Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)
General Physics I Part 2 (lecture 12-19) MECHANICS 2009/2010 To insert this slide into your presentation Save this template as a presentation (.ppt file) on your computer. Open the presentation that will contain the image slide. On the Slides tab, place your insertion point after the slide that will precede the image slide. (Make sure you don't select a slide. Your insertion point should be between the slides.) On the Insert menu, click Slides from Files. In the Slide Finder dialog box, click the Find Presentation tab. Click Browse, locate and select the presentation that contains the image slide, and then click Open. In the Slides from Files dialog box, select the image slide. Select the Keep source formatting check box. If you do not select this check box, the copied slide will inherit the design of the slide that precedes it in the presentation. Click Insert. Click Close. Instructor Tamer A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Energy and Energy Transfer
Lecture 12 Energy and Energy Transfer Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics 1, Lec 12, By/ T.A. Eleyan 2 2

Work Done by a Constant Force
When an object undergoes displacement under force then work is said to be done by the force and the amount of work (W) is defined as the product of the component of force along the direction of displacement times the magnitude of the displacement. where is the angle between the direction of the force and the direction of the motion. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics 1, Lec 12, By/ T.A. Eleyan 3

Work has only magnitude, no direction.
The SI unit of work is (N×m) which is given the name (Joule). 1 Newton×meter = 1 Joule Work is a scalar Work has only magnitude, no direction. If I push on a wall and the wall does not move (no displacement), the work is (0J) because the displacement is (0m). Note that if F is in the same direction as d, then  = 0, and W= Fd Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics 1, Lec 12, By/ T.A. Eleyan 4

Work is done in lifting the box (why?)
No work is done on the bucket to move horizontally (why?) General Physics 1, Lec 12, By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 5

Negative Work and Total Work
Work can be positive, negative or zero depending on the angle between the force and the displacement. If there is more than one force, each force can do work. The total work is calculated from the total (or net) force: Wtotal = Ftotald cosq Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics 1, Lec 12, By/ T.A. Eleyan 6

Example : Suppose I pull a package with a force of 98 N at an angle of 55° above the horizontal ground for a distance of 60m. What is the total work done by me on the package? Solution: Note that (F cosq) is the component of the force along the direction of motion. (Along the direction of the package's displacement.) Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics 1, Lec 12, By/ T.A. Eleyan 7

Problem: A force F = (6ˆi - 2ˆj) N acts on a particle that undergoes a displacement r = (3ˆi +ˆj) m. Find the work done by the force on the particle Example : A 0.23 kg block slides down an incline of 25° at a constant velocity. The block slides 1.5 m. What is the work done by the normal force, by gravity, and by friction? What is the total work done on the block? General Physics 1, Lec 12, By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 8

The friction Force The Normal Force
General Physics 1, Lec 12, By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 9

Work done by the Normal Force:
Now what is the work done by each individual force. Work done by the Normal Force: Work done by the Frictional Force: Work done by the Gravitational Force Now, to finally determine the "net-Work" Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics 1, Lec 12, By/ T.A. Eleyan 10

Force - Displacement Graph
The work done by a force can be found from the area between the force curve and the x-axis (remember, area below the x-axis is negative): Constant force Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics 1, Lec 12, By/ T.A. Eleyan 11

Work done by a varying force

Example: A force acting on a particle varies with x, as shown in Figure .Calculate the work done by the force as the particle moves from x = 0 to x = 6.0 m. Solution The work done by the force is equal to the area under the curve from x= 0 to x= 6.0 m. This area is equal to the area of the rectangular section from 0 to 4 plus the area of the triangular section from 4 to 6. The area of the rectangle is (5.0 N)(4.0 m) = 20 J, and the area of the triangle is ½(2m)(5N)=5J. Therefore, the total work =25J General Physics 1, Lec 12, By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 13

Problem: An object is acted on by the force shown in the Figure
Problem: An object is acted on by the force shown in the Figure. What is the work from 0 to 1.00m General Physics 1, Lec 12, By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 14

Work Done by a Spring Where k is a positive constant called (the spring constant) General Physics 1, Lec 12, By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 15

Kinetic Energy An object in motion has kinetic energy: m = mass
v = speed (magnitude of velocity) The unit of kinetic energy is Joules (J). Kinetic energy is a scalar (magnitude only) Kinetic energy is non-negative (zero or positive) General Physics 1, Lec 12, By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 16

Work-Energy Theorem Wtotal = DKE = KEfinal - KEinitial
The net (total) work done on an object by the total force acting on it is equal to the change in the kinetic energy of the object: Wtotal = DKE = KEfinal - KEinitial General Physics 1, Lec 12, By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 17

Example : How much work does it take to stop a 1000 kg car traveling at 28 m/s?
Solution: Problem: A kg particle has a speed of 2.00 m/s at point (A) and kinetic energy of 7.50 J at point (B). What is (a) its kinetic energy at (A)? (b) its speed at (B)? (c) the total work done on the particle as it moves from A to B? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics 1, Lec 12, By/ T.A. Eleyan 18

Example : A 58-kg skier is coasting down a slope inclined at 25° above the horizontal. A kinetic frictional force of 70 N opposes her motion. Near the top of the slope the skier's speed is 3.6 m/s. What is her speed at a point which is 57 m downhill? Solution: the net-work = Wf+Wg : Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics 1, Lec 12, By/ T.A. Eleyan 19

Now use the Work-Energy Theorem:

How far does the bicycle travel if it takes 4.0 s to come to rest?
Example: A 65-kg bicyclist rides his 10.0-kg bicycle with a speed of 12 m/s. How much work must be done by the brakes to bring the bike and rider to a stop? How far does the bicycle travel if it takes 4.0 s to come to rest? What is the magnitude of the braking force? Solution: Friction = only horizontal force Wnet = Kf -Ki <0 Wnet = 0 – (1/2) m vi2 Wnet = - (1/2) (10+65) (12)2 Wnet = kg m 2 /s 2 = J Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics 1, Lec 12, By/ T.A. Eleyan 21

