Slideshow : Chapter 7 Gravitation

Chapter 7 gravitation

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Newton's Three Laws of Motion & WTC7

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VideoBrief: Newton's Laws of Motion illustrated with 3D animations and m...

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Gravitation

Newton ‘s law of gravitation states that the force of attraction between particles is directly proportional to their mass and inversely proportional to the square of distance apart.

                        

                Where G is the universal gravitational constant,

7.2 GRAVITATIONAL FIELD STRENGTH

•      A gravitational field is a region where gravitational force acts on massive bodies. Eg . gravitaional field of the Earth.

•      The gravitational field strength tells us how strong a gravitational field is. The gravitational field strength of the Earth near its surface is 9.81m / s².

•      The gravitational field strength , E at a point is the force of gravity per unit mass exerted on a mass placed.

The variation of the acceleration due to gravity g’ with distance r from the centre of the Earth is illustrated by the graph above.

7.3 Gravitational Potential

•      The strength of the gravitational force at a point in a gravitational field is described by the gravitational field strength E or g is a vector quantity.

•      Another quantity associated with the point in the gravitational field is the gravitational potential. It is a scalar quantity.

•      The gravitational potential V at a point P in a gravitational field is defined as the work done per unit mass to bring a body from infinity to P. The unit for gravitational potential is Jkgˉ¹.

•      The gravitational potential energy U of a body at a point P in a gravitational field is defined as the work done to bring the body from infinity to P. The unit for gravitational potential energy is J.

•      Hence the gravitational potential energy U of a body of mass m at a point where the gravitational potential, V  is given by

                               U = mV

•      On the surface on the Earth, r = R

–    Gravitational potential, V = -

–    Gravitational potential energy, U = -

–    The graph illustrates the variation of the gravitational potential V with distance r from the centre of the Earth.

                                                         

                   

7.4 RELATIONSHIP BETWEEN g AND G

-G is the Universal Gravitational Constant.

- It is a scalar quantity with dimension

             

- g  is the acceleration due to gravity .

- It is a vector quantity with dimension

                    

                   

              

 





Where g = acceleration due to gravity

             R = constant radius of earth

             G = universal gravitational constant

             M = mass of Earth

7.5 Satellite Motion in Circular Orbits

Satellite is a body that revolves round a planet. Satellites can be categorized as natural satellites or man-made satellites. The moon, the planets and comets are examples of natural satellites. Examples of man-made satellites are Sputnik I , Measat I ,II and III which are   communication satellites. In order to launch satellite into orbit , rockets are used. When rocket that carries the satellite reaches the required height , the satellite is launched into circular orbit with a certain velocity v that is tangential to intended orbit.

                         

                                               

                                            

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Frictional Forces

Friction is the force opposing the motion of one body sliding or rolling over the surface of second object. Several aspects of friction are important at low velocities:

• The direction of the force of friction is opposite the direction of motion.
  
• The frictional force is proportional to the perpendicular (normal) force between the two surfaces in contact.
  
• The frictional force is nearly independent of the area of contact between the two objects.
  
• The magnitude of the frictional force depends on the materials composing the two objects in contact.
  

Static friction is the force of friction when there is no relative motion between two objects in contact, such as a block sitting on an inclined plane. 

The magnitude of the frictional force is F_s ≤ μ_s N, where N is the magnitude of the normal force, and the coefficient of friction (μ_s) is the dimensionless proportionality constant. 

The coefficient of friction may be approximately .2 for normal lubricated surfaces and close to 1 for glass sliding on glass. This equation sets the upper limit for the static frictional force. If a greater external force is applied, the situation will no longer be static, and the object will begin to move.

 

Kinetic friction is the force of friction when there is relative motion between two objects in contact. 

The magnitude of the friction force in this case is F _k ≤ μ_k N, where N is the magnitude of the normal force and μ _k is the coefficient of kinetic friction. Note that μ _k is not strictly a constant, but this empirical rule is a good approximation for finding frictional forces. Values given for the coefficients of static and kinetic friction do vary with speed and surface conditions so that it is not necessarily true that static friction exceeds sliding friction.

The following problem highlights the differences between static and kinetic friction.

 

Example 6: A block sits on an inclined plane. What is the maximum angle that which the block remains at rest? First, draw the free-body diagrams and then write out the force equation for each direction of the coordinate system (see Figure 6 ).

Figure 6 A block on an inclined plane, with friction.

