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**Chapter 6 – Circular Motion and Other applications on Newton’s Laws**

Uniform circular motion. Drag forces and terminal speed. Other Applications of Newton’s Laws

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**Uniform Circular Motion**

A force, Fr , is directed toward the center of the circle This force is associated with an acceleration, ac Applying Newton’s Second Law along the radial direction gives

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**Uniform Circular Motion**

A force causing a centripetal acceleration acts toward the center of the circle It causes a change in the direction of the velocity vector If the force vanishes, the object would move in a straight-line path tangent to the circle

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Centripetal Force The force causing the centripetal acceleration is sometimes called the centripetal force This is not a new force, it is a new role for a force It is a force acting in the role of a force that causes a circular motion

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Conical Pendulum The object is in equilibrium in the vertical direction and undergoes uniform circular motion in the horizontal direction (1) Tcosθ = mg (2) Tsinθ = mac= mv2/r

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Conical Pendulum Dividing (2) by (1) and using sinθ/cosθ = tanθ we eliminate T and find: v is independent of m Using r = Lsinθ

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**Consider a conical pendulum with an 80. 0-kg bob on a 10**

Consider a conical pendulum with an 80.0-kg bob on a m wire making an angle of 5.00 with the vertical. Determine (a) the horizontal and vertical components of the force exerted by the wire on the pendulum and (b) the radial acceleration of the bob.

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A 4.00-kg object is attached to a vertical rod by two strings, as in Figure. The object rotates in a horizontal circle at constant speed 6.00 m/s. Find the tension in (a) the upper string and (b) the lower string.

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How fast can it spin? The speed at which the object moves depends on the mass of the object and the tension in the cord. The centripetal force is supplied by the tension: This shows that v increases with T and decreases with larger m.

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**Horizontal (Flat) Curve**

The force of static friction supplies the centripetal force The maximum speed at which the car can negotiate the curve is Note, this does not depend on the mass of the car Fr

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**Banked Curve These are designed with friction equaling zero**

There is a component of the normal force that supplies the centripetal force (1), and component of the normal force that supplies the gravitational force (2). Dividing (1) by (2) gives:

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Loop-the-Loop A pilot in a jet aircraft executes a loop-the loop. This is an example of a vertical circle At the bottom of the loop, (b), the upward constant force, that keeps the pilot moving in a circular path at a constant speed, is greater than its weight, because the normal and gravitational forces act in opposite direction:

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Loop-the-Loop At the top of the circle, (c), the force exerted on the object is less than its weight, because both the gravitational force and the normal force, ntop, exerted on the pilot by the seat act in same direction:

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**A 0. 400-kg object is swung in a vertical circular path on a string 0**

A kg object is swung in a vertical circular path on a string m long. If its speed is 4.00 m/s at the top of the circle, what is the tension in the string there?

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**Non-Uniform Circular Motion**

When the force acting on a particle moving in a circular path has a tangential component , the particles speed changes. The acceleration has a tangential components Fr produces the centripetal acceleration Ft produces the tangential acceleration. The total force is the vector sum or the radial force and tangential force:

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**Vertical Circle with Non-Uniform Speed**

The gravitational force exerts a tangential force on the object Look at the components of Fg The gravitational force resolve into a tangential component mg sinθ and a radial component mg cosθ.

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**Vertical Circle with Non-Uniform Speed**

Applying II NL to the tangential and radial directions: The tension at any point can be found:

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**Tarzan (m = 85. 0 kg) tries to cross a river by swinging from a vine**

Tarzan (m = 85.0 kg) tries to cross a river by swinging from a vine. The vine is 10.0 m long, and his speed at the bottom of the swing (as he just clears the water) will be 8.00 m/s. Tarzan doesn't know that the vine has a breaking strength of N. Does he make it safely across the river?

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**Motion in accelerated Frames**

When Newton’s laws of motion was introduced in Chapter 5, we emphasized that they are valid only for inertial frames of reference. In this section, we will analyze the noninertial frames, that is, on that is accelerating. Example: Let’s consider a hockey puck on a table in a moving train. The train moving with a constant velocity represents an inertial frame. The puck at rest remains at rest, and Newton’s I low is obeyed.

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**Motion in accelerated Frames**

The accelerating train is not an inertial frame. For the observer on the train, there appears to be no visible force on the puck, but it will accelerate from rest toward the back of the train, as the train start to accelerate. The Newton’s I law is violated. The observer on the accelerating train, if he applied the N II law to the puck, might conclude that a force has acted on the puck to cause it to accelerate. We call an apparent force such as this a fictitious force.