Find acceleration first v = v0+at 0 = v0 + at, a= -v0/t
a = -12/4 = -3 m/s2 x= x0+v0 t+(1/2)at2 x= 0 + (12)(4) + (1/2)(-3)(4)2 x= = 24 m Braking force Wnet = Fnet d = FFriction d FFriction = Wnet /d = (-5400)/(24) = -225 N General Physics 1, Lec 12, By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 22

Power Power is defined as the rate at which work is done.
The SI unit of power is "watts" (W). Power can also be written as; Whenever you want to determine power, you must first determine the force and the velocity or the work being done and the time. General Physics 1, Lec 12, By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 23

Additional Questions (Lecture 12) Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[1] The force acting on a particle varies as in Figure . Find the work done by the force on the particle as it moves (a) from x = 0 to x = 8.00 m, (b) from x = 8.00 m to x = 12.0 m, and (c) from x = 0 to x = 12.0 m. [2] A 4.00-kg particle is subject to a total force that varies with position as shown in Figure. The particle start from rest at x = 0. What is its speed at (a) x = 5.00 m, (b) x = 10.0 m, (c) x = 15.0 m? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[3] A kg ball has a speed of 15.0 m/s. (a) What is its kinetic energy? (b) What If? If its speed were doubled, what would be its kinetic energy? [4] A 40.0-kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130 N. If the coefficient of friction between box and floor is 0.300, find (a) the work done by the applied force, (b) the increase in internal energy in the box-floor system due to friction, (c) the work done by the normal force, (d) the work done by the gravitational force, (e) the change in kinetic energy of the box, and (f) the final speed of the box. [5] The electric motor of a model train accelerates the train from rest to m/s in 21.0 ms. The total mass of the train is 875 g. Find the average power delivered to the train during the acceleration. [6] A 700-N Marine in basic training climbs a 10.0-m vertical rope at a constant speed in 8.00 s. What is his power output? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[7] A 1500-kg car accelerates uniformly from rest to a speed of 10 m/s in 3s. Find (a) the work done on the car in this time, (b) the average power delivered the engine in the first 3s, and (c) the instantaneous power delivered by the engine at t = 2s. [8] A mechanic pushes a 2500kg car from rest to a speed v doing 5000J of work in the process. During this time, the car moves 25m. Neglecting friction between the car and the road, (a) What is the final speed, v, of the car? (b) What is the horizontal force exerted on the car? [9] A 200kg cart is pulled along a level surface by an engine. The coefficient of friction between the carte and surface is 0.4. (a) How much power must the engine deliver to move the carte at constant speed of 5m/s? (b) How much work is done by the engine in 3min? [10] A 65kg woman climbs a flight of 20 stairs, each 23 cm high. How much work was done against the force of gravity in the process? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[11] A skier of mass 70.0 kg is pulled up a slope by a motor driven cable. (a) How much work is required to pull him a distance of 60.0 m up a 30.0° slope (assumed frictionless) at a constant speed of 2.00 m/s? (b) A motor of what power is required to perform this task? [12] If an applied force varies with position according to F.= 3x3 - 5, where x is in m, how much work is done by this force on an object that moves from x = 4 m to x = 7 m? [13] A horizontal force of 150 N is used to push a 40-kg box on a rough horizontal surface through a distance of 6m. If the box moves at constant speed, find (a) the work done by the 150-N force, (b) the work done by friction. [14] When a 4-kg mass is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.5cm. If the 4-kg mass is removed, (a) how far will the spring stretch if a 1.5-kg mass is hung on it, and (b) how much work must an external agent do to stretch the same spring 4.0 cm from its unstretched position? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[15] A 650-kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruising speed of 1.75 m/s. (a) What is the average power of the elevator motor during this period? (b) How does this power compare with the motor power when the elevator moves at its cruising speed? [16] A 4.00-kg particle moves along the x axis. Its position varies with time according to x = t + 2.0t 3, where x is in meters and t is in seconds. Find (a) the kinetic energy at any time t, (b) the acceleration of the particle and the force acting on it at time t, (c) the power being delivered to the particle at time t, and (d) the work done on the particle in the interval t = 0 to t=2.00 s. [17] If a man lifts a 20-kg bucket from a well and does 6 kJ of work, how deep is the well? Assume the speed of the bucket remains constant as it is lifted. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Potential Energy Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

There are two types of forces: conservative (gravity, spring force) All microscopic forces are conservative: Gravity, Electro-Magnetism, Weak Nuclear Force, Strong Nuclear Force nonconservative (friction, tension) Macroscopic forces are non-conservative, Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Conservative Forces A force is conservative if the work it does on an object moving between two points is independent of the path taken.  work done depends only on ri and rf Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