Suppose the surface is tilted to θ, at which the block just begins to move. Then, the force down the plane must be equal to the maximum force of static friction; thus f _friction = μ_s N. Therefore, f _friction = mg sin θ = μ_s N = μ_s ( mg cos θ) and solving for the coefficient of friction:

 

Solution: μ_s = tan θ

At a greater angle of tilt, the object accelerates down the surface, and the force of friction is f _k = μ _k N.

 

Example 7: If the surfaces in Figures 4 and 5 were not frictionless, the frictional force parallel to the surface and opposite the direction of motion must be included in the analysis. Pulling a block along a horizontal surface at a constant speed (zero acceleration) is an example of a problem involving friction, and such a block is analyzed in Figure 7 .

Figure 7 Pulling a block on a plane with friction.

In the x direction, T cos θ − f = 0, where f is the friction force and T is the tension in the rope. In the y direction, N + T sin θ − mg = 0; also, f = μ_k N.

Solve the y direction equation for N, substitute the expression into the friction force equation, and then substitute friction into the first equation to obtain the following:

Solution: solving for T.

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Equilibrium of a Rigid Body

The Moment of a Force

The moment M (turning effect) of a force about a point O is the product of the magnitude of the force (F) and the perp. distance (x)to the point of application.

By convention, anti-clockwise moments are positive.

The Principle of Moments

For a rigid body acted upon by a system of coplanar forces, equilibrium is achieved when:

i)  the vector sum of the coplanar forces = 0

ii) there is no net turning effect produced by the forces
(the sum of clockwise & anti-clockwise moments = 0)

Parallel forces acting on a beam

When attempting problems concerning a balance points or fulcrum, remember that there is always an upward force acting.

Example #1   fulcrum near centre

Consider two forces F₁ and F₂ acting vertically downwards at either end of a beam of negligible mass.

When the beam is balanced at its fulcrum O(i.e. horizontal), the sum of the downward forces equals the sum of the upward forces.

If the reaction force at the fulcrum is R,

Example #2   fulcrum at one end

Consider two forces F₁ and F₂ acting on a beam of negligible mass. One force acts vertically downwards near the centre, while the other acts vertically upwards at the end.

When the beam is balanced, the sum of the downward forces equals the sum of the upward forces.

If the reaction force at the fulcrum is R,

Example #3    a typical problem

A beam of negligible weight is horizontal in equilibrium, with forces acting upon it, as shown in the diagram.
Calculate the value of the weights R and W.

Couples

The turning effect of two equal and opposite parallel forces acting about a point equals the product of one force(F) and the perpendicular distance between the forces(d).

couple = F x d

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Equilibrium of Particles

Equilibrium equations for a particle: 

A particle is in equilibrium if the resultant of ALL forces acting on the particle is equal to zero

Equilibrium equations in component form: 

In a rectangular coordinate system the equilibrium equations can be represented by three scalar equations:

 

Triangle of Forces

When 3 coplanar forces acting at a point are in equilibrium, they can be represented in magnitude and direction by the adjacent sides of a triangle taken in order.

Example

Using the results from the previous example, the three forces acting on the 2 kg mass can be represented by a scale diagram.

Our starting point is the 20N force acting downwards. One force acts at 45 deg. to this line and the other at 60 deg. So to find the magnitude of the two forces, draw lines at these angles at each end of the 20N force. Where the lines cross gives a vertex of the triangle. Measuring the lengths of the lines from this to the ends of the 20N force line will give the magnitudes of the required forces.

Polygon of Forces

For equilibrium, forces are represented in magnitude and direct to form a polygon shape.

If a number of forces are acting at a point, then the missing side in the polygon represents the resultant force. Note the arrow direction on this force is in the opposite direction to the rest.

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Rotational Kinetic Energy, Work and Power

Kinetic energy, work, and power are defined in rotational terms as K.  

E=(1/2) Iω², W= tθ, P= tω.

 

Comparison of dynamics equation for linear and rotational motion. 

The dynamic relations are given to compare the equation for linear and rotational motion (see Table 1 ).

TABLE 1

Comparison of Dynamics Equations for Linear and Rotational Motion

	Linear Motion	Rotational Motion
Newton's second law	F = ma	t = Iα
Momentum	p = mv	L = Iω
Work	W= FΔ x or W = ∫ F·dx	W= tΔθ or W = ∫ t·dθ
Kinetic energy	K· E·=½ mv2	K· E·=½ Iω2

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Torque and Angular Momentum

Torque

It is easier to open a door by pushing on the edge farthest from the hinges than by pushing in the middle. It is intuitive that the magnitude of the force applied and the distance from the point of application to the hinge affect the tendency of the door to rotate. This physical quantity, torque, is t = r × F sin θ, where F is the force applied, r is the distance from the point of application to the center of the rotation, and θ is the angle from r to F.

