1301W.600 Lecture 18. November 15, PDF Free Download

1301W.600 Lecture 18 November 15, 2017

You are Cordially Invited to the Physics Open House Friday, November 17 th, 2017 4:30-8:00 PM Tate Hall, Room B20 Time to apply for a major? Consider Physics!! Program Information! Research Opportunities! Alumni Career Speakers! Lab Tours & Current Research Discussions! Faculty Speaker, Professor Vuk Mandic, will be discussing his work with LIGO, the most precise measuring tape in the world, which won the Nobel Prize in Physics this year! A free pizza dinner will be provided for participants!

A bit more about torques and stability

The center of gravity is the same as the center of mass. The center of gravity is the point at which all the torques due gravity vanish.

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Spinning Top There are many forms and shapes of spinning tops, and they are put into motion in an interesting variety of ways. What drives the motion of a Spinning Top?

The body of a top has at least one axis about which it will spin steadily and smoothly. This rotation axis is a symmetry axis of the top, known as a principal axis. For example, the red hoop in the figure below has two unique symmetry axes.

For each unique symmetry axis, the object has a moment of inertia value that determines how it will spin when a torque is applied. The object's spin about the rotation axis gives it an angular momentum, which will remain constant until some outside torque works on it.

An ideal Top Suppose a top is so perfectly fashioned that its principal rotation axis goes through its center of mass. Gravity can't exert a torque on the top about its point, the top will spin at a steady angular velocity almost indefinitely. Sliding friction between its tip and the floor does slow it gradually. But if the point is very sharp, sliding friction there exerts very little torque on the top about its rotational axis. The top will spin on until it slowly gets rid of its angular momentum through sliding friction and air resistance.

A more realistic Top A slight mismatch between the spin axis and the center of mass will guarantee that gravity exerts a torque on the top about its tip. The rapidly spinning top will precess in a direction determined by the torque exerted by its weight. The precession angular velocity is inversely proportional to the spin angular velocity, so that the precession is faster and more pronounced as the top slows down.

Spinning Top A rapidly spinning top will precess in a direction determined by the torque exerted by its weight. The precession angular velocity is inversely proportional to the spin angular velocity, so that the precession is faster and more pronounced as the top slows down.

A tippe Top A tippe top usually has a body shaped like a truncated sphere, with a short narrow stem attached perpendicular to the center of the flat circular surface of truncation.when a tippe top is spun at a high angular velocity, its stem slowly tilts downwards more and more until it suddenly lifts the body of the spinning top off the ground, with the stem now pointing downward. At first glance the top's inversion may mistakenly seem to be a situation where the object spontaneously gains overall energy. This is because the inversion of the top raises the object's center of mass, which means the potential energy has in fact increased. What causes the inversion (and the increase in potential energy) is a torque due to surface friction, which also decreases the kinetic energy of the top, so the total energy does not actually increase.

iclicker questions

iclicker question 1 A ball is released from rest on a no-slip surface (with friction), as shown in the figure. After reaching its lowest point, the ball begins to rise again, this time on a frictionless surface as shown in the figure. When the ball reaches its maximum height on the frictionless surface, it is A: at a greater height as when it was released. B: at a lesser height as when it was released. C: at the same height as when it was released.! D: It is impossible to tell without knowing the mass of the ball.

iclicker question 1 A ball is released from rest on a no-slip surface, as shown in the figure. After reaching its lowest point, the ball begins to rise again, this time on a frictionless surface as shown in the figure. When the ball reaches its maximum height on the frictionless surface, it is A: at a greater height as when it was released. B: at a lesser height as when it was released. C: at the same height as when it was released.! D: It is impossible to tell without knowing the mass of the ball.

