Description of the motion using vectorial quantities

Transcription

1 Description of the motion using vectorial quantities RECTILINEAR MOTION ARBITRARY MOTION (3D) INERTIAL SYSTEM OF REFERENCE Circular motion Free fall

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3 Description of the motion using scalar quantities Let's search for a scalar quantity to describe the motion of a particle How to find such a relationship in a methodic and consistent way?

4 Observation Force is very effective in changing the speed of the ball: v f < v i Force is not effective in changing the speed of the ball: v f = v i

5 Work W,, vector the concept of and the, will be introduced. Later on, we will find a relationship between W and the change of speed of a particle.

6 Work done by a constant force Definition of the work done by a constant force on a particle that undergoes a displacement d W = = F d cosθ Definition of the work W done by a constant force on a particle that undergoes a displacement.

7 Example F=2 N initial final Example initial final Note: This force F is not the total force acting on the block.

8 Example initial final Note: This force F is not the total force acting on the block.

9 Work done on a particle by several forces Work - Kinetic Energy theorem

10 We are looking for a relationship between: the WORK done by constant forces and constant forces the change that such forces cause on the kinetic energy of a particle s To obtain such a relationship, let's assume the mass "m" is constrained to move along the x-axis only Example The particle moves along a tube. The forces acting on the particle by the inner sides of the tube are not shown; only an additional external force Fext is depicted in the figure. F ext Example F ext friction

11 Let s calculate the work done by ALL the constant forces F1, F2,... acting on the particle (only one generic force is drawn in the figure.) Y X No motion along the vertical direction implies a y = 0 Along the horizontal direction we have: a x = (horizontal-component of the TOTAL FORCE) /m = (F 1x + F 2x +... ) /m Constant forces imply constant acceleration a x. Hence, v f 2 = v i 2 +2 a x d Multiplying by m on each side, (constant acceleration motion) (1/2) m v f 2 = (1/2) m v i 2 + (1/2)m 2 a x d

12 (1/2) m v f 2 - (1/2) m v i 2 = m a x d We obtain m a x is the x-component of the total force acting on the mass m (1/2) m v f 2 - (1/2) m v i 2 = (F 1x + F 2x +... ) d Let s work out the term on the right with a bit more detail = (F 1x + F 2x +... ) d = = (F 1 Cos + F 2 Cos +... ) d = (F 1 dcos + F 2 dcos +...) which can be placed in vectorial form = ( F 1. d + F 2. d +... ) = ( F 1 + F ). d = = F TOTAL. d

13 FORCE (1/2) m v 2 f - (1/2) m v 2 i = F TOTAL CONSTANT work done by the TOTAL FORCE acting on the mass m. d work done by the TOTAL CONSTANT FORCE acting on the mass m We define (1/2) m v 2 =K KINETIC ENERGY So, the result above can be expressed as, K f - K i = Change in kinetic energy W Work done by all the forces Alternatively, K f - K i = W(i f) Change in kinetic energy Work done by all the forces (when taking the particle from i to f )

14 We have just derived the Work - Kinetic energy theorem

15 Here d is the displacement d= r f -r i and F is the total constant force acting on the particle This is the scalar relationship we were looking for to quantify how effective a given force F is to change the speed of a particle Although the result above has been obtained for the particular case in which the particle moves along one dimension, the results is also valid if the particle were moving in 3D (as we will justify later.) In summary We have now two ways to describe the motion (the dynamics) of a particle: 1 Vectorial relationship z Applicable when we are interested in a detailed description of the particle's motion. We want to know everything: r = r(t) x y

16 2 = = Scalar relationship Applicable when we are interested only in what is going on at the beginning and at the end of the motion. (What happens in between: "we don't care".) z x y Although the result above has been obtained for the particular case in which each of the applied forces are constant, it is also valid when the force varies with respect to the space variables, as we will justify in the following sections.

17 The concept of Conservative and Non-Conservative forces Example. Work done by the gravitational force the gravitational force friction force More specifically, we want to evaluate the work done by these forces when the particle undergoes a round trip.