 If an object moves in a closed path (ri = rf) then total work done by the force is zero. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Nonconservative Forces
 work done by the force depends on the path  non-conservative forces dissipate energy Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Work Done by Conservative Forces
Potential Energy: Energy associated with the position of an object. For example: When you lift a ball a distance y, gravity does negative work on the ball. This work can be recovered as kinetic energy if we let the ball fall. The energy that was “stored” in the ball is potential energy. Wc = -DU =-[Ufinal – Uinitial] Wc = work done by a conservative force DU = change in potential energy Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Gravitational Potential Energy
Gravitational potential energy U = mg(y-y0), where, y = height U=0 at y=y0 (e.g. surface of earth). Work done by gravity: Wg = -mg Dy = -mg (y- y0) Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Spring Potential Energy
Uf – Ui = - [Work done by spring on mass] Mass m starts at x=0 (Ui =0) and moves until spring is stretched to position x. WorkSpring = - ½ kx2 U(x) – 0 = - (-1/2 kx2) USpring(x) = ½ kx2 x = displacement from equilibrium position F=-kx x Area in triangle = - kx times increment in x = Work done by spring Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Conservation of Energy
Energy is neither created nor destroyed The energy of an isolated system of objects remains constant. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Mechanical Energy (Conservative Forces) Mechanical energy E is the sum of the potential and kinetic energies of an object. E = U + K The total mechanical energy in any isolated system of objects remains constant if the objects interact only through conservative forces: E = constant Ef = Ei  Uf + Kf = Ui+ Ki DU + DK = DE = 0 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Example: A ball of mass m is dropped from a height h above the ground, as shown in Figure (A) Neglecting air resistance, determine the speed of the ball when it is at a height y above the ground. (B) Neglecting air resistance, determine the speed of the ball when it is at the ground. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Example: A 0.5 kg block is used to compresses a spring with a spring constant of 80.0 N/m a distance of 2.0 cm. After the spring is released, what is the final speed of the block? Solution Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Example: A particle of mass m = 5.00 kg is released from point A and slides on the frictionless track shown in Figure. Determine (a) the particle’s speed at points B and C and (b) the net work done by the gravitational force in moving the particle from A to C. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

(a) the particle’s speed at points B Find the particle’s speed at points C then find the work from the relation Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Work Done by Nonconservative Forces
Nonconservative forces change the amount of mechanical energy in a system. Wnc = work done by nonconservative force Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Problem: Children and sled with mass of 50 kg slide down a hill with a height of 0.46 m. If the sled starts from rest and has a speed of 2.6 m/s at the bottom, how much thermal energy is lost due to friction (i.e. what is the work that friction does)? If the hill has an angle of 20° above the horizontal what was the frictional force. Since vi = 0, and hf = 0, Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

The force done by friction is determined from; Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Example: A child of mass m rides on an irregularly curved slide of height h = 2.00 m, as shown in Figure. The child starts from rest at the top. (A) Determine his speed at the bottom, assuming no friction is present. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

(B) If a force of kinetic friction acts on the child, how much mechanical energy does the system lose? Assume that vf =3.00 m/s and m = 20.0 kg. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Example: A block having a mass of 0.80kg is given an initial velocity vA = 1.2m/s to the right and collides with a spring of negligible mass and force constant k = 50N/m, as shown in Figure. (A) Assuming the surface to be frictionless, calculate the maximum compression of the spring after the collision. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

(B) Suppose a constant force of kinetic friction acts between the block and the surface, with = If the speed of the block at the moment it collides with the spring is vA = 1.2 m/s, what is the maximum compression xC in the spring? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Additional Questions (Lecture 13) Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[1] A 1300 kg car drives up a 17.0 m hill. During the drive, two nonconservative forces do work on the car: the force of friction, and the force generated by the car’s engine. The work done by friction is –3.31  105 J; the work done by the engine is  105 J. Find the change in the car’s kinetic energy from the bottom of the hill to the top of the hill. [2] A single conservative force Fx = (2x + 4) N acts on a 5kg particle, where x is in m. As the particle moves along the x axis from x = 1 m to x = 5 m, calculate (a) the work done by this force, (b) the change in the potential energy of the particle, and (c) its kinetic energy at x = 5 m if its speed at x =1m is 3 m/s. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[3] Use conservation of energy to determine the final speed of a mass of 5.0kg attached to a light cord over a massless, frictionless pulley and attached to another mass of 3.5 kg when the 5.0 kg mass has fallen (starting from rest) a distance of 2.5 m as shown in Figure [4] A 5kg block is set into motion up an inclined plane as in Figure with an initial speed of 8 m/s. The block comes to rest after travelling 3 m along the plane, as shown in the diagram. The plane is inclined at an angle of 30' to the horizontal. (a) Determine the change in kinetic energy. (b) Determine the change in potential energy. (c) Determine the frictional force on the block. (d) What is the coefficient of kinetic friction? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[5] A block with a mass of 3 kg starts at a height h = 60 cm on a plane with an inclination angle of 30', as shown in Figure. Upon reaching the bottom of the ramp, the block slides along a horizontal surface. If the coefficient of friction on both surfaces is ,Uk = 0.20, how far will the block slide on the horizontal surface before coming to rest? [6] Initially sliding with a speed of 1.7 m/s, a 1.7 kg block collides with a spring and compresses it 0.35 m before coming to rest. What is the force constant of the spring? [7] A horizontal force of 100 N is applied to move a 45-kg cart across a 9.0-m level surface. What work is done by the 100-N force? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[8] A Hooke’s law spring is compressed 12.0 cm from equilibrium and the potential energy stored is 72.0 J. What is the spring constant in this case? [9] An object of mass m = 2.0 kg is released from rest at the top of a frictionless incline of height 3 m and length 5 m. Taking g = 10 m/s, use energy considerations to find the velocity of the object at the bottom of the incline. [10] A block of mass 1.0 kg is placed at the top of an incline of length 125 m and height 62.5 m. The plane has a rough surface. When the block arrives at the bottom of the plane it has a velocity of 25 m/ss. What is the magnitude of the constant frictional force acting on the block? Take g = 10m/ss Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[11] A 0.2-kg pendulum bob is swinging back and forth. If the speed of the bob at its lowest point is 0.65 m/s, how high does the bob go above its minimum height? [12] Two objects are connected by a light string passing over a light frictionless pulley as shown in Figure. The object of mass 5.00 kg is released from rest. Using the principle of conservation of energy, (a) determine the speed of the 3.00-kg object just as the 5.00-kg object hits the ground. (b) Find the maximum height to which the 3.00-kg object rises. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[13] A 5.00-kg block is set into motion up an inclined plane with an initial speed of 8.00 m/s. The block comes to rest after traveling 3.00m along the plane, which is inclined at an angle of 30.0° to the horizontal. For this motion determine the change in the block’s kinetic energy, (b) the change in the potential energy of the block–Earth system, (c) the friction force exerted on the block (assumed to be constant). (d) What is the coefficient of kinetic friction? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[14] A block of mass kg is placed on top of a light vertical spring of force constant N/m and pushed downward so that the spring is compressed by m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise? [15] A single constant force acts on a 4.00-kg particle. (a) Calculate the work done by this force if the particle moves from the origin to the point having the vector position . Does this result depend on the path? Explain. (b) What is the speed of the particle at r if its speed at the origin is 4.00 m/s? (c) What is the change in the potential energy? [16] A potential-energy function for a two-dimensional force is of the form U = 3x3y + 7x. Find the force that acts at the point (x, y) Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[17] The coefficient of friction between the 3.00-kg block and the surface in Figure is The system starts from rest. What is the speed of the 5.00kg ball when it has fallen 1.50 m [18] A 5-kg mass is attached to a light string of length 2m to form a pendulum as shown in Figure. The mass is given an initial speed of 4m/s at its lowest position. When the string makes an angle of 37o with the vertical, find (a) the change in the potential energy of the mass, (b) the speed of the mass, and (c) the tension in the string. (d) What is the maximum height reached by the mass above its lowest position? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[19] A 20.0-kg block is connected to a 30.0-kg block by a string that passes over a light frictionless pulley. The 30.0-kg block is connected to a spring that has negligible mass and a force constant of 250 N/m, as shown in Figure. The spring is unstretched when the system is as shown in the figure, and the incline is frictionless. The 20.0-kg block is pulled 20.0 cm down the incline (so that the 30.0-kg block is 40.0 cm above the floor) and released from rest. Find the speed of each block when the 30.0-kg block is 20.0 cm above the floor (that is, when the spring is unstretched). Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[20] find the particle’s speed at points B, where the particle released from point A and slides on the frictionless track [21] Find the distance x. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[22] An object of mass m starts from rest and slides a distance d down a frictionless incline of angleθ . While sliding, it contacts an unstressed spring of negligible mass as shown in Figure. The object slides an additional distance x as it is brought momentarily to rest by compression of the spring (of force constant k). Find the initial separation d between object and spring. [23] A 1.9-kg block slides down a frictionless ramp, as shown in the Figure. The top of the ramp is 1.5 m above the ground; the bottom of the ramp is h = 0.25 m above the ground. The block leaves the ramp moving horizontally, and lands a horizontal distance d away. Find the distance d. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[24] A skateboard track has the form of a circular arc with a 4.00 m radius, extending to an angle of 90.0° relative to the vertical on either side of the lowest point, as shown in the Figure. A 57.0 kg skateboarder starts from rest at the top of the circular arc. What is the normal force exerted on the skateboarder at the bottom of the circular arc? What is wrong with this picture? At bottom, a = v2/r Which direction? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Discussion Lecture 14 General Physics 1, Lec14, By/ T.A. Eleyan
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics 1, Lec14, By/ T.A. Eleyan 65 65