iclicker question 2 A metal bar is hanging from a hook in the ceiling when it is suddenly struck by a ball that is moving horizontally (see figure). The ball is covered with glue, so it sticks to the bar. During this collision A: the angular momentum of the system (ball and bar) is conserved about the hook because only gravity is acting on the system. B: the angular momentum of the system (ball and bar) is not conserved because the hook exerts a force on the bar. C: the angular momentum of the system (ball and bar) is conserved about the hook because neither the hook nor gravity exerts any torque on this system about the hook. D: both the angular momentum of the system (ball and bar) and its kinetic energy are conserved.

iclicker question 3 You are using a wrench and trying to loosen a rusty nut. Which of the arrangements shown is most effective in loosening the nut? List in order of descending efficiency the following arrangements: A: 2,1=4,3 B: 2=4, 1,3 C: 1,2,3,4 D: 1,2,4,3

A: 2/3I iclicker question 4 Two unequal masses m and 2m are attached to a thin massless bar that rotates about an axis perpendicular to the bar. When m is a distance 2d from the axis and 2m is a distance d from the axis, the moment of inertia of this combination is I. If the masses are interchanged, what is the new moment of inertia? B: I C: 3/2I D: 2I

5 min break

Chapter 13: Gravity Chapter Goal: Develop the tools that allow us to understand the universal nature of gravitational interaction. This concept unifies the kinematics and dynamics of objects on Earth and celestial objects.

Universal gravity The force of gravity is an attraction between all objects that have mass, which is a measure of the quantity of material in an object. Mass is the property of an object that determines the strength of its gravitational interaction with other objects. In the late 17 th century, Isaac Newton stated that the gravitation force is a universal attractive force between all objects in the universe. The force that holds celestial bodies in orbit is the gravitational force. This is the same force that causes objects near Earth s surface to fall.

Newton s law of gravity Every point mass attracts every other point mass with a force proportional to each of the two masses, and inversely proportional to the square of the distance between them. The force acts along the line joining the centers of the two masses. ~F 12 = G m 1m 2 r 2 12 ~r 12 Universal gravity constant G =6.67 10 11 Nm 2 /kg 2

Gravitational potential energy The gravitational force exerted on an object 2 of mass m2 by object 1 of mass m1 is F G = G m 1 m 2 12x x 2 The work done by object 1 on object 2 can be written as W = x f x i F x (x)dx x 1 W = Gm m f 1 2 x i x 2 x dx = Gm 1 m 1 f 1 2 = Gm m 1 x 1 2 x x f i x i

Because no other energy associated with the system changes, energy conservation gives us, K = W Now if we consider the closed system of the two interacting objects, then energy conservation gives us, U G = K ΔU G 1 = Gm m 1 1 2 x x i f

If we let object 2 move from x = infinity to an arbitrary position x, we get ΔU G = U G (x) U G ( ) = U G (x) 0 = Gm 1 m 2 0 1 x U G (x) = G m 1 m 2 x gravitational potential energy All values of the potential energy are negative, approaching the value of zero as R approaches infinity

Principle of superposition Given a group of objects with masses, the total gravitational force acting on a chosen object is the sum of the gravitational forces exerted on that object by all the others. This is known as the Principle of superposition. ~F 2,net = F ~ 12 + F ~ 32 = G m 1m 2 (2d) 2 + Gm 2m 3 (d) 2 To calculate gravitational forces in more than one dimension, the force vectors must be broken up into their x, y, and z components, then the principle of superposition can be applied.

Example1: compute the gravitational force due to two equal masses at an equidistant point P. First of all by symmetry, the force at point P must point along the x-axis, so all we have to do is to compute the strength of the x-component of the gravitational force from one mass, and then double it.

Example: compute the gravitational force due to a ring of mass M and radius a.

Where g comes from? R E h

where g comes from F G = Gm E m o G m E m Gm o = m E Eo r 2 R 2 o R 2 Eo E E (near Earth s surface) However, we know from our study of gravitational force that F G = mg, mo and we get g = Gm E R E 2 (near Earth s surface)

Weighing Earth Cavendish is said to have weighed Earth because his determination of G provided the first value for the planet s mass me. Given that the radius of Earth is about 6400 km and given the value of G and g, determine me. ( G =6.67 10 11 Nm 2 /kg 2 ) g = Gm E R E 2 (near Earth s surface) m E = gr E 2 /G, m E = (9.8 m/s2 )(6.4 10 6 m) 2 (6.6738 10 11 N m 2 /kg 2 ) = 6.0 1024 kg.

