IB Physics/Motion in Fields

9.1.1
On an inclined plane, application of Newton's laws is somewhat complicated. Gravity acts downwards, but the plane exerts a force back on the object at an angle. This can be used to build a right angled triangle, since the resultant force down the slope angles to the normal force, and gravity will be the hypotenuse. If there is friction acting, this will be in the opposite direction to the motion down the plane.

9.1.2
Friction is a force which opposes motion, thus if there is no motion, then there will be no force caused by friction. Friction can never make an object move, it can only slow an object down, and eventually stop it.

For two solid surfaces moving over each other, the friction will be affected by the nature (roughness etc) of the two surfaces, however the surface area and velocity of the object do not affect the friction. There are also two types of friction of solid surfaces, static and kinetic friction. Static friction is that which stops objects from beginning to move, and kinetic is that which slows objects down when they are moving.

These two types are defined individually by their constants &micro;s and &micro;k respectively. &micro;s > &micro;k in all cases (fairly obvious if you think about it).

The force of friction caused by each type is also dependent on the normal force the surface is applying, thus Ffr =< &micro;sFn for objects which aren't moving, and Ffr = &micro;<SUB>s</SUB>F<SUB>n</SUB> for objects which are moving. In the first case, the frictional force only exists if there is a force being applied, thus the less than or equal to sign. Friction will counteract all forces up to this point.

Objects falling through fluid (or indeed, air) also experience friction which opposes the force of gravity. This friction increases with velocity, and so there eventually comes a point where the force of gravity is being counteracted by friction, and the object falls at constant speed. This is known as terminal velocity.

9.2.1
When objects in projectile motion are launched at an angle, the horizontal and vertical components must first be calculated. From the vertical, we should be able to calculate its peak height, and the time taken to reach this height. From there, we can calculate the time required to reach the ground and using all these times, and the horizontal component, the horizontal distance travelled.

9.3.1
Simple harmonic motion is like that of a pendulum, where the object moves away from, and returns to the equilibrium point, and the restoring force is proportional to the extension (i.e. the further from the center the pendulum is, the bigger the force pulling it back). Starting from a fully extended position, displacement follows a cos curve, velocity a -sin curve, and acceleration a -cos curve (we're differentiating with respect to t each time : cos -> -sin -> -cos). As can be seen from this, a displacement vs acceleration graph will be a straight line with a negative slope (cos and -cos).

9.3.2
The period in simple harmonic motion (SHM) is the time required for the displacement to return to its original position (complete one cycle). The frequency is <SUP>1</SUP>/<SUB>period</SUB>, thus the number of cycles per second. The amplitude is the maximum displacement from the mean (or central) position. Period is also defined as T = <SUP>(2 x Pi)</SUP>/<SUB>w</SUB> where w = 2 x &pi; x f

Graphing displacement vs time : This graph ranges from r (the amplitude) to -r, and follows a cos curve (assuming we're starting at the extended position). It will be a maximum at the points of extension and zero when the object passes through the mean position.

Graphing velocity vs time : This graph ranges from rw to -rw (where w = 2 x &pi; x f ). Velocity will be a maximum when the object is an the mean position, and zero when the extension is at a maximum. Thus it will follow a sin curve, with velocity being a maximum and when displacement is zero and vice versa.

9.3.3
The total energy for an object in SHM is constant. When displacement is maximum, all the energy is potential (since it is being held up, compressed or whatever). When the displacement is zero, all the energy is kinetic (because it is moving, and at its lowest point/no compression or extension etc.), thus the two follow parabolic curves where the sum of the two curves is always equal.

9.3.4
As an example, we have SHM for a mass on a spring. When the mass is pulled downwards to an extension of x, and then released, the mass undergoes vertical oscillations which obey SHM.

Since the spring obeys Hooke's law ( F = kx, where k is the spring constant ) we know that the restoring force = kx. Newton's second law. We also have F = ma, so combining the two we get a = <SUP>kx</SUP>/<SUB>m</SUB>. This fits the SHM formula a = -w<SUP>2</SUP>x producing a = -<SUP>k</SUP>/<SUB>m</SUB> x (if acceleration is in the same direction as x, i.e. away from mean position) Therefore, w<SUP>2</SUP> = <SUP>k</SUP>/<SUB>m</SUB> and so, T = 2 x &pi; x (&radic;(<SUP>m</SUP>/<SUB>k</SUB>)) since T = <SUP>(2 x &pi;)</SUP>/<SUB>w</SUB> (this is as given in the data book).

If the elastic limit of the spring is exceeded, then it will no longer follow Hooke's law, and no longer follow SHM.