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**Motion in Accelerated Frames**

A fictitious force results from an accelerated frame of reference A fictitious force appears to act on an object in the same way as a real force, but you cannot identify a second object for the fictitious force

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“Centrifugal” Force From the frame of the passenger (b), a force appears to push her toward the door From the frame of the Earth, the car applies a leftward force on the passenger The outward force is often called a centrifugal force It is a fictitious force due to the acceleration associated with the car’s change in direction

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If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is μs = 0.800, how fast can you drive on a horizontal roadway around a right turn of radius 30.0 m before the cup starts to slide? If you go too fast, in what direction will the cup slide relative to the dashboard?

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**A person stands on a scale in an elevator**

A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 591 N. As the elevator later stops, the scale reading is 391 N. Assume the magnitude of the acceleration is the same during starting and stopping, and determine (a) the weight of the person, (b) the person's mass, and (c) the acceleration of the elevator.

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“Coriolis Force” This is an apparent force caused by changing the radial position of an object in a rotating coordinate system The result of the rotation is the curved path of the ball

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**The Earth rotates about its axis with a period of 24. 0 h**

The Earth rotates about its axis with a period of 24.0 h. Imagine that the rotational speed can be increased. If an object at the equator is to have zero apparent weight, (a) what must the new period be? (b) By what factor would the speed of the object be increased when the planet is rotating at the higher speed? Note that the apparent weight of the object becomes zero when the normal force exerted on it is zero.

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**Fictitious Forces, examples**

Although fictitious forces are not real forces, they can have real effects Examples: Objects in the car do slide You feel pushed to the outside of a rotating platform The Coriolis force is responsible for the rotation of weather systems and ocean currents

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**Fictitious Forces in Linear Systems**

The inertial observer (a) sees The noninertial observer (b) sees

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**Fictitious Forces in a Rotating System**

According to the inertial observer (a), the tension is the centripetal force The noninertial observer (b) sees

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**Motion with Resistive Forces**

Motion can be through a medium Either a liquid or a gas The medium exerts a resistive force, R, on an object moving through the medium The magnitude of R depends on the medium The direction of R is opposite the direction of motion of the object relative to the medium R nearly always increases with increasing speed

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**Motion with Resistive Forces**

The magnitude of R can depend on the speed in complex ways We will discuss only two R is proportional to v Good approximation for slow motions or small objects R is proportional to v2 Good approximation for large objects

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**R Proportional To v The resistive force can be expressed as R = - b v**

b depends on the property of the medium, and on the shape and dimensions of the object The negative sign indicates R is in the opposite direction to v

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**R Proportional to v, Example**

Analyzing the motion results in

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**R Proportional to v Initially, v = 0 and dv/dt = g**

As t increases, R increases and a decreases The acceleration approaches 0 when R → mg At this point, v approaches the terminal speed of the object

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**Terminal Speed To find the terminal speed, let a = 0**

Solving the differential equation gives t is the time constant and t = m/b

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R Proportional to v2 For objects moving at high speeds through air, the resistive force is approximately equal to the square of the speed R = ½ DrAv2 D is a dimensionless empirical quantity that called the drag coefficient - r is the density of air A is the cross-sectional area of the object v is the speed of the object

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R Proportional to v2 Analysis of an object falling through air accounting for air resistance

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**R Proportional to v2, Terminal Speed**

The terminal speed will occur when the acceleration goes to zero Solving the equation gives

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Some Terminal Speeds

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**Process for Problem-Solving**

Analytical Method The process used so far involves the identification of well-behaved functional expressions generated from algebraic manipulation or techniques of calculus

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**Analytical Method Apply the method using this procedure:**

Sum all the forces acting on the particle to find the net force, SF. Use this net force to determine the acceleration from the relationship a =SF/m Use this acceleration to determine the velocity from the relationship dv/dt = a Use this velocity to determine the position from the relationship dx/dt = v

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**Analytic Method, Example**

Applying the procedure: Fg = may = - mg ay = -g and dvy/dt = -g vy(t) = vyi – gt y(t) = yi + vyi t – ½ gt2

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Numerical Modeling In many cases, the analytic method is not sufficient for solving “real” problems Numerical modeling can be used in place of the analytic method for these more complicated situations The Euler method is one of the simplest numerical modeling techniques

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Euler Method In the Euler Method, derivatives are approximated as ratios of finite differences Dt is assumed to be very small, such that the change in acceleration during the time interval is also very small

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**Equations for the Euler Method**

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Euler Method It is convenient to set up the numerical solution to this kind of problem by numbering the steps and entering the calculations into a table Many small increments can be taken, and accurate results can be obtained by a computer

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Euler Method Set Up

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Euler Method Final One advantage of the method is that the dynamics are not obscured The relationships among acceleration, force, velocity and position are clearly shown The time interval must be small The method is completely reliable for infinitesimally small time increments For practical reasons a finite increment must be chosen A time increment can be chosen based on the initial conditions and used throughout the problem In certain cases, the time increment may need to be changed within the problem