[1] A force F = (4xî + 3yˆj) N acts on an object as the object moves in the x direction from the origin to x = 5.00 m. Find the work done on the object by the force. [2] A 3.00-kg object has a velocity (6.00î ˆj) m/s. (a) What is its kinetic energy at this time? (b) Find the total work done on the object if its velocity changes to (8.00î ˆj) m/s. General Physics 1, Lec14, By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 66

the force of friction, and the force generated by the car’s engine.
[3] A 1300 kg car drives up a 17.0 m hill. During the drive, two nonconservative forces do work on the car: the force of friction, and the force generated by the car’s engine. The work done by friction is –3.31  105 J; the work done by the engine is  105 J. Find the change in the car’s kinetic energy from the bottom of the hill to the top of the hill. General Physics 1, Lec14, By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 67

[4] A 1500-kg car accelerates uniformly from rest to a speed of 10 m/s in 3s. Find (a) the work done on the car in this time, (b) the average power delivered the engine in the first 3s, and a) b) Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics 1, Lec14, By/ T.A. Eleyan 68

[5] When a 4kg mass is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.5cm. If the 4kg mass is removed, (a) how far will the spring stretch if a 1.5kg mass is hung on it, and (b) how much work must an external agent do to stretch the same spring 4.0 cm from its unstretched position? a) يحل عن طريق نسبة وتناسب b) Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics 1, Lec14, By/ T.A. Eleyan 69

[6] A single conservative force Fx = (2x + 4) N acts on a 5kg particle, where x is in m. As the particle moves along the x axis from x = 1 m to x = 5 m, calculate (a) the work done by this force, (b) the change in the potential energy of the particle, and (c) its kinetic energy at x = 5 m if its speed at x =1m is 3 m/s. a) b) c) Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics 1, Lec14, By/ T.A. Eleyan 70

[7] Use conservation of energy to determine the final speed of a mass of 5.0kg attached to a light cord over a massless, frictionless pulley and attached to another mass of 3.5 kg when the 5.0 kg mass has fallen (starting from rest) a distance of 2.5 m as shown in Figure Substitute and find v General Physics 1, Lec14, By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 71

[8] A 5kg block is set into motion up an inclined plane as in Figure with an initial speed of 8 m/s. The block comes to rest after travelling 3 m along the plane, as shown in the diagram. The plane is inclined at an angle of 30' to the horizontal. (a) Determine the change in kinetic energy. (b) Determine the change in potential energy. (c) Determine the frictional force on the block. (d) What is the coefficient of kinetic friction? General Physics 1, Lec14, By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 72

a) b) c) d) General Physics 1, Lec14, By/ T.A. Eleyan
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics 1, Lec14, By/ T.A. Eleyan 73