Angular Momentum

Angular momentum is rotational momentum that is conserved in the same way that linear momentum is conserved. 

For a rigid body, the angular momentum (L) is the product of the moment of inertia and the angular velocity: L = Iω. 

For a point of mass, angular momentum can be expressed as the product of linear momentum and the radius ( r): L = mvr. L is measured in units of kilograms-meters² per second or more commonly joule-seconds.

The law of conservation of angular momentum can be stated that the angular momentum of a system of objects is conserved if there is not external net torque acting on the system.

Analogous to Newton's law (F = Δ( mv)/Δ t) there is a rotational counterpart for rotational motion: t = Δ L/Δ t, or torque is the rate of change of angular momentum. 

Consider the example of a child who runs tangential to the edge of a playground merry-go-round with a velocity v_o and jumps on while the merry-go-round is at rest. 

The only external forces are that of gravity and the contact forces provided by the support bearings, neither of which causes a torque because they are not applied to cause a horizontal rotation. 

Treat the child's mass as a point of mass and the merry-go-round as a disc with a radius R and mass M. 

From the conservation law, the total angular momentum of the child before the interaction is equal to the total angular momentum of the child and merry-go-round after the collision: mrv_o = mrv′ + Iω, where r is the radial distance from the center of the merry-go-round to the place where the child hits. If the child jumps on the edge, (r = R) and the angular velocity for the child after the collision can be substituted for the linear velocity, mRv_o = mR( Rω)+(1/2) MR². If the values for the masses and the initial velocity of the child are given, the final velocity of the child and merry-go-round can be calculated.

A single object may have a change in angular velocity due to the conservation of angular momentum if the distribution of the mass of the rigid body is altered. 

For example, when a figure skater pulls in her extended arms, her moment of inertia will decrease, causing an increase in angular velocity. According to the conservation of angular momentum, I _o (ω _o ) = I _f (ω _f ) where I _o is the moment of inertia of the skater with arms extended, I _f is her moment of inertia with her arms close to her body, ω_o is her original angular velocity, and ω _f is her final angular velocity.

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Moment of Inertia

Substitute Newton's second law into the definition for torque with θ of 90 degrees (a right angle between F and r) and use the relationship between linear acceleration and tangential angular acceleration to obtain t = r F = rma = mr² ( a/ r) = mr²α. The quantity mr² is defined as moment of inertia of a point mass about the center of rotation. 

Imagine two objects of the same mass with different distribution of that mass. The first object might be a heavy ring supported by struts on an axle like a flywheel. The second object could have its mass close to the central axis. Even though the masses of the two objects are equal, it is intuitive that the flywheel will be more difficult to push to a high number of revolutions per second because not only the amount of mass but also the distribution of the mass affects the ease in initiating rotation for a rigid body. The general definition of moment of inertia, also called rotational inertia, for a rigid body is I = ∑ m_i r_i² and is measured in SI units of kilogram-meters². 

The moments of inertia for different regular shapes are shown in Figure 2 .

Figure 2  Moments of inertia for various regular shapes.

Mechanics problems frequently include both linear and rotation motions.

Example 1: Consider Figure 3 , where a mass is hanging from a rope wrapped around a pulley. The falling mass (m) causes the pulley to rotate, and it is no longer necessary to require the pulley to be massless. Assign mass ( M) to the pulley and treat it as a rotating disc with radius (R). What is the acceleration of the falling mass, and what is the tension of the rope?

Figure 3  A hanging mass spins a pulley.

The force equation for the falling mass is T − mg = − ma. The tension of the rope is the applied force to the edge of the pulley that is causing it to rotate. Thus, t = Iα, or TR = (1/2) MR²(^a/R), which reduces to T = (1/2) Ma, where angular acceleration has been replaced by^a/R because the cord does not slip and the linear acceleration of the block is equal to the linear acceleration of the rim of the disk. Combining the first and last equation in this example leads to

Solution:

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Rotational Kinematics

Rotational motion is more complicated than linear motion, and only the motion of rigid bodies will be considered here. A rigid body is an object with a mass that holds a rigid shape, such as a phonograph turntable, in contrast to the sun, which is a ball of gas. Many of the equations for the mechanics of rotating objects are similar to the motion equations for linear motion. 

