Chapter 7 Rotational Motion and the Law of Gravity

Rotational Motion and the Law of Gravity  Describing Rotational Motion  A Fraction of One Revolution can be Measured in grads, degrees, or radians.  A Grad is 1/400 of a Revolution  A Degree is 1/360 of a Revolution  A Radian is the Linear Distance around the Circle Equal to the Radius  A Full Revolution is Equal to 2  r (Circumference)

Rotational Motion and the Law of Gravity  Describing Rotational Motion

Rotational Motion and the Law of Gravity  Angular Position (  )  Radian  “Angle where the arc length (s) of a circle is equal to the radius of the circle.”  Remember that  must be in Radians

Rotational Motion and the Law of Gravity  Angular Position (  )  Angle Measured from a Reference Line  Units  Radians  Convention  Positive  CCW  Negative  CW

Rotational Motion and the Law of Gravity  Angular Displacement (s)  For rotation through an angle, θ, a point at a distance, r, from the center moves a distance given by s = rθ.  Remember that  must be in Radians

Rotational Motion and the Law of Gravity  Angular Velocity (  )  Velocity is  d/  t  Angular Velocity is the Change in Angular Position (  ) During an Elapsed Time

Rotational Motion and the Law of Gravity  Angular Velocity (  )  Units  Radians / second  Convention  Positive  CCW  Negative  CW

Rotational Motion and the Law of Gravity  Angular Acceleration (  )  Acceleration  Angular Acceleration

Rotational Motion and the Law of Gravity  Angular Acceleration (  )  Units  radians/sec/sec (rad/s 2 )  Convention   f >  i,  is Positive   f <  i,  is Negative  Rotational Speed  If  and  have the Same Sign, Speed Increases  If  and  have Different Signs, Speed Decreases

Rotational Motion and the Law of Gravity  Rotational Kinematics  Let’s Compare Linear Kinematics Angular Kinematics d  v  a 

Rotational Motion and the Law of Gravity  Rotational Kinematics Linear KinematicsAngular Kinematics

Rotational Motion and the Law of Gravity  Problem A wheel is rotated so that a point on the edge moves through 1.50m. The radius of the wheel is 2.50m. Through what angle (in radians) is the wheel rotated?

Rotational Motion and the Law of Gravity  Solution  s = 1.50m  r = 2.50m

Rotational Motion and the Law of Gravity  Problem A steering wheel is rotated through 128°. Its radius is 22cm. How far would a point on the steering wheel’s edge move?

Rotational Motion and the Law of Gravity  Solution   = 128 o  r = 22cm

Rotational Motion and the Law of Gravity  Problem A propeller slows from 475rev/min to 187rev/min in 4.00s. What is its angular acceleration?

Rotational Motion and the Law of Gravity  Solution   i = 475rev/min   f = 187rev/min  t = 4.00s

Rotational Motion and the Law of Gravity Homework –Page 269 Problems –6 (4.1m) –8 (-1.2rad/s 2 ) –12 (41 rad/s 2 )

Rotational Motion and the Law of Gravity  Tangential Speed (v T )  Linear Velocity of an Object in a Circular Path  Units  m/s

Rotational Motion and the Law of Gravity  Tangential Acceleration (a T )  Linear Acceleration of an Object in a Circular Path  Units  m/s 2

Rotational Motion and the Law of Gravity  Centripetal Acceleration  Recall that for acceleration…

Rotational Motion and the Law of Gravity  Centripetal Acceleration  Velocity is a Vector  It has Magnitude and Direction  Either can Change the Velocity  So Using Rotational Kinematics

Rotational Motion and the Law of Gravity  Centripetal Acceleration (a c )  Of an Object on a Circular Path  Units  m/s 2

Rotational Motion and the Law of Gravity  Centripetal Acceleration (a c )  Direction  Always Toward the Center of Rotation

Rotational Motion and the Law of Gravity  Forces of Rotational Motion  Centripetal Force

Rotational Motion and the Law of Gravity  Forces of Rotational Motion  Centripetal Force

Rotational Motion and the Law of Gravity  Forces of Rotational Motion  Centripetal Force

Rotational Motion and the Law of Gravity  Forces of Rotational Motion  Centripetal Force  Direction  Always toward the Center of Rotation

Rotational Motion and the Law of Gravity  Problem A pebble breaks loose from the treads of a tire with a radius of 23cm. If the pebble’s angular speed is 213 rad/s, what is the tire’s tangential speed?

Rotational Motion and the Law of Gravity  Solution  r = 23cm   = 213rad/s

Rotational Motion and the Law of Gravity  Problem A washing machine’s two spin cycles are 328rev/min and 542rev/min. The diameter of the drum is 0.43m. What is the ratio of the centripetal accelerations for the fast and slow spin cycles?

Rotational Motion and the Law of Gravity  Solution   slow = 328rev/min   fast = 542rev/min  d= 0.43m (r = 0.215m)

Rotational Motion and the Law of Gravity  Problem A washing machine’s two spin cycles are 328rev/min and 542rev/min. The diameter of the drum is 0.43m. What is the ratio of the linear velocity of an object at the surface of the drum for the fast and slow spin cycles?

Rotational Motion and the Law of Gravity  Solution   slow = 328rev/min   fast = 542rev/min  d= 0.43m (r = 0.215m)

Rotational Motion and the Law of Gravity  Problem A laboratory ultracentrifuge is designed to produce a centripetal acceleration of 3.5x10 5 g at a distance of 2.50cm from the axis. What angular velocity in rev/min is required?