[9] A block with a mass of 3 kg starts at a height h = 60 cm on a plane with an inclination angle of 30', as shown in Figure. Upon reaching the bottom of the ramp, the block slides along a horizontal surface. If the coefficient of friction on both surfaces is ,Uk = 0.20, how far will the block slide on the horizontal surface before coming to rest? General Physics 1, Lec14, By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 74

And find v And find s2 General Physics 1, Lec14, By/ T.A. Eleyan
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 75

[10] Initially sliding with a speed of 1. 7 m/s, a 1
[10] Initially sliding with a speed of 1.7 m/s, a 1.7 kg block collides with a spring and compresses it 0.35 m before coming to rest. What is the force constant of the spring? General Physics 1, Lec14, By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 76

[11]The ambulance (mass 3000kg) shown in the Figure(2) slides (wheels locked) down a frictionless incline that is 10 m long. It starts from rest at point A, and continues along a rough surface until it comes to a complete stop at point C. If the coefficient of kinetic friction between the ambulance and the horizontal rough surface is 0.1. (a) Calculate the speed of the ambulance at point B.(b) Compute the distance d the ambulance slides on the horizontal rough surface before stopping. General Physics 1, Lec14, By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 77

Lecture 15 Linear Momentum and Collisions
General Physics I, Lec By/ T.A. Eleyaan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 78

Linear Momentum Linear momentum is defined as: p = mv
Momentum is given by mass times velocity. Momentum is a vector. The units of momentum are (no special unit): [p] = kg·m/s Since p is a vector, we can also consider the components of momentum: px = mvx, py = mvy, pz = mvz Note: momentum is “large” if m and/or v is large. (define large, meaning hard for you to stop). Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics I, Lec By/ T.A. Eleyaan 79

Recall that Another way of writing Newton’s Second Law is
F = dp/dt= rate of change of momentum This form is valid even if the mass is changing. This form is valid even in Relativity and Quantum Mechanics. General Physics I, Lec By/ T.A. Eleyaan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 80

Impulse of a Force Exercise: Show that impulse and momentum have the same units. General Physics I, Lec By/ T.A. Eleyaan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 81

Impulse & Momentum The impulse of a force during a certain interval of time is the change in momentum that the force produces General Physics I, Lec By/ T.A. Eleyaan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 82

F-t The Graph Impulse is a vector quantity
The magnitude of the impulse is equal to the area under the force-time curve Dimensions of impulse are M L / T Impulse is not a property of the particle, but a measure of the change in momentum of the particle General Physics I, Lec By/ T.A. Eleyaan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 83

F-t The Graph The impulse can also be found by using the time averaged force I = Dt This would give the same impulse as the time varying force does General Physics I, Lec By/ T.A. Eleyaan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 84

Example: In a particular crash test, a car of mass 1500 kg collides with a wall, as shown in Figure. The initial and final velocities of the car are , respectively. If the collision lasts for 0.15 s, find the impulse caused by the collision and the average force exerted on the car. 1) The impulse 2) The average force Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics I, Lec By/ T.A. Eleyaan 85

The force acting on the golf ball can be determined from
Example: A 100-g ball is dropped from 2.00 m above the ground. It rebounds to a height of 1.50 m. What was the average force exerted by the floor if the ball was in contact with the floor for 1×10-2 s. The force acting on the golf ball can be determined from General Physics I, Lec By/ T.A. Eleyaan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 86

What is the change in momentum of the hockey puck during the bounce?
Problem: A 0.3-kg hockey puck moves on frictionless ice at 8 m/s toward the wall. It bounces back away from the wall at 5 m/s. The puck is in contact with the wall for 0.2 s. What is the change in momentum of the hockey puck during the bounce? What is the impulse on the hockey puck during the bounce? What is the average force of the wall on the hockey puck during the bounce? General Physics I, Lec By/ T.A. Eleyaan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 87

Example: A 325-gm ball with a speed v of 6
Example: A 325-gm ball with a speed v of 6.22 m/s strikes a wall at an angle of and then rebounds with the same speed and angle. It is in contact with the wall for 10.4 ms. (a) What impulse was experienced by the ball ? What was the average force exerted by the ball on the wall ? x y General Physics I, Lec By/ T.A. Eleyaan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 88

(a) (b) يوثر على الx وy كلو موجب يوثر على –xو+y
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics I, Lec By/ T.A. Eleyaan 89

Principle of Conservation of Linear Momentum
When the net external force acting on a system is zero, the total linear momentum of the system remains constant. Proof (Two-Body Systems) However, Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics I, Lec By/ T.A. Eleyaan 90

In the absence of a net external force,

Look at two billiard balls colliding:
Before collision: After collision: From the law of the conservation of momentum. It states: The Total Momentum of an isolated system of bodies remains constant. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics I, Lec By/ T.A. Eleyaan 92

The third part has a mass of 100-34-25 = 41g.
Example: A firecracker with a mass of 100g, initially at rest, explodes into 3 parts. One part with a mass of 25g moves along the x-axis at 75m/s. One part with mass of 34g moves along the y-axis at 52m/s. What is the velocity of the third part? The third part has a mass of = 41g. Along x-direction: Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics I, Lec By/ T.A. Eleyaan 93

So the magnitude of the velocity of the third piece is;
Along y-direction: So the magnitude of the velocity of the third piece is; And the direction of v3 is given by Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics I, Lec By/ T.A. Eleyaan 94

Collisions In general, a “collision” is an interaction in which two objects strike one another the net external impulse is zero or negligibly small (momentum is conserved) Examples: Two billiard balls colliding on a billiards table An alpha particle colliding with a heavy atom Two galaxies colliding General Physics I, Lec By/ T.A. Eleyaan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 95

