Acceleration in Uniform Circular Motion
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1 Acceleration in Uniform Circular Motion The object in uniform circular motion has a constant speed, but its velocity is constantly changing directions, generating a centripetal acceleration: a c v r 2 Section 2.4
2 Determine the magnitude of the centripetal acceleration of a car going 12 m/s on a circular track with a radius of 50 m. Solution: Finding Centripetal Acceleration Example a c 12 m/s 2 2 v r 50 m 2.9 m/s 2 Section 2.4
3 James T. Shipman Jerry D. Wilson Charles A. Higgins, Jr. Omar Torres Chapter 3 Force and Motion
4 Inertia Inertia: is the tendency of an object to resist change in motion The greater the mass of an object, the greater its inertia The smaller the mass of an object, the smaller its inertia Section 3.2
5 Force Vector Force: is a vector quantity capable of producing motion A force is capable of changing an object s velocity and thereby producing acceleration. m Acceleration Force Section 3.1
6 Balanced Forces No Motion F 1 = F 2 Equal in magnitude but in opposite directions Section 3.1
7 Unbalanced Forces Result in Motion F 1 F 2 Net force to the right Net Force Section 3.1
8 Newton s First Law of Motion Newton s 1 st Law of Motion An object will remain at rest or in uniform motion in a straight line unless acted on by an external, unbalanced force Section 3.2
9 Newton s Second Law of Motion F m a a.if we double the force we double the acceleration b.if we double the mass we halve the acceleration Section 3.3
10 Newton s Second Law of Motion F = ma Force = mass acceleration = kg m/s 2 = N N = kg m/s 2 = newton this is the metric system (SI) unit of force. Where: F is the net force, which is the sum of all forces applied, and m is the total mass & a is the acceleration of the whole system Section 3.3
11 Net Force and Total Mass Example Forces are applied to blocks connected by a string (weightless) resting on a frictionless surface. Mass of each block = 1 kg; F 1 = 5.0 N; F 2 = 8.0 N. What is the acceleration of the system? Section 3.3
12 Net Force and Total Mass Example Forces are applied to blocks connected by a string (weightless) resting on a frictionless surface. Mass of each block = 1 kg; F 1 = 5.0 N; F 2 = 8.0 N. What is the acceleration of the system? a F net F 8.0 N 5.0 N 1.5 m/s m m m 1.0 kg kg Section 3.3
13 Mass & Weight Mass: amount of matter present (a measure of inertia) Weight: is a force due to gravity (w = mg ). Where m is the mass of the object, and g is the gravitational acceleration Example: What is the weight of a 2.45-kg mass on (a) Earth, and (b) the Moon (g Moon =1/6 g)? a. Earth: w = mg = (2.45 kg)(9.8 m/s 2 ) = 24.0 N b. Moon: w = mg = (2.45 kg)[(9.8 m/s 2 )/6] = 4.0 N Section 3.3
14 Newton s Third Law of Motion w = mg w = mg For every action there is an equal and opposite reaction F 1 ma 1 1 F 2 ma 2 2
15 1. What keeps the moon orbiting the earth? 2. Can the moon ever fall back onto the earth?
16 How does gravitational force vary with mass? F F F F m 1 m 2 m 1 m 2 2F 2F 2F 2F 2m 1 m 2 m 1 2m2 F α m 1 F α m 2 F α m 1. m 2
17 Force Arbitrary unit(n) How does gravitational force vary with distance? F F d m 1 m 2 1 3/4 F/4 F/4 2d m 1 m 2 1/2 F/9 F/9 1/4 1/4 3d m 1 m 2 1/9 1/16 1/25 F α 1/d Distance (arbitrary unit)
18 Newton s Universal Law of Gravitation F F F = - F Every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The gravitational forces of attraction between particles are equal but have opposite direction Section 3.5
19 Formulating Newton s Law of Gravitation F d m 1 m 2 Gm m Gravitational force: F d F is measured in Newton N. m 1, m 2 are measured in Kg. d is measured in meters. F G is called the universal gravitational constant, and is equal to G = N. m 2 /kg 2
20 Example Two objects with masses of 1.0 kg and 2.0 kg are 1.0 m apart. What is the magnitude of the gravitational force between the masses? Section 3.5
21 Solution: F F F Gm m d N m /kg 1.0 kg 2.0 kg N 1.0 m 2 Section 3.5
22 Can the moon ever fall back onto the earth? The moon is actually falling toward Earth but has great enough tangential velocity to stay in orbit.
23 Weightlessness in space is the result of both the astronaut and the spacecraft falling to Earth at the same rate. Section 3.5
24 Calculating the mass of Earth M E The force of gravity on object with mass m: F G m M 2 R E E This force is just the object s weight (W = mg) mg g M E G m M R GM R 2 E E 2 E E, cancel out m 2 gr E, rearrange for M E G (9.81 m / s )( m) 2 11 N m Kg kg Section 3.5
25 Linear Momentum m,v m,v Stopping a moving bullet is difficult because of its high speed Stopping a slow moving car is difficult because of its large mass. In addition to velocity and acceleration, there is a new variable that can also describe motion, called Momentum Linear Momentum= mass x velocity
26 Linear momentum = mass velocity Linear momentum: is a vector quantity that is equal to the product of mass and velocity p = mv m Velocity, v Momentum, p = m.v If we have a system of masses, the linear momentum is the sum of all individual momentum vectors. p = p 1 + p 2 + p 3 + Section 3.7
27 The Conservation of Linear Momentum Linear Momentum is conserved as long as there are no external unbalanced forces. p i = p f = 0 (for man and boat) Section 3.7
28 The Conservation of Linear Momentum Two masses at rest on a frictionless surface. When the string is burned, the two masses fly apart due to the release of the compressed (internal) spring (v 1 = 1.8 m/s). Find v 2? P f = P i = 0 P f = p 1 + p 2 = 0 p 1 = p 2 m 1 v 1 = m 2 v 2 Section 3.7
29 The Conservation of Linear Momentum m 1 v 1 = m 2 v 2 v 2 mv m kg 1.8 m/s 2.0 kg 0.90 m/s Section 3.7
30 The Conservation of Linear Momentum
31 Angular Momentum L = mvr L = angular momentum, m = mass, v = velocity, and r = distance to center of motion L 1 = L 2 m 1 v 1 r 1 = m 2 v 2 r 2 Section 3.7
32 Conservation of Angular Momentum The angular momentum of an object remains constant if there is no external, unbalanced torque acting on it Figure Skater by pulling the arms in the skater spins faster m 1 v 1 r 1 = m 2 v 2 r 2, m is constant; r decreases; v increases Section 3.7
33 Conservation of Angular Momentum Example A comet at its farthest point from the Sun is 900 million miles, traveling at 6000 mi/h. What is its speed at its closest point of 30 million miles away? mv 1 r 1 = mv 2 r 2 mv 1 r 1 = mv 2 r 2 v vr mi/h mi r mi = mi/h Section 3.7
34 How can massive objects float?
35 Buoyancy ρ object < ρ water ρ object > ρ water Weight Buoyancy weight Buoyancy weight weight
36 Principle of Buoyancy An object placed in a liquid experiences a buoyant force equal to the weight of the displaced liquid
37 Principle of Buoyancy An object placed in a liquid experiences a buoyant force equal to the weight of the displaced liquid F B W = m. g A = object cross sectional area d = liquid density g = gravity acceleration h = object submerged length h W=mg d
38 Pressure Definition Pressure Force Area 2 ( N / m )
39 Chapter 3 Homework:
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