LECTURE 10- EXAMPLE PROBLEMS. Chapter 6-8 Professor Noronha-Hostler Professor Montalvo

Transcription

1 LECTURE 10- EXAMPLE PROBLEMS Chapter 6-8 Professor Noronha-Hostler Professor Montalvo

2 TEST!!!!!!!!! Thursday November 15, :40 11:00 PM Classes on Friday Nov. 16th NO CLASSES week of Thanksgiving (Nov )

3 TEST LOCATIONS

4 TEST SUBJECTS Chapters Uniform Circular Motion (Chapter 4.5) Topics: Uniform Circular Motion, Forces, Kinetic Energy, Potential Energy, Work, Power, Thermal Energy Read your chapters too!

5 TODAY Inclined plane with Forces, Work-Kinetic Energy Theory, Conservation of Energy Pendulum Potential energy vs. Kinetic energy Loop the loop Uniform Circular Motion, Conservation of energy Stopping a go kart with a spring Conservation of Energy, Hook s law

6 INCLINED PLANE Jennifer is pushing a heavy box up a rough inclined surface at a constant speed by applying a horizontal force as shown in the drawing. The coefficient of kinetic friction for the box on the inclined surface is µk. Which one of the following expressions correctly determines the normal force on the box? A.) B.) F N F = mg tan θ µ FN = F µ kmg tan θ k C.) F N = F cos θ µ k mg sin θ D.) FN = F cos θ µ kmg sin θ E.) F N = F sin θ µ mg k cos θ

7 INCLINED PLANE+FRICTION

8 SPEED AT THE BOTTOM OF THE HILL? Girl on a sled starts at rest down a hill of length L, what is her velocity at the bottom of the hill? L a θ

9 SPEED AT THE BOTTOM OF THE HILL? Girl on a sled starts at rest down a hill of length L, what is her velocity at the bottom of the hill? F N Draw the Free body diagram f k mg

10 SPEED AT THE BOTTOM OF THE HILL? Girl on a sled starts at rest down a hill of length L, what is her velocity at the bottom of the hill? y+ F N Forces along axes mg sin θ x+ mg θ f k mg cos θ

11 SPEED AT THE BOTTOM OF THE HILL? Girl on a sled starts at rest down a hill of length L, what is her velocity at the bottom of the hill? F N Newton s Laws mg sin θ mg θ f k mg cos θ F N = mg cos θ mg sin θ f k = ma

12 SPEED AT THE BOTTOM OF THE HILL? Find acceleration mg sin θ f k = ma Recall f k = μ k F N mg sin θ μ k F N = ma Recall F N = mg cos θ mg sin θ μ k mg cos θ = ma a = g sin θ μ k g cos θ

13 SPEED AT THE BOTTOM OF THE HILL? Find velocity a = g sin θ μ k g cos θ v 2 = v0 2 v 0 = 0 + 2ad d = L v 2 = 2aL v 2 = 2L(g sin θ μ k g cos θ) v 2 = 2Lg(sin θ μ k cos θ) v = 2Lg(sin θ μ k cos θ)

14 SPEED AT THE BOTTOM OF THE HILL? Work-Kinetic Energy Theorem L a θ

15 SPEED AT THE BOTTOM OF THE HILL? Work-Kinetic Energy Theorem W = ΔK = K f K i K f = 1 2 mv2 f K i = 0 W = Fd cos ϕ θ

16 SPEED AT THE BOTTOM OF THE HILL? Work-Kinetic Energy Theorem W = 1 2 mv2 f W = Fd cos ϕ Angle between F and d, NOT the same as angle of the inclined plane! Also, d = L θ W = F x L

17 SPEED AT THE BOTTOM OF THE HILL? Work-Kinetic Energy Theorem F x L = 1 2 mv2 f Recall, the force is in the x direction is F x = mg sin θ f k L(mg sin θ f k ) = 1 2 mv2 f L(mg sin θ μ k mg cos θ) = 1 2 mv2 f v f = 2Lg(sin θ μ k cos θ)

18 SPEED AT THE BOTTOM OF THE HILL? Conservation of Energy L a h θ

19 SPEED AT THE BOTTOM OF THE HILL? Conservation of Energy for Isolated System ΔE = ΔE mec + ΔE th = 0 ΔE mec = (U f + K f ) (U i + K i ) ΔE th = f k L = μ k LF N = μ k Lmg cos θ

20 SPEED AT THE BOTTOM OF THE HILL? Mechanical Energy ΔE mec = (U f + K f ) (U i + K i ) Initial Final U i = mgh U f = 0 ΔE mec = 1 2 mv2 f mgh K i = 0 K f = 1 2 mv2 f Need to find h