Comparing gravitational pulls Compare the gravitational force exerted by Earth on you with (a) that exerted by a person standing 1 m away from you and (b) that exerted by Earth on Pluto. The average mass of the student is about 70 kg, and the masses of Earth and Pluto are m E =6 10 24 kg m P =1.3 10 22 kg The gravitational force between two objects with masses m1 and m2 separated by a distance r is proportional to the factor m1m2/r 2. Thus, to compare gravitational forces, we should compare these factors for each given situation.

You stand on the surface of Earth: m 1 m E R 2 E = (70 kg)(5.97 1024 kg) (6.38 10 6 m) 2 = 1.0 10 13 kg 2 / m 2.

Two 70-kg people separated by 1 m: m 1 m 2 r 2 12 = (70 kg)2 (1 m) 2 = 4.9 10 3 kg 2 / m 2. This gravitational force is (4.9 x10 3 kg 2 /m 2 )/(1.0 x10 13 kg 2 /m 2 ) = 4.9 x10 10 times the gravitational force exerted by Earth on you.

For Earth and Pluto m E m p r 2 EP = (5.97 1024 kg)(1.36 10 22 kg) (5.9 10 12 m) 2 = 2.3 10 21 kg 2 /m 2. This is (2.3 x10 21 kg 2 /m 2 )/(1.0 x10 13 kg 2 /m 2 ) = 2.3 x10 8 times greater than the attraction between Earth and you.

Inertial mass vs gravitational mass Inertial mass -causes objects to resist changes in their motion -Appears in Newton s second law: F=ma e.g. inertial mass prevented the incoming ship from stopping in time Gravitational mass -responsible for the gravitational force -Appears in Newton's law of gravity: FG= Gm1m2/r12 e.g. gravitational mass of containers makes them be attracted towards the Earth (and thus falling)

Weight and apparent weightlessness The weight of an object as the magnitude of the force of gravity acting on it. Why do astronauts on the space shuttle feel apparent weightlessness?

Because the shuttle + astronauts are always in free fall. Space shuttle orbiting around Earth Any object in free fall that is, any object subject to only a force of gravity experiences weightlessness.

If you were to travel in a vertical circle, at what point in the circle would you be most likely to experience weightlessness? To experience weightlessness, the only force acting on you is the gravitational force exerted by Earth, which points downward. This downward force provides the centripetal acceleration needed for traveling in a circle. The only position from which the center of the circle is downward is the top of the circle.

Now let s see what happened when you ride an elevator.

What must the elevator s acceleration be for an individual to experience weightlessness in an elevator? If a person experiences weightlessness, the only force acting on her or him is the gravitational force. When you stand on a scale, the scale s reading is a measure of your weight. It is actually recording the Normal force the scale has to exert to support your weight.

ΣF = ma According to Newton s second law, the forces acting on you are: ΣF = N mg N mg = ma N = ma + mg N = m(a + g) The scale reads the Normal force, so it reports your weight as greater than it was at rest.

ΣF = -ma According to Newton s second law, the forces acting on you are: ΣF = N mg N mg = -ma N = mg ma N = m(g a) The scale will read the Normal force which is less than it was at rest. You will experience weightlessness when a=g. This is not surprising: If the person is in free fall, the elevator must also be in free fall.

iclicker question 5 You stand on a spring scale placed on the ground and read your weight from the dial. You then take the scale into an elevator. Does the dial reading increase, decrease, or stay the same when the elevator accelerates downward as it moves upward? A: Increase B: Decrease C: Stay the same

thank you

1301W.600 Lecture 18. November 15, 2017