Rotational Motion and the Law of Gravity  Solution  a c = 3.5x10 5 g  r = 2.50cm

Rotational Motion and the Law of Gravity  Problem A bicycle wheel has an angular acceleration of 1.5rad/s 2. If a point on it’s rim has a tangential acceleration of 48cm/s 2, what is the radius of the wheel?

Rotational Motion and the Law of Gravity  Solution   = 1.5rad/s 2  a t = 48cm/s 2

Rotational Motion and the Law of Gravity  Problem Mr. Funchess twirls a set of keys in a circle at the end of a cord. If the keys have a centripetal acceleration of 145m/s 2 and the cord has a length of 0.34m, what is the tangential speed of the keys?

Rotational Motion and the Law of Gravity  Solution  r = 0.34m  a c = 145m/s 2

Rotational Motion Homework –Page Problems –22 (1.8x10 -3 rad/s ) –24 (0.023m) –26 (2.7m/s) –37 (a, 515kg b, 12.1m/s)

Rotational Motion and the Law of Gravity  Newton’s Law of Universal Gravitation  Newton found that the magnitude of the force, F, on a planet due to the Sun varies inversely with the square of the distance, r, between the centers of the planet and the Sun.  That is, F is proportional to 1/r 2. The force, F, acts in the direction of the line connecting the centers of the two objects.

Rotational Motion and the Law of Gravity  Newton’s Law of Universal Gravitation  The sight of a falling apple made Newton wonder if the force that caused the apple to fall might extend to the Moon, or even beyond.  He found that both the apple’s and the Moon’s accelerations agreed with the 1/r 2 relationship.

Rotational Motion and the Law of Gravity  Newton’s Law of Universal Gravitation  According to his own third law, the force Earth exerts on the apple is exactly the same as the force the apple exerts on Earth.  The force of attraction between two objects must be proportional to the objects’ masses, and is known as the gravitational force.

Rotational Motion and the Law of Gravity  Newton’s Law of Universal Gravitation  The law of universal gravitation states that objects attract other objects with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them.

Rotational Motion and the Law of Gravity  Newton’s Law of Universal Gravitation  The gravitational force is equal to the universal gravitational constant, times the mass of object 1, times the mass of object 2, divided by the square of the distance between the centers of the objects.

Rotational Motion and the Law of Gravity  Inverse Square Law  According to Newton’s equation, F is inversely related to the square of the distance.

Rotational Motion and the Law of Gravity  Universal Gravitation and Kepler’s Third Law  Newton stated his law of universal gravitation in terms that applied to the motion of planets about the Sun. This agreed with Kepler’s third law and confirmed that Newton’s law fit the best observations of the day.

Rotational Motion and the Law of Gravity  Universal Gravitation and Kepler’s Third Law  Consider a planet orbiting the Sun. Newton's second law of motion, F net = ma, can be written as F net = m p a c.  F net is the gravitational force

Rotational Motion and the Law of Gravity  Universal Gravitation and Kepler’s Third Law

Rotational Motion and the Law of Gravity  Universal Gravitation and Kepler’s Third Law  For an elliptical orbit, a c = 4π 2 r/T 2. This means that F net = m p a c may now be written as F net = m p 4π 2 r/T 2.  T is the time required for the planet to make one complete revolution about the Sun.  Period

Rotational Motion and the Law of Gravity  Universal Gravitation and Kepler’s Third Law

Rotational Motion and the Law of Gravity  Universal Gravitation and Kepler’s Third Law  The period of a planet orbiting the Sun can be expressed as follows.  The period of a planet orbiting the Sun is equal to 2 times the square root of the orbital radius cubed, divided by the product of the universal gravitational constant and the mass of the Sun.

Rotational Motion and the Law of Gravity  Universal Gravitation and Kepler’s Third Law  Cavendish’s Experiment  Often is called “weighing Earth,” because his experiment helped determine Earth’s mass. Once the value of G is known, not only the mass of Earth, but also the mass of the Sun can be determined.

Rotational Motion and the Law of Gravity  Why is “g” 9.8m/s 2 ?  Newton’s Law of Universal Gravitation  “G” = Universal Gravitational Constant  x N*m 2 /kg 2

Rotational Motion and the Law of Gravity  Weight  Equals Gravitational Force on Earth  Force Toward the Center of the Earth  (Center of Mass)

Rotational Motion and the Law of Gravity  Mass vs. Weight

Rotational Motion and the Law of Gravity  Universal Gravitation and Kepler’s Third Law  Cavendish’s Experiment

Gravitation  Problem Two balls have their centers 2.0m apart. One ball has a mass of 8.0kg. The other has a mass of 6.0kg. What is the gravitational force between them?

Gravitation  Solution  m 1 = 8.0kg  m 2 = 6.0kg  r = 2.0m

Gravitation  Problem Assume that you have a mass of 50.0 kg. Earth has a mass of 5.97x10 24 kg and a radius of 6.38x10 6 m. What is the force of gravitational attraction between you and Earth?

Gravitation  Solution  m 1 = 5.97x10 24 kg  m 2 = 50.0kg  r = 6.38x10 6 m

Gravitation  Problem Assume that you have a mass of 50.0 kg. Earth has a mass of 5.97x10 24 kg and a radius of 6.38x10 6 m. What is your weight?

Gravitation  Solution  m 1 = 5.97x10 24 kg  m 2 = 50.0kg  r = 6.38x10 6 m

Gravitation  Problem Two spheres are placed so that their centers are 2.6m apart. The force between the two spheres is 2.75x N. What is the mass of each sphere if one sphere is twice the mass of the other sphere?

Gravitation  Solution  F = 2.75x N  r = 2.6m

Rotational Motion Homework –Page Problems –39 (2.50m) –40 (0.10nm) –50