Homework 6. problems: 8.-, 8.38, 8.63

Transcription

1 Homework 6 problems: 8.-, 8.38, 8.63

2 Problem A circus trapeze consists of a bar suspended by two parallel ropes, each of length l. allowing performers to swing in a vertical circular arc. Suppose a performer with mass m holds the bar and steps off an elevated platform, starting from rest with the ropes at an angle θ i with respect to the vertical. Assume the size of the performer s body is small compared to the length l, she does not pump the trapeze to swing higher, and air resistance is negligible. (a) Show that when the ropes make an angle θ with the vertical, the performer must exert a force mg ( 3 cosθ - cosθi ) so as to hang on. (b) Determine the angle θ i for which the force needed to hang on at the bottom of the swing is twice as large as the gravitational force exerted on the performer. T θ l θ W v h a) According to Newton s third law of motion, the performer exert a force on the rope opposite to the force that the rope exerts on the performer. Motion of the performer is affected by two forces the gravitational force exerted by the earth and the tension force exerted by the ropes. The tension force adjusts itself to such a value that the radial component of the net force results in the centripetal acceleration of the performer. The radial component of the tension force is equal to tension while radial component of the gravitational force is equal to the projection of the weigh on the radial direction

3 () mac Fc T mgcosθ In circular motion, the centripetal acceleration depends on the speed of the performer and the radius of the path v () a c l It can be determined from the work-energy theorem. With the reference for potential energy at the lowest point, the her potential energy at an arbitrary point of the path is (3) U g mgh mg( l - l cosθ) Since the work due to tension is zero (at any point the tension force is perpendicular to the differential displacement), mechanical energy on the performed is conserved (4) mg( - l cosθ ) + 0 mg( l - l cosθ) i + l. (Note. One can also relate the change in kinetic energy to the net work (4 ) 0 ΔW net ΔW g mg( l - l cosθ ) mg( l - l cosθ) ) The rest is algebra. Solving (4) for square of speed v gl (cosθ cosθi) Substituting square of speed in equation (), the centripetal acceleration at the considered location of the performer is a c g(cosθ cosθ Finally solving equation () for the unknown tension T ma c i ) + mgcosθ 3mgcosθ mgcosθ b) At the lowest point θ 0. For tension to be twice as large as the weight of the performer, the initial angle satisfies the following equation mg 3mg mgcosθi from which θi aeccos 60 i i

4 Problem 8.38 A 659-kg elevator starts from rest. It moves upward for 3.0 s with constant acceleration until it reaches its cruising speed of.75 m/s. (a) What is the average power of the elevator motor during this time? (b) How does this power compare with the motor power when the elevator moves at its cruising speed? a) Solution. Using the definition of the average value of a function, the average power delivered by the motor over a certain time interval is related to the work done in this time interval t t t dw P dt M M dt dwm t dt t t ΔW P M av.m t t t t t t Δt There are two forces exerted on the elevator: the earth exerts the gravitational force and the cords (motor) exert the tensile force. According to the work-energy theorem, the resultant net work is equal to the change in the kinetic energy of the elevator. The gravitational work can be found from the displacement and weight of the elevator a( Δt) Δv Δt Δ Wg mgδh mg mg Hence Δv Δt 0 Pav,M Δt mg and P m.75 v v s m Δ g 659kg + t Δ 3s.75 m s av,m m s 5987W T W

5 Solution From the differential form of the work-energy theorem, we can determine the instantaneous value of the power delivered by the motor at an instant when the elevator has speed v and accelerates at rate a dwm + dwg dk from which dw dw dk d d P M g M + ( mgh) a + mgv dt dt dt dt dt From the definition of the average value of a function t t t dv ( a + mgv) dt m v dt + mg vdt t t dt t Pav,M t t t t Δt mgδh + Δt Δt ( Δt) mga + Δt Δt mgv + Δt The motor delivers power to balance the negative gravitational work and to increase the kinetic energy of the elevator. b) When the elevator reaches its cruising speed the motor is working at a constant power represented by an earlier function m m PM mgv 659kg W s s Despite at constant speed, of the elevator does not increase the kinetic energy of the elevator, at the higher speed the elevator requires higher power to increase its gravitational potential energy at higher rate.

6 Problem 8.63 A 0-kg block is released from point A as shown in Figure The track is frictionless except for the portion between B and C, which is 6 m long. The block travels down the track, hits a spring of force constant k,50 N/m and compresses the spring 0.3 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between B and C. and the block. A h 3m B l 6m C x Let's assume that the block is a particle. During the motion, the block interacts with the earth (weight), the surface (friction), and the spring (tensile-compressive stress). Solution. We could use the first version (for the kinetic energy) of the work-energy theorem. The change in the kinetic energy of the block is zero. According to the work-energy theorem it must be equal to the work done by all forces exerted on the block. From the definition of potential energy, we can find the gravitational work and the work done by the spring. ) W ΔU ( 0 mgh) mgh g g ) Ws ΔUs 0 Directly from the definitions of work, the work due to the frictional force can be expressed in terms of the length of the rough part of the track. r r 3) W f d fl μnl μmgl f track

7 From the first version of the work-energy theorem we obtain 4) mgh μmgl 0 The rest is math. Solving for the coefficient of kinetic friction we find N mgh 50 ( 0.3m) 5) m μ mgl h l mg 6m 3m m 0kg 9.8 s 0.33 Solution. Two of the exerted forces, the gravitational and the normal (elastic), are conservative. The frictional force is not conservative. I will include the conservative interactions into the mechanical energy and use the second version (concerning the mechanical energy of the particle) of the work-energy theorem in the solution. Which position we choose for the reference of potential energy is arbitrary. I will choose the level of the horizontal part of the track as the reference potential energy for the gravitational interaction and a relaxed spring for the elastic interaction. We should have no problem determining the change in the mechanical energy between the two positions when the block's speed is zero. At point A, the kinetic energy and the potential energy due to the spring are zero. Therefore the mechanical energy is equal to the gravitational potential energy at this position 6) E K + U + U mgh A A g,a s, A The kinetic and gravitational potential energy at point D, where the block rests momentarily, are zero. The mechanical energy is equal to the elastic potential energy of the spring. 7) ED KD + Ug,D + Us,D where x is the compression of the spring from the relaxed position. There is only one nonconservative force exerted on the block during its motion, the frictional force. From the definition of work and

8 the nature of frictional force, we can relate the frictional work with the length of the rough part of the track. r r 8) W f d fl μnl μmgl f track I used the fact that the vertical component of acceleration is zero, therefore the magnitude of the normal force is equal to the magnitude of the weight. Now I can relate the expression for the work done by the frictional force with the change in the mechanical energy 9) W W ΔE f nc The rest is math. Combining all three equations, I can find the unknown friction coefficient: mgh Wf ΔE 0) μ h mg mg mg l l l l mg