9.3.5
Assume a mass is on a pendulum, with string length l, and is displaced by x. This forms an angle of &Oslash; between the mean position of the string, and the displaced position. The force of gravity towards the origin = mg sin&Oslash; ( this can be shown with a right angle triangle, mg is the hypotenuse, mg sin&Oslash; is the side at a tangent to the curve, because it is a similar triangle to the one made by the mass, the origin and the top of the string ). For small angles &Oslash;, sin &Oslash; = <SUP>x</SUP>/<SUB>l</SUB>(If we assume once again, that the curve between the origin and the mass is actually a straight line, which is valid for small &Oslash;). Thus, the force towards O = <SUP>mgx</SUP>/<SUB>l</SUB>. Since F = ma, acceleration towards O = <SUP>gx</SUP>/<SUB>l</SUB>. Thus acceleration in x direction (away from O) = -<SUP>gx</SUP>/<SUB>l</SUB>. So as above, w<SUP>2</SUP> = <SUP>g</SUP>/<SUB>l</SUB>, and so, T = 2 x Pi x (square root of <SUP>l</SUP>/<SUB>g</SUB>) (once again, this is given in the data book).

9.4.1
Angular displacement : The angle from the centre of a circle (in radians) around which an object has moved. It's symbol is &Oslash;, and it is defined as &Oslash; = <SUP>s</SUP>/<SUB>r</SUB>where s is the length travelled around the circumference, and r is the radius. It is equivalent to displacement in linear motion.

Angular velocity : Uses the symbol w. <SUP>angular displacement</SUP>/<SUB>time</SUB> (or w = <SUP>&Delta;&Oslash;</SUP>/<SUB>&Delta;t</SUB>). Its linear equivalent is velocity. This is no longer included in the syllabus.

9.4.2
s = r&Oslash; : The length covered on the circumference is the radius x the angle (in radians) covered. This allows the angular displacement to be converted into a length.

v = rw : The velocity of an object on the circumference is the radius times the angular velocity (as defined above, in radians s<SUP>-1</SUP>)

9.4.3
That an object rotating in a circle has an acceleration towards the centre can be shown by taking two vectors at tangents to different points on the circle, and subtracting one from the other. The resulting vector will be directed towards the centre of the circle.

If we take two such vectors (of length V) from close together, so the angular displacement of their starting points is small, and draw them from the same origin point, then the vector between their ends is the centripetal acceleration. Using similar triangles, take the angle between the two vectors to be &Oslash; (the same as the angular displacement), and let the vector between them be &Delta;V. If a<SUB>c</SUB> is centripetal acceleration, then it equals <SUP>&Delta;V</SUP>/<SUB>&Delta;t</SUB>. From the triangle, &Oslash; = <SUP>&Delta;V</SUP>/<SUB>V</SUB>, so &Delta;V = V&Oslash;. Subbing into the acceleration formula, we get a<SUB>c</SUB> = <SUP>V&Oslash;</SUP>/<SUB>&Delta;t</SUB>. From above, w = <SUP>&Oslash;</SUP>/<SUB>&Delta;t</SUB>, thus &Oslash; = wt, and so a<SUB>c</SUB> = vw. This can be rearranged by subbing v = rw to get a<SUB>c</SUB> = rw<SUP>2</SUP> = V<SUP>2</SUP>/<SUB>r</SUB> (as in the data book).

9.4.4
For an object to travel in a circular path, constant acceleration is required. F=ma (newton's second law) in circular motion F = |ma<SUB>c</SUB>| = |m v<SUP>2</SUP>/<SUB>r</SUB>| = |4 x &pi;<SUP>2</SUP> x <SUP>mr</SUP>/<SUB>T</SUB>2| = |mrw<SUP>2</SUP>|.

9.4.5
Bodies in circular motion include planets orbiting ( gravity provides a<SUB>c</SUB> ), buckets being spun around on ropes ( rope provides a<SUB>c</SUB> ) and cars going around banked curves ( The 'down the slope' component of the normal force provides a<SUB>c</SUB> ). Problems involving these can then be solved by finding a<SUB>c</SUB> and then working back through the above equations.

9.4.6
Masses moving in a vertical circle and affected by gravity : At every point in the circle, the force keeping the object moving in a circle (ie towards the center) must be constant, and can be calculated with the equations above. At the top of the circle, the force total keeping the object in a circle is F<SUB>c</SUB> - F<SUB>g</SUB> (since gravity is providing the force, Fc is smaller). At the bottom, F<SUB>t</SUB> = F<SUB>c</SUB> + F<SUB>g</SUB>. At the sides, F<SUB>t</SUB> = F<SUB>c</SUB> (since no component of gravity is acting towards the center).

9.4.7
Centripetal force does not change the kinetic energy of a body because it only affects the direction of the velocity, not the magnitude, and E<SUB>k</SUB> = <SUP>1</SUP>/<SUB>2</SUB>mv<SUP>2</SUP> is only conserved with the magnitude.

9.5.1
Newton's law of universal gravitation : F = (Gm<SUB>1</SUB>m<SUB>2</SUB>) / (R<SUP>2</SUP>), where G is the gravitational constant (on earth, 6.67 x 10<SUP>-11</SUP> N m<SUP>2</SUP> kg<SUP>-2</SUP>), m<SUB>1</SUB> and m<SUB>2</SUB> are the two masses in question, and R is the distance between the centers of the two masses. This must be applied both as a source of centripetal and gravitational force (in the pulling objects down sense). (This equation is in the data book)

9.5.2
Gravitational field strength is the gravitational force per unit mass at a particular point in the gravitational field. In simple terms, it's the force on one kg of mass, which is 9.8 N on Earth.

