## Inertia

### The magnitude of the linear acceleration produced by an unbalanced Force will depend on the Mass of the object, which is also known as the object's Inertia. The word Inertia can be simply described as "An object's Resistance to a change in motion". For example, a large Mass object is difficult to start to move, then difficult to stop once moving.

## Moment of Inertia

### As seen in the previous section ( Torque ), the "Moment of ..." means the Angular equivalent. In this case, Moment of Inertia is the "Resistance to a change in Angular motion".

### The Moment of Inertia ( I ) depends on both the Mass and the distribution of Mass around the axis of rotation.

### As the Moment of Inertia depends on the distribution of Mass, there is not one, but many different Moment of Inertia formulae, each used to describe a different distribution of Mass.

### Note - For more than one Mass, treat each separately, then sum together to find the overall Moment of Inertia.

### The video below shows an example of the effect of Moment of Inertia on several rotating objects.

## Point Mass at a fixed Radius

### For a Point Mass ( a 1 dimensional mass ) at a fixed radius from the axis of rotation, the Moment of Inertia can be found using the formula :-

### I = mr^{2}

### Where :-

### I = Moment of Inertia ( kgm^{2} )

### m = Mass ( kg )

### r = Radius ( m )

### Two Identical Point Masses, each of Mass 2 kg are attached together by a light fixed rod of length 3 m. If the rod rotated about a fixed point as shown above, calculate the total Moment of Inertia of the system.

### m_{1} = m_{2} = 2 kg

### r = ^{L}/_{2} = 1.5 m

### Total Moment of Inertia ( I ) = Σmr^{2} = m_{1}r^{2} + m_{2}r^{2}

### Total Moment of Inertia ( I ) = ( 2x1.5^{2 }) + ( 2x1.5^{2 })

### Total Moment of Inertia ( I ) = 9 kgm^{2}

### Note - In the above example, note the key phrase "light rod". This use of "light" means that the rod's Mass and therefore its Moment of Inertia is negligible and is therefore ignored.

## Ring at a Fixed Radius

### If the use of a Point Mass at a fixed radius calculation above is continued to include a large range of Point Masses at a fixed radius, the Moment of Inertia for a ring can be found using the following formula :-

### I = MR^{2}

^{Where :-}

### I = Moment of Inertia ( kgm^{2} )

### m = Total Mass of Ring ( kg )

### R = Radius of Ring ( m )

### By assuming that all the mass is situated at a fixed radius, we can ignore the mass of spokes as negligible, and calculate the Moment of Inertia of a wheel using this formula.

### What is the Moment of Inertia of a bicycle wheel of Mass 1,630 g with a diameter of

### 622 mm, assuming that the Mass of the spokes is negligible ?

### M = 1630 g = 1.63 kg

### R = ^{D}/_{2} = 311 mm = 0.311 m

### I = MR^{2}

### I = 1.63 x 0.311^{2}

### I = 0.158 kgm^{2}

## Moment of Inertia of a Disc

### So far, only Point Masses have been considered. In most situations, however, the Mass is spread out over a large region, and as such the above formulae do not apply. In order to calculate the Moment of Inertia for object with extended Mass, the object must be considered to consist of many Point masses, each which sum together to give the Moment of Inertia for the whole object.

### In order to complete these calculations from scratch, Calculus is involved. This is, however, beyond the scope of the Advanced Higher course, and as such all required formulae for different shapes are given in the relationship sheet.

### Below is a list of the formulae required for Advanced Higher Physics :-

## Resultant Torque and Angular Acceleration

### In earlier courses, applying a unbalanced Force gave an Acceleration, and the same is seen with Torque. When an Unbalanced Torque is applied to an object, an Angular Acceleration is experienced. The magnitude of this Angular Acceleration depends upon the Moment of Inertia of the object.

### T = I α

### Where :-

### T = Unbalanced Torque ( Nm )

### I = Moment of Inertia ( kgm^{2} )

### α = Angular Acceleration ( rad s^{-2} )

## Example 3 - "AB" Level Question

### In the Film "Moonraker" James Bond enters a centrifuge used to train Astronauts. Once inside, Cheng tries to kill him by applying a Force equivalent to 13G to Bond, but he escapes by destroying the machine with a wrist dart...

### The centrifuge works by placing the occupant at the end of a long rotating arm, the faster the rotation, the greater the Centripetal Force required to keep the occupant within the circle. This increase in Centripetal Force is used to simulate changes in Gravity.

### If the arm has a length of 9.1 m, has a mass of 3,200 kg and has a maximum angular Velocity of 5.24 rad s^{-1}, which it takes 35 seconds to reach, calculate :-

### The Torque provided by the Centrifuge

### Maximum Centripetal Force Experienced by Bond

### Mass of Rodger Moore whilst filming "Moonraker" = 79 kg

### Torque :-

### To find Torque use the formula T = I α, therefore must first find I and α.

### ω = ω_{0} + αt

### α = ( ω - ω_{0} ) / t

### α = ( 5.24 - 0 ) / 35

### α = 0.15 rad s^{-2}

### Moment of Inertia - Rod about end = ^{1}/_{3 }ml^{2}

### ( Assumption is that Bond's Mass is negligible at this stage )

### I = 1/3 x 3200 x 9.1^{2}

### I = 88331 kgm^{2}

### Torque required ( T ) = 88331 x 0.15

### Torque required ( T ) = 13250 Nm

### Max Centripetal Force :-

### F = mω^{2}r

### F = 79 x 5.24^{2} x 9.1

### F = 19739 N

### Note - By dividing the above Force by Rodger Moore's Weight, we can find the equivalent G experienced. In this case of the centrifuge at maximum velocity, Bond would have experienced 25G of Force, Bond would not have survived...

### The video below shows a summary of Moment of Inertia