Newton s Three Laws. F = ma. Kinematics. Gravitational force Normal force Frictional force Tension More to come. k k N

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2 Newton s Three Laws F = ma Gravitational force Normal force Frictional force Tension More to come Kinematics f s,max = µ sfn 0 < fs µ sfn k k N f = µ F

3 Rules for the Application of Newton s Laws of Motion 1. Choose the system to be studied. 2. Make a simple sketch of the system. 3. Choose a convenient coordinate system. 4. Identify all the forces that act on the system. Label them on the diagram. 5. Apply Newton s laws of motion to the system.

4 Problem : The figure shows three blocks attached by cords that loop over frictionless pulleys. Block B lies on a table; the masses are m A =6.00 kg, m B =8.00 kg and m C =10.kg. What is the tension is the rope at the right? What is the acceleration of block B if (a) there is no friction between block B and the table, and (b) there is kinetic friction with μ k = 0.25? Two Tensions T L and T R. Three different Force Equations with up positive on A and C and left to right positive for B. Draw these! T L M A g = M A a T R T L = M B a M C g T R = M C a Find a: a = M C M A g = ( 0.167)9.8 = 1.63m / s 2 M A + M B + M C Find T R : T R = M ( C g a) = 10( 8.13) = 81.7N

5 Problem: The figure shows, as a function of time t, the force component F x that acts on a 3.00 kg ice block that can move only along the x axis. At t=0 s, the block is moving in the positive direction of the axis, with a speed of 3.0 m/s. What are its (a) speed and (b) direction of travel at t=11s. (a) The acceleration (which equals F/m in this problem) is the derivative of the velocity. Thus, the velocity is the integral of F/m, so we find the area in the a = graph dv t (15 units) and divide by the mass (3) to obtain v v o = 15/3 = 5. Since dt v = o = F 3.0 m v( t) v(0)= ( F m)dt m/s, then 0 t 0 (b) Our v( t) positive = 1 answer in part (a) implies points in the +x direction. m Fdt area in F(t) graph + v(0)= + v(0) m

6 Drag Force and Terminal Speed When an object moves through a fluid (gas or liquid) it experiences an opposing force known as drag. Under certain conditions (the moving object must be blunt and must move fast so the flow of the liquid is turbulent) the magnitude of the drag force is given by the expression 1 D = Cρ Av 2 2 Here C is a constant, A is the effective crosssectional area of the moving object, ρ is the density of the surrounding fluid, and v is the object s speed. Consider an object (a cat of mass m in this case) that starts moving in air. Initially D = 0. As the cat accelerates D increases and at a certain speed v t D = mg. At this point the net force and thus the acceleration become zero and the cat moves with constant speed v t known as the terminal speed. 1 2 D = Cρ Avt = mg 2 v t = 2mg Cρ A

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8 What drives circular motion?

9 Centripetal Force C In Chapter 4 we saw that an object that moves on a circular path of radius r with constant speed v has an acceleration a. The direction of the acceleration vector always points toward the center of rotation C (thus the name centripetal). Its magnitude is constant and is given by the equation a = v r 2. If we apply Newton s law to analyze uniform circular motion we conclude that the net force in the direction that points toward C must have 2 mv magnitude: This force is known as centripetal force. F = r.

10 Question: What provides force for centripetal acceleration? A centripetal force accelerates a body by changing the direction of the body s velocity without having to change the body s speed F cent = ma cent = mv2 r because v and r are constant, magnitude of F cent (& a cent ) is constant. direction of F cent (& a cent ) towards center (constantly changing direction!!) The notion of centripetal force may be confusing at times. A common mistake is to invent this force out of thin air. Centripetal force is not a new kind of force. It is simply the net force that points from the rotating body to the rotation center C. - tension (points along the direction of string/rope) - friction (points parallel to surface - opposes motion) - gravity (points downward with magnitude g) - normal (points perpendicular to surface)

11 Recipe for Problems That Involve Uniform Circular Motion of an Object of Mass m on a Circular Orbit of Radius r with Speed v C. y r Draw the force diagram for the object. x m v Choose one of the coordinate axes (the y-axis in this diagram) to point toward the orbit center C. Determine F net, y. Set F net y mv r 2, =.

12 C x y Sample problem 6-8: The Rotor is a large hollow cylinder of radius R that is rotated rapidly around its central axis with a speed v. A rider of mass m stands on the Rotor floor with his/her back against the Rotor wall. Cylinder and rider begin to turn. When the speed v reaches some predetermined value v min, the Rotor floor abruptly falls away. The rider does not fall but instead remains pinned against the Rotor wall. The coefficient of static friction μ s between the Rotor wall and the rider is given. Find v min. We draw a free-body diagram for the rider using the axes shown in the figure. The normal reaction F N is the centripetal force. 2 mv Fx,net = FN = ma = (eq. 1) R F = f mg = 0, f = µ F mg = µ F (eqs. 2) y,net s s s N s N 2 mv 2 Rg Rg min R µ s µ s If we combine eq. 1 and eqs. 2 we get: mg = µ v = v =. s

13 Turning a Corner You turn a corner with your car --> How do you do that? (your speed and radius of motion are constant) what friction is needed for the car not to slip? Friction provides the centripetal acceleration ˆ x : ˆ y : ' ) R ( " F x = # f S = #ma x = m # v 2 & " F y = F N # mg = ma y = 0 $ % F N = mg " m v 2 % $ ' = f S # R & f S max If we use then we can extract the maximum v that we can go: " m v 2 % $ ' = f max S = µ S F N = µ S mg # R & v max = µ S gr

14 Vertical Circular Motion What happens in vertical circular motion? What is going on with the centripetal acceleration? a c There must always be a centripetal acceleration pointed toward the axis of rotation. In a cyclist doing a loop-de-loop, the SUM OF FORCES provides the a c. The two forces that provide a c include the WEIGHT and the NORMAL FORCE. Since the F N is always to the surface, and W points down, the magnitude of F N varies around the loop. The same thing applies if you are swinging a pail of water holding on to the handle Your TENSION T will vary around the orbit.

15 circular motion problem What minimum velocity (constant) must the car have at the top of the loop to remain on the track? Use Newton s 2nd Law knowing the forces at the top of the loop # & y ˆ : " N " F g = m %" v2 ( $ r ' car If car just remains on track, N 0 or # & y ˆ :"F g = "( mg) = m %" v2 ( or v = gr $ r ' so minimum velocity is v min = gr If constant v, how does N change around loop? At TOP # & " y ˆ : N = m% v2 ( ) mg $ r ' N small At SIDES # & " x ˆ : ± N = m % ± v2 ( $ r ' At BOTTOM # & " y ˆ : N = m% v2 ( + mg $ r ' N large

16 Swing a pail of Water over your Head Why doesn t it fall out? What does a free body diagram look like for this? If the pail isn t moving, water falls out on your head Tension T mg When you swing it around, you apply an additional tension T, and when the pail is above your head, the free body diagram here is complete. WHY DOESN T IT FALL ON YOUR HEAD?? These forces provide the centripetal acceleration for the circular motion the force equation is: m v 2 r = T + W The slowest speed you can swing it is the point where T=0 (this is tension, not period!). Then you must have a velocity that is: m v 2 min r = mg an arm of 60cm gives a period of the orbit 2 v min = rg v = (0.6m)(9.8 m ) = 2.4 m s 2 s T = 2"r v =1.5s

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