is the radius of the mo'on. This is called the centripetal accelera'on. Both magnitude and direc'on can be combined in a vector rela'on v2 ˆ a=− r r r Consider a ball moving uniformly along a circle. The centripetal accelera'on measures the constant devia'on from the otherwise straight line mo'on (Newton’s first law). The force causing the centripetal accelera'on, i.e., the force needed to keep the ball in a circle, is called the centripetal force. By Newton’s second law it is given by mv 2 Fr = ma = − r ˆ r mv 2 Magnitude of this force is Fr = r . Radial and tangen'al accelera'on, non ­uniform circular mo'on If a par'cle moves along a smooth curved path where the velocity changes both in magnitude and direc'on, its accelera'on in general has not just the radial part. It possesses the transverse part as well. The radial component is perpendicular to the tangen'al component. The total accelera'on vector is the vector the sum of the two: Note that the tangen'al accelera'on results from the change in magnitude of velocity while the radial component results from the change in the direc'on. We recognize that the radial component is the centripetal accelera'on: 2 ar = − v r at = dv dt a = ar + at It is clear that non ­uniform circular mo'on, not just mo'on along any curve, possesses both radial and tangen'al components and accelera'on. v t ar at a Example: the conical pendulum The figure on the right shows the mo'on of a conical pendulum. Let’s figure out how fast the pendulum moves round the circle (the bobom of the cone). From the free ­body diagram of the ball, we can get the following equa'ons from Newton’s second law: ∑ Fy = T cosθ − mg = 0 ⇒ T cosθ = mg..................(1) x L T θ r mg ∑F Tcosθ θ Tsinθ mg = T sin θ = max mv 2 ⇒ T sin θ = ..................(2 ) r From (1) and (2): v2 tan θ = ⇒ v = rg tan θ rg ⇒ v = Lg sin θ tan θ Example: Maximum speed of a car around a turn A 1500kg car moving on a flat road turns through a curve. If the radius of the curve is 35.0m and the coefficient of kine'c fric'on is 0.523, what is the car’s max speed without sliding (or driding)? n ⇒ µ k n = mac ..............(1) ∑F y x = f = ma. FG Ff ∑F = 0 → n − mg = 0 → n = mg.............(2 ) 2 vmax (1) and (2) → µ k mg = m r → vmax = µ k gr vmax = 13.4 m / s y Example: Car about a banked turn At what angle should the curve of the road be banked so that on ice (mk=0) the car wont slip, if v=13.4 m/s, r=35.0 m? x ∑F x = nx = max → − n sin(θ ) = − mac 2 vmax ⇒ n sin(θ ) = m ...................(1) r ∑F Dividing (1) by (2) gives: y = 0 → ny − mg = 0 → n cos(θ ) = mg.....................(2 ) Example  ­ Tony Hawk A skateboarder goes through a loop ­the ­loop with radius R=2.5m. How fast does he have to go in order to make a full loop, without falling off ? mg + n = ma If the skateboard is just about to fall off, n = 0. Minimum speed: What if he goes faster than that ? n=m v nR − mg > 0 ⇒ v = ( + g R) > gR R m 2 a mg n y i.e. gravity is not sufficient to keep him on a 'ght circular path by itself. Need the normal force from track as well. Determine the force exerted by the skateboard on the skater at the top and at the boAom of vt the loop. n radial Examining the tangent components: ∑ Ft = mat → mg sin(θ ) = mat → at = g sin(θ ) θ mg tangen'al mg mg n vt n ∑F Considering the radial components: r = − mg cosθ + nr = mac → n − mg cosθ = mac The normal force at the top (θ = 180): ⎛ v2 ⎞ mv 2 → n = mg cos(θ ) + → n = mg ⎜ + cos(θ )⎟ r ⎝ gr ⎠ The normal force at the bobom (θ = 0): Fic''ous forces Forces that are “felt” as a result of a non ­iner'al reference frame. Examples: •  when turning a corner, you feel pulled towards the outside edge of the turn (= centrifugal force) •  air hockey table on a train. As the train accelerates forwards, the air hockey puck appears to be accelera'ng backwards by some force. •  Coriolis Force: the Coriolis force arises when a rota'ng reference frame accelerates underneath objects moving in a straight line (seen by an observer in an iner'al reference frame). Merry ­go ­round, Earth Coriolis Force Mo'on in presence of resis've forces R Feather falling in air At first it accelerates and gains speed downward. But it faces upward air drag , which increases with the velocity. Ul'mately a point comes when the upward air drag balances its weight, the net force is zero. Accelera'on thus reduces to zero and the the cobon ball coast down with constant velocity. This constant velocity is called the terminal velocity. This is an example of mo'on in presence of resis've force. Here the resis've force is the air drag. The direcBon of resisBve force is always opposite to the direcBon of moBon of the body relaBve to the medium (air in this example). mg Fnet = R – mg = 0 a = 0 v = constant Skydiver Mo'on in presence of resis've forces (contd…) Magnitude of a resis've depends on the speed of the object. The dependence can be complex. We usually take the following two simplified models: Model 1: Resis've force propor'onal to velocity Model 2: Resis've force propor'onal to speed squared The first is normally used to model a very small object moving slowly in the medium, while the second is used to model large objects and/or fast moving objects in the medium. Example of the might be a small ball falling in a tall jar of glycerin. A high speed base ball in air might be the example of the second type. Model 1 R R = −bv b ⎛ dv ⎞ ⎛ dv ⎞ Fy = mg − bv = ma → mg − bv = m ⎜ ⎟ → ⎜ ⎟ = g − v ∑ ⎝ dt ⎠ ⎝ dt ⎠ m dv b → + v−g= 0 dt m − bt mg →v= (1 − e m ) b mg At terminal velocity (vT), dv/dt = 0 mg/b = vT. v = vT (1 − e Note that the Bme constant τ = m/b. When t = τ ,v = 0.632 vT. − bt m ) Mo'on in presence of resis've forces (contd…) Magnitude of a resis've depends on the speed of the object. The dependence can be complex. We usually take the following two simplified models: Model 1: Resis've force propor'onal to velocity Model 2: Resis've force propor'onal to speed squared The first is normally used to model a very small object moving slowly in the medium, while the second is used to model large objects and/or fast moving objects in the medium. Example of the might be a small ball falling in a tall jar of glycerin. A high speed base ball in air might be the example of the second type. Model 1 R R = −bv b ⎛ dv ⎞ ⎛ dv ⎞ Fy = mg − bv = ma → mg − bv = m ⎜ ⎟ → ⎜ ⎟ = g − v ∑ ⎝ dt ⎠ ⎝ dt ⎠ m dv b → + v−g= 0 dt m − bt mg →v= (1 − e m ) b mg At terminal velocity (vT), dv/dt = 0 mg/b = vT. v = vT (1 − e Note that the Bme constant τ = m/b. When t = τ ,v = 0.632 vT. − bt m ) Example 6.8 (textbook): sphere falling in oil A small sphere of mass 2.00 g is released from rest in a large vessel filled with oil, where it experiences a resis've force propor'onal to its speed. The sphere reaches a terminal speed of 5.00 cm/s. Determine the 'me constant and the 'me it takes the sphere to reach 90% of its terminal speed. Model 2 Resis've force propor'onal to speed squared For objects moving with high speed in a medium, the resis've force can be wriben as: R= 1 2 DρAv 2 Where, D = dragg coeffcient ( a dimensionless empirical quuan'ty ρ = density of the medium (eg., air) A = cross sec'onal area of the moving object measured in a plane perpendicular its velocity In this case Newton’s second law gives a = g − ⎜ ⎝ vT = ⎛ D ρA ⎞ 2 ⎟v 2m ⎠ 2 mg D ρA ...
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