18 Example. Work done by the gravitational force (near the surface of the Earth) a) A ball is thrown up with initial speed v i What is the maximum height reached by the ball? Assume m= 1 Kg. v f = 0 Final Gravitational force F = mg v i = 10 m/s Initial K = K f K i = 0 (1/2) m (v i ) 2 = (1/2) m (v i ) 2 negative number During the travel going up, the work done by the gravitational force on the mass m is given by W up : Negative W = F d cos(180 o UP = ) = - F d = - mgd < 0 work The magnitude of d in this case is the max height h MAX reached by the mass m. d = h MAX. Since the gravitational force is the only force acting on the particle, then W up is the total work done on the mass m. W up = W total Using K = W total one obtains, - (1/2) m (v i ) 2 = - m g d which gives, d = (v i ) 2 / 2g = 100/20 = 5 m. d = h MAX = 5 m

19 b) The ball reaches its highest height and then it starts its return trip W DOWN = = F d cos(0 o ) = F d = mgd > 0 Positive work Notice also, = W DOWN > 0 W UP = - mg d W DOWN = mg d Total work = = 0 UP DOWN

20 Notice: W UP = W DOWN = TOTAL WORK = W UP + W DOWN = 0 (closed path cycle) This results leads us to the concept of CONSERVATIVE FORCES: ThE NET WORK done by a conservative force on a particle during a closed path cycle is ZERO

21 Example. A block of mass m = 1 Kg is released with an initial speed v i = 2 m/s along an inclined ramp. The block reaches its maximum height and returns to its initial position Find: - The maximum height hmax - The total work done by the gravitational force during the whole trip (up and down) Solution ( Case of no friction) a) Block moves UP F = mg is decomposed in two components Work done by Work done by N is zero (why?) is zero (why?) F // Then, we have to evaluate only the work done by F // The force expressed as the sum of two vectors:

22 Work done by all the forces W UP = = (F // ) d cos(180 o ) = - (F // ) d = - ( ) W UP = d in meters W UP in Joules (1) = zero when d = d max = = = (2)

23 Using (2) in (1) one obtains, h max (3) Work done by the gravitational force b) Block moves DOWN Let's consider now the return trip: (use the same free-body diagram shown two pages above). initial final d Work done by the gravitational force (4) Total work done by the gravitational force = W up + W down = -2J + 2 J = 0

24 Exercise In the previous exercise, prove that when the block returns to the floor the magnitude of ist velocity is again 2 m/s Inital (Because the velocity is zero when the block is at the top position) Final But we know that Using

25 The previous exercises indicate that under the action of the gravitational force: a) The block comes back to its initial position with the same kinetic energy it had when the trip started. b) The total work done in a complete cycle (up first and down later) is zero. This type of results are typical of CONSERVATIVE FORCES. We have just found that the gravitational force is a conservative force

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28 Exercise. Solve the same problem addressed above but in the case that friction is taken into account. Find: - The maximum height reached by the block. - What is the kinetic energy when the block returns to its initial position? - What is the work done by the friction force (in the entire cycle up and down motion)?

29 A) We will work out the details in the next section. But before that, we can draw some conclusions about the effects of the friction force. We know there are three forces acting on the block. But, for now, let's focus ONLY in the frictional force When the block is on its way UP When the block is on its way DOWN

30 Work done by a friction force during a whole closed cycle path (round trip path) is, W friction = W UP + W DOWN < 0 (close path cycle) Work done by a friction force B) Now the details Friction force is NOT conservative

31 K = W gravity + W normal force + W friction!!! W gravity d Wgravity = -(mg Sin30 o ) d = - (1Kg)(10 m/s2)(0.5)d = -5d mg 30 o W normal-force W normal-force = 0 d N N = mg Cos30 o W friction N d f friction = k N = (0.1) mg Cos 30 o = 0.87 Newtons W friction = 0.87 d

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33 Example: The motion of a particle is described by a velocity vs time graph (as shown in the figure below. For each of the intervals indicated in the figure, indicate whether the change in the kinetic energy of the particle, caused by a variable total force acting on the particle, is positive, zero, or negative Velocity B C A W( A B) > 0 W( B C) = 0 D F E W( C D) < 0 W( D E) > 0 W( E F) < 0