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Free fall with air resistance Forces which resist motion play an important part in everyday life: –Brakes in cars; friction in many forms, damped vibrations Laws established by experiments: –For very low speed (<1 meter per second), ignore air resistance. –For low speeds ( up to about 25 meters per second ): the magnitude of the force of resistance in air is directly proportional to the speed. –For higher speed ( up to about 300 meters per second ): force of resistance in air is directly proportional to the square of the speed. –Near speed of sound ( 340 meters per second ): force increases dramatically and cannot be expressed as some power of speed, sound barrier –Supersonic flow: force is directly proportional to the speed.

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Free fall with air resistance Point-mass (or particle) motion: the motion of a body is considered as an entire unit without any regard for the dimensions of the body or rotation about its center of mass. The particle is situated at the center of mass of the body. No mass change in the motion. Position coordinate (or displacement): Suppose at any given instant time t the particle will occupy a certain position on the straight line. To define the position P of the particle, choose a fixed origin O on the straight line and a positive direction along the line. Measure the distance from O to P and record it with a plus or minus sign as y, where the plus or minus sign is assigned according to whether P is reached from O by moving along the line in a positive or negative direction. Velocity: –v>0: move forward; v<0: move backward

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Free fall with air resistance Speed: |v| Acceleration: –a>0: speed up; a<0: speed down (deceleration or retardation) Resultant force: the sum of all forces act on a particle Physical law: Isaac Newton’s second law of mechanics (1687) Model for free fall with air resistance:

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Free fall with air resistance –F 0 >0: constant force, e.g. gravitational force –P(v)>=0: force of air resistance, P(0)=0 & P(v) increases when v increases Terminal velocity: (V: P(V)=F 0 ) –Consider v(0)=0: since F 0 >0, the acceleration a(0)>0 will be positive initially, for t>0 and small, the velocity v(t) increases and p(v) also increases, the resultant force decreases until it becomes zero. Hence v will increase until a speed V is reached. As v tends to V, the acceleration tends to zero. –If v(0)= V, then v(t) = V for all time t !!! –If 0

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Free fall with air resistance An example: a man drops from an aeroplane at low or intermediate speed Assumptions: –The air resistance is directly proportional to the speed –The velocity of the man and the aeroplane is zero when the man leaves the aeroplane so that only a vertical motion resulted (no horizontal component in the velocity or displacement of the motion of the man) –Starting point is the origin O and downwards is the positive direction!! Variables: –t: time; v(t): velocity at time t; y(t): position at time t Model:

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Free fall with air resistance Parameters: –g: gravitational acceleration of the earth –k: constant for air resistance Terminal velocity: Solution of the velocity: Solution of the displacement:

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Free fall with air resistance An example: a man drops from an areoplane at an altitude of 9577 meters with a parachute. He reaches the ground with 116 seconds. Determine the distance the man had fallen during the free fall at any time. Questions: y(t)=???? Solution: Known data: when t=116, y(t)=9577. Plug into the solution: Solution:

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Free fall with air resistance Questions: –Set the ground is the origin O and upwards is the positive direction, find the ODE equation and find the solution of the problem. Compare the solution with the previous setup –Suppose the air resistance is proportional to the square of the velocity, find the ODE equation and find the solution of the problem.

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Rockets Motion problems –No change in mass or change is so small (mass is constant and Newton’s second law is valid) Motion of a bullet (no mass lost) Car is driven over a long distance (gasoline consumed is so small) –Mass changes substantially (Newton’s second law of mechanics is invalid) Motion of a rocket (mass lost by exhaust gases out of the engines) Some notations: –Displacement, Velocity, acceleration –Momentum (linear): product of the mass and the velocity (i.e. m v), to describe an impact.

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Rockets Interpretation: when a ball hits you, it is not only the mass of the ball which is important, but also the velocity at which it was thrown. Examples: –A basketball with 50 gram gram at 2 meters per second –A golfball with 10 gram at 25 meters per second Newton’s second law in momentum –If m is constant, it reduces

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Rockets Impulse: The impulse of a force F during a given time interval t in [a,b] is defined as the integral –The shock up your arm when you hit a fast ball in baseball is due to the impulse transmitted to your hand –The amount of damage when a car hits a wall depends on the impulse received by the car. –An impulse need not be short, e.g. when a lever is pulled. Impulse-momentum law: The impulse of the resultant force on a particle over a given interval is equal to the change in momentum of the particle in the same time interval

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Rockets Building a model: motion with changing mass –A variable mass m(t) moving at a velocity v(t) at time t –In a small time interval, changes in mass and velocity are – Assume the amount of mass being added may have a velocity relative to the mass m(t) and denote u(t). Its velocity is u(t)+v(t)!!! –Change in momentum in the time interval –Denote F(t) the total external force (the resultant force) on m(t), then the impulse is

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Rockets –Impulse-Momentum law –Divide by and let, we obtain –Special cases Case 1: u(t)=0, it is the standard one Case 2: u(t)= -v(t), the velocity of the particle which joins m(t) is zero

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Rockets The problem: a rocket of mass M kg is fired vertically upwards from the launching pad. The engines blow exhaust gases out at a constant rate of meters per second, while the fuel is consumed at a constant rate of kg per second. At the start the fuel in the rocket is n M kg where 0

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Rockets Constructing the model –Choose a y-axis vertically upwards with the origin at the launching pad –Denote t: time ( with t=0 at the moment of blast-off) y(t): altitude (in meters) (with y(0)=0 at the launching pad) v(t): velocity (in meters per second) (with v(0)=0 at the launching pad) m(t): the mass (kg) (with m(0)=M at the launching pad) –The total external force: –The equation for motion

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Rockets –The equation for mass change –Motion equations For the first part: For the second part: Solve the ODE:

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Rockets –For the first part: (linear first order ODE) Integration factor Solution Integrate again, we can get the displacement. –For the second part (free fall with air resistance) Solutions Terminal velocity

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