Types of Collisions [1] In an elastic collision, momentum and kinetic energy are conserved Perfectly elastic collisions occur on a microscopic level In macroscopic collisions, only approximately elastic collisions actually occur [2] In an inelastic collision, kinetic energy is not conserved although momentum is still conserved If the objects stick together after the collision, it is a perfectly inelastic collision In an inelastic collision, some kinetic energy is lost, but the objects do not stick together Elastic and perfectly inelastic collisions are limiting cases, most actual collisions fall in between these two types Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics I, Lec By/ T.A. Eleyaan 96

Elastic Collisions Both momentum and kinetic energy are conserved

Perfectly Inelastic Collisions
Since the objects stick together, they share the same velocity after the collision General Physics I, Lec By/ T.A. Eleyaan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 98

Example: A ball with a mass of 1. 2 kg moving to the right at 2
Example: A ball with a mass of 1.2 kg moving to the right at 2.0 m/s collides with a ball of mass 1.8 kg moving at 1.5 m/s to the left. If the collision is an elastic collision, what are the velocities of the balls after the collision? General Physics I, Lec By/ T.A. Eleyaan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 99

We can apply conservation of momentum and conservation of
kinetic energy: Solve (1)&(2) to find velocities of the balls after the collision? General Physics I, Lec By/ T.A. Eleyaan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 100

Example: A 3kg sphere makes a perfectly inelastic collision with a second sphere initially at rest. The composite system moves with a speed equal to 1/3rd the original speed of the 3 kg sphere. What is the mass of the second sphere? General Physics I, Lec By/ T.A. Eleyaan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 101

we can apply conservation of momentum:

Lecture 16 Linear Momentum and Collisions
General Physics I, Lec By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 103

Example: A 1 500-kg car traveling east with a speed of 25
Example: A kg car traveling east with a speed of 25.0 m/s collides at an intersection with a 2500-kg van traveling north at a speed of 20.0 m/s, as shown in Figure. Find the direction and magnitude of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (that is, they stick together) From the x axis component General Physics I, Lec By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 104

From the y axis component

a) the velocity of the bullet and block just after impact and
Example: In the ballistic pendulum experiment, a bullet of mass .06 kg is fired horizontally into a wooden block of mass .2 kg. The wooden block is suspended from the ceiling by a long string as shown in the diagram. The collision is perfectly inelastic and after impact the bullet and the block swing together until the block is 12 m above it's initial position. Find a) the velocity of the bullet and block just after impact and b) the velocity of the bullet just before impact Energy is conserved between points B and C B Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics I, Lec By/ T.A. Eleyan 106

Momentum is conserved between A and B

Example: Two metal spheres, suspended by vertical cords, initially just touch, as shown in Fig. Sphere 1, with mass m1=30 g, is pulled to the left to height h1=8.0cm, and then released from rest. After swinging down, it undergoes an elastic collision with sphere 2, whose mass m2=75 g. What is the velocity v1f of sphere 1 just after the collision? General Physics I, Lec By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 108

where General Physics I, Lec 16 By/ T.A. Eleyan
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 109

Example: The figure below shows an elastic collision of two pucks on a frictionless air table. Puck A has mass mA = kg, and puck B has mass mB = kg. Puck A has an initial velocity of 4.00 m⁄s in the positive x−direction and a final velocity of 2.00 m ⁄s in an unknown direction. Puck B is initially at rest. We want to find the final speed VB2 of puck B and the angles α and β as illustrated in the figure. Because the collision is elastic, the initial and final kinetic energies are equal General Physics I, Lec By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 110

Conservation of the x− component of total momentum gives
Conservation of the y− component of total momentum gives There are two simultaneous equations for α and β Solve to find the angles Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics I, Lec By/ T.A. Eleyan 111

Center of Mass The center of mass (CM) of an object or a group of objects (system) is the “average” location of the mass in the system. The system behaves as if all of its mass were concentrated at the center of mass. General Physics I, Lec By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 112

Center of mass – point objects – 1 D

Center of mass – point objects – 2 D

Center of mass – solid objects – 1 D

Center of mass – solid objects – 2 D

A method for finding the center of mass of any object.
Hang object from two or more points. Draw extension of suspension line. Center of mass is at intercept of these lines. General Physics I, Lec By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 117

Problem: Three particles of masses mA = 1.2 kg, mB = 2.5 kg, and mC = 3.4 kg form an equilateral triangle of edge length a = 140 cm. Where is the center of mass of this three particle system? Problem: Three masses located in the xy plane have the following coordinates: 2 kg at (3,-2) 3 kg at (-2,4) 1 kg at (2,2) Find the location of the center of mass Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics I, Lec By/ T.A. Eleyan 118

Motion of the Center of Mass
A system of objects behaves as if all its mass were located at the center of mass. Velocity of CM: Acceleration of CM: Newton's Laws for a System of Particles MAcm = Fnet,ext General Physics I, Lec By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 119

Problem: The three particles in Fig. a are initially at rest. Each experiences an external force due to bodies outside the three-particle system. The directions are indicated, and the magnitudes are FA=6 N , FB=12 N , and FC=14 N. What is the magnitude of the acceleration of the center of mass of the system, and in what direction does it move? General Physics I, Lec By/ T.A. Eleyan Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 120