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**Accuracy of the Euler Method**

The size of the time increment influences the accuracy of the results It is difficult to determine the accuracy of the result without knowing the analytical solution One method of determining the accuracy of the numerical solution is to repeat the solution with a smaller time increment and compare the results If the results agree, the results are correct to the precision of the number of significant figures of agreement

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**Euler Method, Numerical Example**

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**Euler Method, Numerical Example**

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**II. Drag force and terminal speed**

Fluid: anything that can flow. Example: gas, liquid. Drag force: D - Appears when there is a relative velocity between a fluid and a body. Opposes the relative motion of a body in a fluid. - Points in the direction in which the fluid flows.

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*** Body is blunt (baseball). * Fast relative motion turbulent air.**

Assumptions: * Fluid = air. * Body is blunt (baseball). * Fast relative motion turbulent air. C = drag coefficient (0.4-1). ρ = air density (mass/volume). A = effective body’s cross sectional area area perpendicular to v -Terminal speed: vt - Reached when the acceleration of an object that experiences a vertical movement through the air becomes zero Fg=D

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**III. Uniform circular motion**

Centripetal acceleration: v, a = const, but direction changes during motion. A centripetal force accelerates a body by changing the direction of the body’s velocity without changing its speed. Centripetal force: a, F are directed toward the center of curvature of the particle’s path.

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49. A puck of mass m slides on a frictionless table while attached to a hanging cylinder of mass M by a cord through a hole in the table. What speed keeps the cylinder at rest? Mg mg T N

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**33E. Calculate the drag force on a missile 53cm in diameter**

cruising with a speed of 250m/s at low altitude, where the density of air is 1.2kg/m3. Assume C=0.75

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32. The terminal speed of a ski diver is 160 km/h in the spread eagle position and 310 km/h in the nose-dive position. Assuming that the diver’s drag coefficient C does not change from one point to another, find the ratio of the effective cross sectional area A in the slower position to that of the faster position.

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**assume that the cord between B and **

Block B weighs 711N. The coefficient of static friction between the block and the table is 0.25; assume that the cord between B and the knot is horizontal. Find the maximum weight of block A for which the system will be stationary. N FgB T1 f T1 T2 T3 T3 FgA

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**Block A Block B Light block A leads**

Two blocks of weights 3.6N and 7.2N, are connected by a massless string and slide down a 30º inclined plane. The coefficient of kinetic friction between the lighter block and the plane is 0.10; that between the heavier block and the plane is Assuming that the lighter block leads, find (a) the magnitude of the acceleration of the blocks and (b) the tension in the string. (c) Describe the motion if, instead, the heavier block leads. Block A Block B NA NB A B FgA FgB NA NB T fk,B fk,A Movement FgxA T FgxB fkB fkA T FgyA FgyB Light block A leads

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**determine whether the block moves, the direction of motion, **

74. A block weighing 22N is held against a vertical wall by a horizontal force F of magnitude 60N. The coefficient of static friction between the wall and the block is 0.55 and the coefficient of kinetic friction between them is A second force P acting parallel to the wall is applied to the block. For the following magnitudes and directions of P, determine whether the block moves, the direction of motion, and the magnitude and direction of the frictional force acting on the block: (a) 34N up, (b) 12N up, (c) 48N up, (d) 62N up, (e) 10N down, (f) 18N down. F=60N N P mg=22N N=F=60N 22N f

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**28. Blocks A and B have weights of 44N and 22N, respectively**

28. Blocks A and B have weights of 44N and 22N, respectively (a) Determine the minimum weight of block C to keep A from sliding if μs between A and the table is 0.2. (b) Block C suddenly is lifted of A. What is the acceleration of block A if μk between A and the table is 0.15? N f T Wc WA=44N T WB=22N

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**Fmin required to keep m from sliding down?**

29. The two blocks (with m=16kg and m=88kg) shown in the figure below are not attached. The coefficient of static friction between the blocks is: μs=0.38 but the surface beneath the larger block is frictionless. What is the minimum value of the horizontal force F required to keep the smaller block from slipping down the larger block? N f F’ F’ mg Fmin required to keep m from sliding down? Mg Treat both blocks as a single system sliding across a frictionless floor Movement

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**The force of the boom on the car is capable of pointing any direction**

44. An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weigh of the car and riders is 5kN, and the radius of the circle is 10m. What are the magnitude and the direction of the force of the boom on the car at the top of the circle if the car’s speed is (a) 5m/s (b) 12m/s? The force of the boom on the car is capable of pointing any direction y FB W

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