Angular velocity and angular acceleration

The angular displacement of a rotating wheel is the angle between the radius at the beginning and the end of a given time interval. The SI units are radians. The average angular velocity (ω, Greek letter omega), measured in radians per second, is

The angular acceleration (α, Greek letter alpha) has the same form as the linear quantity

and is measured in radians/second/second or rad/s².

The kinematics equations for rotational motion at constant angular acceleration are

ω f = ω 0 α t Angular velocity as a function of time Angular displacement as a function of velocity and time Angular displacement as a function of time ω f 2 = ω 0 2 + 2αθ Angular velocity as a function of displacement

Consider a wheel rolling without slipping in a straight line. The forward displacement of the wheel is equal to the linear displacement of a point fixed on the rim. As can be shown in Figure 1 , d = S = rθ

Figure 1 A wheel rolling without slipping.

In this case, the average forward speed of the wheel is v = d/ t = ( rθ)/ t = rω, where r is the distance from the center of rotation to the point of the calculated velocity. The direction of the velocity is tangent to the path of the point of rotation.

The average forward acceleration of the wheel is a_T = r(ω f − ω o)/ t = rα. This component of the acceleration is tangential to the point of rotation and represents the changing speed of the object. The direction is the same as the velocity vector.

The radial component of the linear acceleration is a_r = v²/ r = ω² r.

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Centripetal Force

Centripetal force is a force that makes a body follow a curved path: it is always directed orthogonal to the velocity of the body, toward the instantaneous center of curvature of the path.

In simple terms, centripetal force is defined as a force which keeps a body moving with a uniform speed along a circular path and is directed along the radius towards the centre. The mathematical description was derived in 1659 by Dutch physicist Christiaan Huygens. Isaac Newton's description was: "A centripetal force is that by which bodies are drawn or impelled, or in any way tend, towards a point as to a centre."

The magnitude of the centripetal force on an object of mass m moving at a speed v along a path with radius of curvature r is:

where is the centripetal acceleration. The direction of the force is toward the center of the circle in which the object is moving, or the osculating circle, the circle that best fits the local path of the object, if the path is not circular. This force is also sometimes written in terms of the angular velocity ω of the object about the center of the circle:

Expressed using the period for one revolution of the circle, T, the equation becomes:

Mass performing vertical circular motion under gravity

Consider a mass m performing circular motion under gravity, the circle with radius r .
The centripetal force on the mass varies at different positions on the circle.

Conical pendulum

Problems concerning the conical pendulum assume no air resistance and that the string has no mass and cannot be stretched.
Solution of problems involves resolving forces on the mass vertically and horizontally. In this way the speed of the mass, the tension in the string and the period of revolution can be ascertained. 

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Uniform Circular Motion

In physics, uniform circular motion describes the motion of a body traversing a circular path at constant speed. The distance of the body from the axis of rotation remains constant at all times. Though the body's speed is constant, its velocity is not constant: velocity, a vector quantity, depends on both the body's speed and its direction of travel. This changing velocity indicates the presence of an acceleration; this centripetal acceleration is of constant magnitude and directed at all times towards the axis of rotation. This acceleration is, in turn, produced by a centripetal force which is also constant in magnitude and directed towards the axis of rotation.

In the case of rotation around a fixed axis of a rigid body that is not negligibly small compared to the radius of the path, each particle of the body describes a uniform circular motion with the same angular velocity, but with velocity and acceleration varying with the position with respect to the axis.



Figure above illustrates velocity and acceleration vectors for uniform motion at four different points in the orbit. Because the velocity v is tangent to the circular path, no two velocities point in the same direction. Although the object has a constant speed, its direction is always changing. This change in velocity is caused by an acceleration a, whose magnitude is (like that of the velocity) held constant, but whose direction also is always changing. The acceleration points radially inwards (centripetally) and is perpendicular to the velocity. This acceleration is known as centripetal acceleration.
For a path of radius r, when an angle θ is swept out, the distance traveled on the periphery of the orbit is s = rθ. Therefore, the speed of travel around the orbit is

,

where the angular rate of rotation is ω. (By rearrangement, ω = v/r.) Thus, v is a constant, and the velocity vector v also rotates with constant magnitude v, at the same angular rate ω.



The left-hand circle in Figure above is the orbit showing the velocity vectors at two adjacent times. On the right, these two velocities are moved so their tails coincide. Because speed is constant, the velocity vectors on the right sweep out a circle as time advances. For a swept angle dθ = ω dt the change in v is a vector at right angles to v and of magnitude v dθ, which in turn means that the magnitude of the acceleration is given by

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