CentralForces - University of Colorado Boulder
... What about the outward "centrifugal force"? A person on a merry-go-round (or twirled on a rope by a giant) "feels" an outward force. This is an illusion! There is no outward force on the person. Our intuition is failing us. Our intuition about forces was developed over a lifetime of experiences in i ...
... What about the outward "centrifugal force"? A person on a merry-go-round (or twirled on a rope by a giant) "feels" an outward force. This is an illusion! There is no outward force on the person. Our intuition is failing us. Our intuition about forces was developed over a lifetime of experiences in i ...
m: mass, v: velocity
... Linear momentum • Pi = Pf = 0 (for man and boat) • When the man jumps out of the boat he has momentum in one direction and, therefore, so does the boat, but in the opposite direction. • Their momenta must cancel out! ...
... Linear momentum • Pi = Pf = 0 (for man and boat) • When the man jumps out of the boat he has momentum in one direction and, therefore, so does the boat, but in the opposite direction. • Their momenta must cancel out! ...
Newton`s Laws and Forces
... Example: you in accelerating car where cup moved but not by any force. Most reference frames on Earth are assumed to be inertial (neglect spin) ...
... Example: you in accelerating car where cup moved but not by any force. Most reference frames on Earth are assumed to be inertial (neglect spin) ...
Problem set 13
... 2. h5i Euler angles θ, φ, ψ, were defined in the lecture (see the lecture notes). Express the generalized velocities θ̇, φ̇, ψ̇ in terms of the angular velocity components Ω1 , Ω2 , Ω3 . 3. h6i Consider force free rotational motion of a symmetric top ( I1 = I2 , I3 ) described in terms of Euler angl ...
... 2. h5i Euler angles θ, φ, ψ, were defined in the lecture (see the lecture notes). Express the generalized velocities θ̇, φ̇, ψ̇ in terms of the angular velocity components Ω1 , Ω2 , Ω3 . 3. h6i Consider force free rotational motion of a symmetric top ( I1 = I2 , I3 ) described in terms of Euler angl ...
document
... measure its period, T, the time needed for the object to make one complete revolution. During this time, the object travels a distance equal to the circumference of the circle, 2πr. The object’s speed, then, is represented by v = 2πr/T ...
... measure its period, T, the time needed for the object to make one complete revolution. During this time, the object travels a distance equal to the circumference of the circle, 2πr. The object’s speed, then, is represented by v = 2πr/T ...
Newton`s Second Law:
... Use Newton’s 2nd Law to find the x and y component of the puck’s acceleration magnitude ...
... Use Newton’s 2nd Law to find the x and y component of the puck’s acceleration magnitude ...
Lecture 18
... then be pulled back so that the string makes an angle theta with the vertical. The bob will then be released and allowed to swing. Predict what will happen to the reading on the spring scale at the bottom of the swing (more, less or same as when the object is at rest). ...
... then be pulled back so that the string makes an angle theta with the vertical. The bob will then be released and allowed to swing. Predict what will happen to the reading on the spring scale at the bottom of the swing (more, less or same as when the object is at rest). ...
1 - HCC Learning Web
... 1. Two ropes are attached to a 40-kg object. The first rope applies a force of 25 N and the second, 40 N. If the two ropes are perpendicular to each other, what is the resultant acceleration of the object? a. 1.2 m/s2 b. 3.0 m/s2 c. 25 m/s2 d. 47 m/s2 2. Two blocks, joined by a string, have masses o ...
... 1. Two ropes are attached to a 40-kg object. The first rope applies a force of 25 N and the second, 40 N. If the two ropes are perpendicular to each other, what is the resultant acceleration of the object? a. 1.2 m/s2 b. 3.0 m/s2 c. 25 m/s2 d. 47 m/s2 2. Two blocks, joined by a string, have masses o ...
( )N ( )m ( )N
... 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 ...
... 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 ...
Word
... amount of centripetal force needed (mv²/r) is equal to the amount of friction available (µF N). Fc = F f mv²/r = µ FN = µ (mg) v= ...
... amount of centripetal force needed (mv²/r) is equal to the amount of friction available (µF N). Fc = F f mv²/r = µ FN = µ (mg) v= ...
Exam 2 study guide
... Banked roads - road is sloped so that normal force points inwards, supplying the centripetal force necessary for a turn. This reduces the need for friction forces. The banking angle is v2 = rg tanθ. Newton’s law of gravitation - gravitational force between objects 1 and 2 is in the direction between ...
... Banked roads - road is sloped so that normal force points inwards, supplying the centripetal force necessary for a turn. This reduces the need for friction forces. The banking angle is v2 = rg tanθ. Newton’s law of gravitation - gravitational force between objects 1 and 2 is in the direction between ...
Chapter 2: Forces and the Momentum Principle
... ❑ Now, we begin the study of the second part of mechanics – dynamics - which does address the cause of motion - that cause ! is a force, a push or pull ❑ Force, F, is a vector, has magnitude and direction ...
... ❑ Now, we begin the study of the second part of mechanics – dynamics - which does address the cause of motion - that cause ! is a force, a push or pull ❑ Force, F, is a vector, has magnitude and direction ...
Fall Semester Review
... Newton’s 1st Law: an object with no (net) force on it moves with constant velocity. Newton’s 2nd Law: F=ma Newton’s 3rd Law: When one object exerts a force on a second object, the second exerts a force on the first that is equal in magnitude but opposite in direction. There are all kinds of Forces, ...
... Newton’s 1st Law: an object with no (net) force on it moves with constant velocity. Newton’s 2nd Law: F=ma Newton’s 3rd Law: When one object exerts a force on a second object, the second exerts a force on the first that is equal in magnitude but opposite in direction. There are all kinds of Forces, ...