Additional Questions (Lecture 15,16) Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[1] A 3.00-kg particle has a velocity of (3.00i ˆ j ˆ ) m/s. (a) Find its x and y components of momentum. (b) Find the magnitude and direction of its momentum. [2] An estimated force–time curve for a baseball struck by a bat is shown in Figure. From this curve, determine (a) the impulse delivered to the ball, (b) the average force exerted on the ball, and (c) the peak force exerted on the ball.. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[3] A 3.00-kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60.0° with the surface. It bounces off with the same speed and angle (Fig. ). If the ball is in contact with the wall for s, what is the average force exerted by the wall on the ball ? [4] A 10.0-g bullet is fired into a stationary block of wood (m = 5.00 kg). The relative motion of the bullet stops inside the block. The speed of the bullet-plus-wood combination immediately after the collision is m/s. What was the original speed of the bullet? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[5] Two blocks are free to slide along the frictionless wooden track ABC shown in Figure. The block of mass m1 = 5.00 kg is released from A. Protruding from its front end is the north pole of a strong magnet, repelling the north pole of an identical magnet embedded in the back end of the block of mass m2 =10.0 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[6] A 12.0-g wad of sticky clay is hurled horizontally at a 100-g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50 m before coming to rest. If the coefficient of friction between the block and the surface is 0.650, what was the speed of the clay immediately before impact? [7] (a) Three carts of masses 4.00 kg, 10.0 kg, and 3.00 kg move on a frictionless horizontal track with speeds of 5.00 m/s, 3.00 m/s, and 4.00 m/s, as shown in Figure . Velcro couplers make the carts stick together after colliding. Find the final velocity of the train of three carts. (b) What If? Does your answer require that all the carts collide and stick together at the same time? What if they collide in a different order? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[8] Two automobiles of equal mass approach an intersection. One vehicle is traveling with velocity 13.0 m/s toward the east, and the other is traveling north with speed v2i. Neither driver sees the other. The vehicles collide in the intersection and stick together, leaving parallel skid marks at an angle of 55.0° north of east. The speed limit for both roads is 35 mi/h, and the driver of the northward-moving vehicle claims he was within the speed limit when the collision occurred. Is he telling the truth? [9] A billiard ball moving at 5.00 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves, at 4.33 m/s, at an angle of 30.0° with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball’s velocity after the collision. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[10] Two particles with masses m and 3m are moving toward each other along the x axis with the same initial speeds vi . Particle m is traveling to the left, while particle 3m is traveling to the right. They undergo an elastic glancing collision such that particle m is moving downward after the collision at right angles from its initial direction. (a) Find the final speeds of the two particles. (b) What is the angle & at which the particle 3m is scattered? [11] Four objects are situated along the y axis as follows: a 2.00 kg object is at 3.00 m, a 3.00-kg object is at 2.50 m, a 2.50-kg object is at the origin, and a 4.00-kg object is at m. Where is the center of mass of these objects?. [12] A 2.00-kg particle has a velocity (2.00ˆi ˆ j) m/s, and a 3.00-kg particle has a velocity (1.00ˆi +6.00ˆ j) m/s. Find (a) the velocity of the center of mass and (b) the total momentum of the system. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[13] A 0.2 kg baseball is traveling at 40 m/s. After being hit by a bat, the ball's velocity is 50 m/s in the opposite direction. Find a) the impulse and b) the average force exerted by the bat if the ball and bat are in contact for 0.002 s [14] A 1000 kg car traveling 22 m/s hits a concrete bridge support and comes to a stop in .5 s. a) Find the average force acting on the car and b) if the bridge support had been cushioned so that the stopping time was increased to 3 s, what would have been the average force? [15] A 1.0 kg object traveling at 1.0 m/s collides head on with a 2.0 kg object initially at rest. Find the velocity of each object after impact if the collision is perfectly elastic [16] Car A has a mass of 1000 kg and is traveling to the right at 5 m/s. Car B has a mass of 5000 kg and is traveling to the right at 3 m/s. Car A read-ends Car B and they stick together. What is the velocity of the center of mass after the collision? (Before the collision, the velocity of the CM is 3.33 m/s) Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[17] Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them (Fig.). A cord initially holding the blocks together is burned; after this, the block of mass 3M moves to the right with a speed of 2.00 m/s. (a) What is the speed of the block of mass M? (b) Find the original elastic potential energy in the spring if M = kg. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Discussion Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[1] An estimated force–time curve for a baseball struck by a bat is shown in Figure. From this curve, determine (a) the impulse delivered to the ball, (b) the average force exerted on the ball, and (c) the peak force exerted on the ball.. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[2] Two blocks are free to slide along the frictionless wooden track ABC shown in Figure. The block of mass m1 = 5.00 kg is released from A. and makes a head on elastic collision with a block of mass m2 =10.0 kg, initially at rest at point B. Calculate the maximum height to which m1 rises after the elastic collision.. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[4]As shown in Figure , a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length L and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[5] Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them (Fig.). A cord initially holding the blocks together is burned; after this, the block of mass 3M moves to the right with a speed of 2.00 m/s. (a) What is the speed of the block of mass M? (b) Find the original elastic potential energy in the spring if M = kg. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[6] Show that the center of mass of a rod of mass M and length L lies in midway between its ends, assuming the rod has a uniform mass per unit length. The formula for CM of a continuous object is Since the density of the rod (l) is constant; The mass of a small segment Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Therefore Find the CM when the density of the rod non-uniform but varies linearly as a function of x, l=a x Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Lecture 18 Rotational Motion
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Angular Position, θ For circular motion, the distance (arc length) s, the radius r, and the angle  are related by: q > 0 for counterclockwise rotation from reference line Note that  is measured in radians: Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

The angular displacement is defined as the angle the object rotates through during some time interval Every point on the disc undergoes the same angular displacement in any given time interval Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Angular Velocity, ω Notice that as the disk rotates,  changes. We define the angular displacement, , as:  = f - i which leads to the average angular speed wav Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Instantaneous Angular Velocity
As usual, we can define the instantaneous angular velocity as: Note that the SI units of  are: rad/s = s-1 w > 0 for counterclockwise rotation < 0 for clockwise rotation If v = speed of a an object traveling around a circle of radius r w = v / r Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

