If a massis tied to a string which is then wound around the pulley, then let loose, and the pulley is free to turn, then the massapplies a torqueto the system.

The moment of inertia of the whole system is

The angular accelerationis given by

Another way to show this is using energy.

The kinetic energy of the system is

This will be equal to the gravitational potential energy lost by the massas it falls a distanceEquating these – gives

Now useSince the system starts from rest

]]>Example: Solve the equation {jatex options:inline}\frac{d^2 \mathbf{r}}{dt^2}+2 \mathbf{i}=0{/jatex} with the initial conditions - where {jatex options:inline}t=0{/jatex}, {jatex options:inline}\mathbf{v}=2 \mathbf{i}+3 \mathbf{j}{/jatex} m/s, {jatex options:inline}\mathbf{r}= \mathbf{j}{/jatex} m

With {jatex options:inline}\mathbf{r}= x \mathbf{i} + y \mathbf{j}{/jatex}, the equations of the components are

{jatex options:inline}\frac{d^2x}{dt^2}+2=0{/jatex} (1_

{jatex options:inline}\frac{d^2y}{dt^2}=0{/jatex} (2)

(1) has solution {jatex options:inline}x=-t^2+At+B{/jatex}.

(2) has solution {jatex options:inline}y=Ct+D{/jatex}.

Then {jatex options:inline}\mathbf{r} =(-t^2+At+B) \mathbf{i}+ (Ct+D) \mathbf{j}{/jatex}.

Now use the initial conditions.

{jatex options:inline}\mathbf{v}=(-2t+A) \mathbf{i} + (C) \mathbf{j}{/jatex}

{jatex options:inline}\mathbf{v}(0)=2 \mathbf{i}+3 \mathbf{j} \rightarrow 2=(-2(0)+A), \; 3 =C \rightarrow A=2, C=3{/jatex}

{jatex options:inline}\mathbf{r}(0)= \mathbf{j}= (B) \mathbf{i}+ (D) \mathbf{j} \rightarrow B=0, \; D=1{/jatex}.

Then {jatex options:inline}\mathbf{r} =(-t^2-t) \mathbf{i}+ ( 3t+1) \mathbf{j}{/jatex}.

Then {jatex options:inline}x=-t^2+2t, \; y=3t+1{/jatex}.

From the second equation, {jatex options:inline}t=\frac{y-1}{3}{/jatex} so {jatex options:inline}x=(\frac{y-1}{3})^2+2(\frac{y-1}{3}){/jatex} ]]>

Suppose that a disc of massand radiusis rotating about an axis through the edge of the disc, perpendicular to the plane of the disc. The moment of inertia of the disc about an axis through its centre, perpendicular to the plane of the disc isThe parallel axis theorem gives the moment of inertia about the pivot point on the circumference as

The disc is rotating at two revolutions per second (so that), and is hit by a particle of masstravelling with a speedwhich strikes the disc a perpendicular distancefrom the centre.

The angular momentum of the disc isand the angular momentum of the particle about the pivot isThe total angular momentum is

The moment of inertia of disc plus particle isso if the angular velocity of particle plus disc after the collision isthen

]]>If the mass of the pulley isthen the moment of inertia of the pulley about its centre is

Applying the equationto the pulley gives

Applyingto the mass gives

We have the simultaneous equations

Adding these gives

Substituting this into (1) above gives

]]>The moment of inertia of a rod of lengthand mass M about its centre isso the moment of inertia of the rod above about its centre is

The parallel axis theorem states that if it is pivoted about a parallel axis a distance(equal toin this case) from the first, the moment of inertia about that point is

The kinetic energy of the rad at an time is

When the rod has turned through an anglethe centre of mass has fallen a distance(see the diagram which shows the changing position of the centre of mass).

The increase in kinetic energy is equal to the loss in gravitational potential energy.

(1)

Taking moments about the hinge give

whereThis is the same equation as obtained by differentiating (1) with respect to t.

Divide byand putThe result is

]]>For example, the moment ofacting atabout the pointis

Properties:

The moment is at right angles to bothandThis is a consequence of the cross product. In the above diagram the moment is directed out of the paper, since the cross product defines a right handed coordinate system.

If a force acts at a point, the moment of the force about that point is zero.

The total moment of a system of forces is equal to the sum of the moments:

The Total Moment is a vector, with magnitude given bywhereis the acute angle betweenand

]]>In general a body is extended in space and has a continuous mass distribution. We must use integration to find the moment of inertia using the formula

For example to find the moment of inertia of the uniform lamina illustrated below.

About theaxis the moment of inertia is

by breaking the area into vertical strips and the limits of integration are then

About theaxis the moment of inertia is

as before by breaking the area into vertical strips and the limits of integration are still

]]>since that would mean the solution is trivial (and does not satisfy any boundary conditions) andso

Ifwhenthen

If instead we have a first order differential equation with vectors, say(2) withwhenwe can still assume a solution of the formbut nowis also a vectorwhich we have to find given the initial conditions.

IfthenSubstituting into the differential equation (2) gives

As before, we can factorise withto give

Again as beforeandsohence(3)

To finduse the initial conditionswhenhenceso

We can write this solution in vector form as

More complicated equations may include constant terms added. For example(4)

withwhen

We find the solution in two parts. The first part will be the solution toThis is (3) above. The second part is any solution to (4). We can see that ifis a constant thenso putinto (4) then

The general solution is the sum ofandsoWe can find using the initial conditionswhen

Hence

We can write this as

]]>Different types of forces often occur. The force may depend

on position,as when a particle is moving up or down a slope. If as is often the casewhich represents a restoring force towards an equilibrium position (and this is approximately true for small oscillations for many systems).

on the speed of the particle, as when the particle experiences a drag force. Often the drag force,is proportional to the speed (and always in the opposite direction to the motion of the particle), so(since).

the force may be a drive forceThis is an externally applied force, and represents an external source of energy. This may result in the system vibrating wildly, if the frequency of the forcing term is close to the natural frequency of the system.

Substitute these intoto get

Now useandto get

This is often written as

]]>then assumingis constant during the time interval

Diving bygives

Now letto give the differential equation

If the mass of the raindrop is increasing at the constant rate C with respect to x thenandso

Write

The equation becomes

Now divide byand write:

We can solve this equation using the integrating factor method:

The equation now becomes

Hence

Ifwhenthen

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