PHYS 1441 – Section 002 Lecture #16AnnouncementsSlide 3Reminder: Special ProjectReminder: Special Project IILinear MomentumImpulse and Linear MomentumImpulseBall Hit by a BatEx. A Well-Hit BallExample 7.6 for ImpulseExample 7.6 cont’dLinear Momentum and ForcesConservation of Linear Momentum in a Two Particle SystemLinear Momentum ConservationMore on Conservation of Linear Momentum in a Two Body SystemWednesday, Nov. 10, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu1PHYS 1441 – Section 002Lecture #16Wednesday, Nov. 10, 2010Dr. Jaehoon Yu•Linear Momentum•Linear Momentum and Impulse•Linear Momentum and Forces•Linear Momentum Conservation•Linear Momentum Conservation in a Two -body SystemAnnouncements•Quiz #5–Beginning of the class, Wednesday, Nov. 17–Covers: Ch. 6.5 – what we finish next Monday, Nov. 15•Term exam results–Class Average: 46/102•Equivalent to 45/100•Exam 1 average: 49/100•Very consistent!–Top score: 94/102•Thanksgiving Wednesday–We will not have the class on Wednesday, Nov. 24•Colloquium today–Dr. Picraux on nano-wiresWednesday, Nov. 10, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu2Reminder: Special Project1. A ball of mass M at rest is dropped from the height h above the ground onto a spring on the ground, whose spring constant is k. Neglecting air resistance and assuming that the spring is in its equilibrium, express, in terms of the quantities given in this problem and the gravitational acceleration g, the distance x of which the spring is pressed down when the ball completely loses its energy. (10 points)2. Find the x above if the ball’s initial speed is vi. (10 points)3. Due for the project is Wednesday, Nov. 17.4. You must show the detail of your OWN work in order to obtain any credit.Wednesday, Nov. 10, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu4Wednesday, Nov. 10, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu5Reminder: Special Project IIA ball of mass m is attached to a light cord of length L, making up a pendulum. The ball is released from rest when the cord makes an angle θA with the vertical, and the pivoting point P is frictionless. A) Find the speed of the ball when it is at the lowest point, B, in terms of the quantities given above.B) Determine the tension T at point B in terms of the quantities given above. Each of these problem is 10 point. The due date is Wednesday, Nov. 17.mgmmθALTh{BPWednesday, Nov. 10, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu6Linear MomentumThe principle of energy conservation can be used to solve problems that are harder to solve just using Newton’s laws. It is used to describe motion of an object or a system of objects.A new concept of linear momentum can also be used to solve physical problems, especially the problems involving collisions of objects. pur≡Linear momentum of an object whose mass is m and is moving at the velocity of v is defined as What can you tell from this definition about momentum?What else can use see from the definition? Do you see force?The change of momentum in a given time intervalptD=Dr0mv mvt-=Dr r( )0m v vt-=Dr rvmtD=DrF�rma =r1. Momentum is a vector quantity.2. The heavier the object the higher the momentum3. The higher the velocity the higher the momentum4. Its unit is kg.m/s mvrJur≡FurΔtWednesday, Nov. 10, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu7Impulse and Linear Momentum The quantity impulse is defined as the change of momentumEffect of the force F acting on an object over the time interval Δt=tf-ti is equal to the change of the momentum of the object caused by that force. Impulse is the degree of which an external force changes an object’s momentum.The above statement is called the impulse-momentum theorem and is equivalent to Newton’s second law. pFtD=DrrNet force causes change of momentum Newton’s second lawSo what do you think an impulse is?What are the dimension and unit of Impulse? What is the direction of an impulse vector? Defining a time-averaged force 1iiF F tt� DD�r rImpulse can be rewritten Jur≡FurΔtIf force is constant Impulse is a vector quantity!!p F tD = Drrf ip p- =r rpD =r rJ =0fmv mv-r rWednesday, Nov. 10, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu8There are many situations when the force on an object is not constant. ImpulseWednesday, Nov. 10, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu9=ar=�Fr=�Fr rF∑( )Δt =J=rt-Df ov vr rmarm mt-Df ov vr rm m-f ov vr rBall Hit by a BatMultiply either side by ΔtWednesday, Nov. 10, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu10Ex. A Well-Hit BallA baseball (m=0.14kg) has an initial velocity of v0=-38m/s as it approaches a bat. We have chosen the direction of approach as the negative direction. The bat applies an average force F that is much larger than the weight of the ball, and the ball departs from the bat with a final velocity of vf=+58m/s. (a) determine the impulse applied to the ball by the bat. (b) Assuming that the time of contact is Δt=1.6x10-3s, find the average force exerted on the ball by the bat.What are the forces involved in this motion? The force by the bat and the force by the gravity.( )0.14 58 0.14 38 13.4kg m s= � - �- =+ �(a) Using the impulse-momentum theorem Jur=Since the force by the bat is much greater than the weight, we ignore the ball’s weight.F tDrpD =r0fmv mv-r r(b)Since the impulse is known and the time during which the contact occurs are know, we can compute the average force exerted on the ball during the contact J =rF =rW =rF =rJt=Dr313.484001.6 10N-+=+�How large is this force?mg =0.14 9.8 1.37N� =84001.37W =r6131 WrWednesday, Nov. 10, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu11Example 7.6 for Impulse(a) Calculate the impulse experienced when a 70 kg person lands on firm ground after jumping from a height of 3.0 m. Then estimate the average force exerted on the person’s feet by the ground, if the landing is (b) stiff-legged and (c) with bent legs. In the former case, assume the body moves 1.0cm during the impact, and in the second case, when the legs are bent, about 50 cm.We don’t know the force. How do we do this?Obtain velocity of the person before striking the ground.KE =212mv =( )img y y- - =imgyv =Solving the above for velocity v, we obtain2igy =2 9.8 3 7.7 /m s� �=Then as the person strikes the ground, the momentum becomes 0 quickly giving the impulse Jur=FurΔt =70 7.7 / 540kg m s j jN s=- � =- �r r Δpur= purf−puri=0 mv- =rPE- DWednesday, Nov. 10, 2010PHYS
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