The period of rotation is the time it takes to complete one revolution. Problem: What is the period of the Earth’s rotation about its own axis? What is the angular velocity of the Earth’s rotation about its own axis? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Angular Acceleration, a We can also define the average angular acceleration aav: And instantaneous angular acceleration The SI units of a are: rad/s2 = s-2 We will skip any detailed discussion of angular acceleration, except to note that angular acceleration is the time rate of change of angular velocity Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Notes about angular kinematics:
When a rigid object rotates about a fixed axis, every portion of the object has the same angular speed and the same angular acceleration i.e. q,w, and a are not dependent upon r, distance form hub or axis of rotation Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Examples: 1. Bicycle wheel turns 240 revolutions/min. What is its angular velocity in radians/second? 2. If wheel slows down uniformly to rest in 5 seconds, what is the angular acceleration? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

3. How many revolution does it turn in those 5 sec? Recall that for linear motion we had: Perhaps something similar for angular quantities? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Analogies Between Linear and Rotational Motion
Rotational Motion About a Fixed Axis with Constant Acceleration Linear Motion with Constant Acceleration Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Relationship Between Angular and Linear Quantities
Displacements Speeds Accelerations Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Tangential and the radial acceleration. Since the tangential speed v is The magnitude of tangential acceleration at is The radial or centripetal acceleration ar is Total linear acceleration is Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Example: (a) What is the linear speed of a child seated 1.2m from the center of a steadily rotating merry-go-around that makes one complete revolution in 4.0s? (b) What is her total linear acceleration? Since the angular speed is constant, there is no angular acceleration. Tangential acceleration is Radial acceleration is Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Calculation of Moments of Inertia
Moments of inertia for large objects can be computed, if we assume that the object consists of small volume elements with mass, Dmi. The moment of inertia for the large rigid object is: It is sometimes easier to compute moments of inertia in terms of volume of the elements rather than their mass Using the volume density, r, replace dm in the above equation with dV. The moments of inertia becomes Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

The moment of inertia of a uniform hoop of mass M and radius R about an axis perpendicular to the plane of the hoop and passing through its center. The moment of inertia is The moment of inertia for this object is the same as that of a point of mass M at the distance R. x y R O dm Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Example for Rigid Body Moment of Inertia
Calculate the moment of inertia of a uniform rigid rod of length L and mass M about an axis perpendicular to the rod and passing through its center of mass. The line density of the rod is so the mass is Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Torque: The ability of a force to rotate a body about some axis. Only the component of the force that is perpendicular to the radius causes a torque. = r (Fsinq) Equivalently, only the perpendicular distance between the line of force and the axis of rotation, known as the moment arm r, can be used to calculate the torque. t = rF = (rsinq)F Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

The net torque about a point O is the sum of all torques about O: Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Problem: Calculate the net torque on the 0.6-m rod about the nail at the left. Three forces are acting on the rod as shown in the diagram. 30° 0.3 m 4 N 5 N 6 N Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Torque & Angular Acceleration
Let’s consider a point object with mass m rotating on a circle. The tangential force Ft is The torque due to tangential force Ft is Torque acting on a particle is proportional to the angular acceleration. Analogs to Newton’s 2nd law of motion in rotation. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

How about a rigid object? The external tangential force dFt is The torque due to tangential force Ft is The total torque is Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Example : A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical plane. The rod is released from rest in the horizontal position. What are the initial angular acceleration of the rod and the initial linear acceleration of its right end? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

The only force generating torque is the gravitational force Mg Since the moment of inertia of the rod when it rotates about one end We obtain Using the relationship between tangential and angular acceleration The tip of the rod falls faster than an object undergoing a free fall. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Rotational Kinetic Energy
Kinetic energy of a rigid object that undergoing a circular motion: Kinetic energy of a mass mi, moving at a tangential speed, vi, is Since a rigid body is a collection of mass, the total kinetic energy of the rigid object is The moment of Inertia, I, is defined as The above expression is simplified to Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Example: In a system consists of four small spheres as shown in the figure, assuming the radii are negligible and the rods connecting the particles are massless, compute the moment of inertia and the rotational kinetic energy when the system rotates about the y-axis at w. x y M l m b O Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Since the rotation is about y axis, the moment of inertia about y axis, Iy, is Why are some 0s? This is because the rotation is done about y axis, and the radii of the spheres are negligible. Thus, the rotational kinetic energy is Find the moment of inertia and rotational kinetic energy when the system rotates on the x-y plane about the z-axis that goes through the origin O. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Discussion Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[1] During a certain period of time, the angular position of a swinging door is described by θ= t t2, where θ is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door (a) at t = 0 (b) at t = 3.00 s. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

[5] A disk 8.00 cm in radius rotates at a constant rate of rev/min about its central axis. Determine (a) its angular speed, (b) the tangential speed at a point 3.00 cm from its center, (c) the radial acceleration of a point on the rim, and (d) the total distance a point on the rim moves in 2.00 s. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Consider a car driving at 20 m/s on a level circular turn of radius 40.0 m. Assume the car’s mass is 1000 kg. What is the magnitude of frictional force experienced by car’s tires? What is the minimum coefficient of friction in order for the car to safely negotiate the turn? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

1. Draw a free body diagram, introduce coordinate frame and consider vertical and horizontal projections 2. Use definition of friction force: Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Consider a car driving at 20 m/s on a 30° banked circular curve of radius 40.0 m. Assume the car’s mass is 1000 kg. What is the magnitude of frictional force experienced by car’s tires? What is the minimum coefficient of friction in order for the car to safely negotiate the turn? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

1. Draw a free body diagram, introduce coordinate frame and consider vertical and horizontal projections 2. Use definition of friction force: Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Example: Uniform Solid Cylinder A uniform solid cylinder has a radius R, mass M, and length L. Calculate its moment of inertia about its central axis The volume density of the shell is so the mass is Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

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Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

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