9.5.3
Beyond the surface of the mass, the force of gravity decreases in a parabolic curve (because of the R<SUP>2</SUP> term it obeys the inverse square law). Inside the mass is messy to work out, but luckily it isn't required.

9.5.4
Gravitational potential ( with the symbol V ) is defined as the potential energy per unit mass at a point in the field. V approaches 0 as the distance between the masses approaches infinity. V = -G<SUP>m</SUP>/<SUB>r</SUB> where m is the mass of the earth (or whatever) and r is the distance from the center of the Earth. (given in the data book)

9.5.5
Escape velocity is the speed at which an object must leave the planet's surface to completely escape its gravitational field. For this, the object must be given enough kinetic energy to go from the surface, where
 * $$V=-G\frac{M}{r}m$$

to where V = 0. Therefore,


 * $$\frac{1}{2}mv^2 - G\frac{M}{r}m = 0$$

Which rearranges to
 * $$v = \sqrt{2G\frac{M}{r}}$$

Note: this one is not in the data book, so you can either memorize it, or remember how to derive it.

9.6.1
The work done on a force displacement graph is the area under the graph. This would normally be found by integrating, but since there's no calculus in this course, in any given question you'll be able to break the area up into triangles and rectangles to find the area.

9.6.2
For a linear spring, the force to extend (or compress) the spring is directly proportional to the displacement. If no force acts on it, the spring will naturally return to its equilibrium position. Unless, of course, the spring is displaced beyond it's elastic, in which case it's broken and probably no longer interesting.

9.6.3
Energy (elastic potential) is stored in a spring by compressing or extending it. When released, the spring will convert this energy into other forms (kinetic, thermal, sound etc.).

9.6.4
Deriving conservation of momentum for two bodies from Newton's laws.

First assume there are two objects, of mass m<SUB>1</SUB> and m<SUB>2</SUB> travelling towards each other at velocities v<SUB>1</SUB> and v<SUB>2</SUB>. First we write Newton's second law in the form 'force is the rate of change of momentum'

$$F =\frac{\Delta P}{\Delta t}$$

$$\begin{align} \therefore F\Delta t &= \Delta P \\ &= mv \end{align} $$

This can then be applied to the first mass giving the equation Ft = m<SUB>1</SUB>v'<SUB>1</SUB> - m<SUB>1</SUB>v<SUB>1</SUB>. Using Newton's third law, we know that the force on m<SUB>2</SUB> will be equal and opposite and so the second mass. -Ft = m<SUB>2</SUB>v'<SUB>2</SUB> - m<SUB>2</SUB>v<SUB>2</SUB>.

These two can then be combined to form m<SUB>1</SUB>v'<SUB>1</SUB> - m<SUB>1</SUB>v<SUB>1</SUB> = -(m<SUB>2</SUB>v'<SUB>2</SUB> - m<SUB>2</SUB>v<SUB>2</SUB>) which rearranges to m<SUB>1</SUB>v<SUB>1</SUB> + m<SUB>2</SUB>v<SUB>2</SUB> = m<SUB>1</SUB>v'<SUB>1</SUB> + m<SUB>2</SUB>v'<SUB>2</SUB>, which is the law of conservation of momentum for two masses.

9.6.5
In two dimensions, the law of conservation of momentum applies in both senses, so the momentum value must be broken up into its vector components. The total amount of vertical (or north/south or whatever) momentum will be conserved, and will the total amount of horizontal momentum. Therefore, a problem involving this can be treated as two separate one dimensional conservation of momentum problems.

9.7.1
Angular acceleration is the rate of change of angular velocity : &alpha; = <SUP>&Delta;w</SUP>/<SUB>&Delta;t</SUB>. This is no longer included in the syllabus.

Torque is the rotational equivalent of force : T = I x &alpha; (moment of inertia x angular acceleration).

Moment of inertia is the rotational equivalent of mass. It is defined as the sum of (mr<SUP>2</SUP>) for every point mass in the system where m = mass and r = the distance form the axis of rotation. This allows dumbbell type arrangements to be calculated, but the following shape formulae should also be known
 * A 'hoop' or cylinder rotating around the center of the hoop : I = MR<SUP>2</SUP>
 * A solid disk rotating around the center of the disk : I = <SUP>1</SUP>/<SUB>2</SUB>MR<SUP>2</SUP>
 * A solid sphere rotating about any axis running through the centre : I = <SUP>2</SUP>/<SUB>5</SUB>MR<SUP>2</SUP>

Angular momentum is the equivalent of linear momentum : L = Iw (angular momentum = moment of inertia x angular velocity).

9.7.2
All of the quantities and equations in rotational motion are equivalent to others in linear motion. The tables below illustrate this equivalence.

All these can be applied as appropriate to problems.