The relationship between displacement, velocity and acceleration is summarised in the following diagram.

Given the displacement, to find the velocity, we differentiate. To find the acceleration, we differentiate twice.

Given the acceleration, to find the velocity, we integrate. To find the displacement, we integrate twice.

Example

A particle P moves on the- axis. At time t seconds, its acceleration isWhen its velocity isWhenthe displacement is 2m. Find expressions in terms of for the velocity and displacement.

When

When

Example:

The displacement of a particle isFind the velocity and acceleration in terms of t. When is the particle at rest for? What are the displacement and acceleration at this time?

When the particle is at rest,

The particle is at rest whenor

The displacement whenis

The acceleration whenis

]]>In two dimensions the displacements in theanddirections are generally expressed as functions ofwith thecomponent labelled withand thecomponent labelled with

If

then velocity is found by differentiation of (the components of) displacement:and the acceleration is found by differentiation of the components of velocity:

Conversely, given the acceleration, the velocity is found by integration, and the displacement is found by integrating the velocity.

Ifwiththen

Then if

]]>Resolving vertically gives

Subtracting these, and using thatgives

Then

Taking moments about A gives

so the road will swing clockwise about A and cannot be in equiliubrium.

It is however, possible for the system to be in equilibrium if the rod is non – uniform.

]]>The horizontal component of velocity is {jatex options:inline}v_x = u cos \: \alpha{/jatex}

The angle the projectile makes with the horizontal is {jatex options:inline}tan \beta = \frac{u_y}{u_x} = \frac{u \: sin \: \alpha -gt}{u \: cos \: \alpha}{/jatex}

Rearranging this gives {jatex options:inline}t= \frac{u sin \: \alpha - u cos \: \alpha \: tan \beta}{g}{/jatex} ]]>

The triangular prism above is on a slope which makes an angle %theta with the horizontal. If a horizontal force is applied to the left at the top of the triangle, assuming the plane is rough enough so that the prism will topple before it slips, it will begin to topple when the vertical through the centre of gravity C passes through A.

The distance from the top of the triangle to the centre of the base isand the distance from C to the centre of the base is one third of this:

Angle CAB is then

The prism has to move through the extra angleand obviously as increases up to the toppling angle, this extra angle will decrease, making the prism easier to topple.

]]>If the ball hits the ground with a speedand the coefficient of restitution between the ball and the ground isthen the ball bounces with a speedIf we ignore air resistance, and the ball is dropped from a heightthen the gravitational potential energy of the ball is initiallyand as the ball falls, this energy is changed into kinetic energy,so that just before it hits the ground, all the gravitational potential energy is converted into kinetic energy. We can write

After the ball hits the ground, it bounces with a speed

We can use the principle of conservation of energy to find the height h-1 reached by the ball:

After the ball hits the ground again, it will rebound with a speedand we can use the principle of conservation of energy to find the height reached by the ball after the second bounce.

In general, after the nth bounce, the ball will rebound with a speedand bounce to a height of

The total distance travelled by the ball is then(since after falling to hit the floor, the ball has to travel the same distance up then down again)

Ignoring the first termthis is a geometric series with first termand common ratioso the total distance travelled can be found using the formula for the infinite sum of a converging geometric series

Now addto get the total distance travelled by the ball from the instant it is dropped to the instant it comes to rest.

]]>When a lamina is made by cutting out a part, we can treat the missing part as having a negative oment. Still we add the moments of the individual parts – some negative – and equate this to the whole moment.

For example, the uniform lamina below has a rectangle cut from one corner. We can position the origin at the lower left corner and take moments about the x and y axes for the lamina and missing part, and equate the sum to the moment of the whole lamina.

The centre of gravity of the whole lamina – missing part included - is atand the mass is 60 units. The centre of gravity of the missing part is atand the mass is 8 units. The centre of mass of the lamina with the missing part taken away is atand the mass is 52 units.

Taking moments about the x – axis

Taking moments about the y – axis

The centre of gravity of the lamina is at

]]>Description

Shape

Area

Triangle

Quarter Circle

Semicircle

0

Sector of Circle

0

One Quarter the Circumference of a Circle in the 1^{st} Quadrant

Half the Circumference of a Cicle Above the x - axis

0

Cone

0

Hemisphere

0

]]>Consider the framework below, in which three rods form a triangle.

Since each rod is uniform, the centre of gravity of each rod is in the centre of each rod. Suppose the mass of AB is 5m, the mass of BC is 4m and the mass of AC is 10m. The centre if mass is G(x,y), relative to A as the origin in the diagram below and the mass of the famework is 19m.

Taking moments about the x axis gives

Taking moments about the y axis gives

The centre of gravity is then at

]]>The lamina below has uniform density, so the centre of gravity of P is at the centre of P and the centre of gravity of Q is at the centre of Q.

Taking moments about AB, we can draw up the table:

Section |
Distance of centre of gravity from AB |
Area |
Distance * Area |

P |
|||

Q |
|||

Lamina |

Taking moments about A'B':

Section |
Distance of centre of gravity from A'B' |
Area |
Distance * Area |

P |
|||

Q |
|||

Lamina |