21 SPEED AT THE BOTTOM OF THE HILL? sin θ = h L h = L sin θ L h θ

22 SPEED AT THE BOTTOM OF THE HILL? Mechanical energy ΔE mec = 1 2 mv2 f mgh h = L sin θ ΔE mec = 1 2 mv2 f mgl sin θ

23 SPEED AT THE BOTTOM OF THE HILL? Conservation of Energy for Isolated System ΔE = ΔE mec + ΔE th = 0 ΔE mec = 1 2 mv2 f mgl sin θ ΔE th = μ k Lmg cos θ 1 2 mv2 f mgl sin θ + μ k Lmg cos θ = 0 v f = 2Lg(sin θ μ k cos θ)

24 A steel ball is whirled on the end of a chain in a horizontal circle of radius R with a constant period T. If the radius of the circle is then reduced to 0.75R, while the period remains T, what happens to the centripetal acceleration of the ball? A.) The centripetal acceleration increases to 1.33 times its initial value. b) The centripetal acceleration increases to 1.78 times its initial value. c) The centripetal acceleration decreases to 0.75 times its initial value. d) The centripetal acceleration decreases to 0.56 times its initial value. e) The centripetal acceleration does not change.

25 PENDULUM

26 SPEED AT LOWEST POINT? A Pendulum of Length L is released with an angle of θ. What is its speed at the lowest point? L

27 SPEED AT LOWEST POINT? θ L L h L L θ h cos θ = L h L θ h = L L cos θ

28 SPEED AT LOWEST POINT? Isolated System, no friction ΔE mec = (U f + K f ) (U i + K i ) = 0 K i = 0 K f = 1 2 mv2 f 1 2 mv2 f = mgh U i = mgh U f = mv2 f = mg(l L cos θ) v f = 2g(L L cos θ)

29 A ball is attached to a string and whirled in a horizontal circle. The ball is moving in uniform circular motion when the string separates from the ball (the knot wasn t very tight). Which one of the following statements best describes the subsequent motion of the ball? a) The ball immediately flies in the direction radially outward from the center of the circular path the ball had been following. b) The ball continues to follow the circular path for a short time, but then it gradually falls away. c) The ball gradually curves away from the circular path it had been following. d) The ball immediately follows a linear path away from, but not tangent to the circular path it had been following. e) The ball immediately follows a line that is tangent to the circular path the ball had been following

30 LOOP THE LOOP

31 MINIMUM HEIGHT TO SURVIVE THE LOOP THE LOOP? 1 U 1 = mgh K 1 = 0 2 U 2 = mg(2r) K 2 = 1 2 mv2 2 3 U 3 = 0 K 3 = 1 2 mv2 3 mgh = 2mgR mv2 2

32 MINIMUM HEIGHT TO SURVIVE THE LOOP THE LOOP? For the minimum h, that means that the normal force from the loop is approaching zero. F N 0 F N F g F N + F g = ma = m v 2 2 R F g = m v2 2 Since R F g = mg mg = m v2 2 R v 2 2 = gr

33 MINIMUM HEIGHT TO SURVIVE THE LOOP THE LOOP? Collecting the Equations mgh = 2mgR mv2 2 And v 2 2 = gr h = 2R + 1 2g v2 2 h = 2R R = 5 2 R

34 MINIMUM HEIGHT TO SURVIVE THE LOOP THE LOOP? h = 5 2 R R = cm h = 5 (11.75 cm) = 30 cm 2

35 A block is in contact with a rough surface as shown in the drawing. The block has a rope attached to one side. Someone pulls the rope with a force, which is represented by the vector in the drawing. The force is directed at an angle θ with respect to the horizontal direction. The magnitude of is equal to two times the magnitude of the frictional force, which is designated f. For what value of θ is the net work on the block equal to zero joules? a) 0 b) 30 c) 45 d) 60 e) Net work will be done in the object for all values of θ.

36 GO KART + SPRING

37 HOW MUCH DOES THE SPRING COMPRESS? v = 10m/s K i = 1 2 mv2 0 K f = 0 W = ΔK = 1 2 mv2 0 W = 1 2 kx2 m = 100kg k = 10,000N/m 1 2 kx2 = 1 2 mv2 0

38 HOW MUCH DOES THE SPRING COMPRESS? 1 2 kx2 = 1 2 mv2 0 x = mv 2 0 k m = 100kg k = 10,000N/m v = 10m/s x = 100kg(10m/s) 2 10,000N/m = 1m

39 Jamal s gravitational potential energy is 1870 J as he sits 2.20 m above the ground in a sky diving airplane. What is his gravitational potential energy when be begins to jump from the airplane at an altitude of 923 m? a) J b) J c) J d) J e) J

40 GOOD LUCK ON THE TEST!