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### Advanced Higher Physics Unit 1

Rotational Dynamics

Using moments

The spanner exert a moment or turning effect on the nut.

Turning point

distance from force to turning point

force

If the moment is big enough, it will unscrew the nut.

If not, there are two ways of increasing the moment.

Using moments-increasing the moment

- Increase the distance from the Force to the pivot-apply a force
- at the end of the spanner or use a longer spanner.

Turning point

distance from force to turning point

force

If the same force is applied over a greater distance, a larger moment

is produced.

Using moments-increasing the moment

2. Increasing the Force applied-push/pull harder or get someone

stronger to do it!

Turning point

distance from force to turning point

force

If a greater force is applied over the same distance,

a larger moment is produced.

Moment of a force

The moment of a force is given by:

F is the force applied, measured in Newton (N)

d is the distance from turning point, measured in metres (m)

The moment of a force is therefore measured in

Newton metres (Nm).

Torque

A force is applied to the rim of a disc which can rotate around its centre axis. In this case the moment of a force is called the Torque.

F

r

In data booklet

F is the force applied, measured in Newton (N)

r is the radius of the circle, measured in metres (m)

T is the Torque associated with force F,

measured in Newton metres (Nm).

Inertia

Inertia can be defined as resistance to change in motion.

In linear motion, MASS is a measure of an object’s inertia

(since a large mass needs a large force to produce an acceleration).

In angular motion, we use MOMENT OF INERTIA.

Moment of inertia

- The moment of inertia of an object is its resistance to change
- in angular motion.
- The moment of inertia depends on:
- The mass of an object
- How the mass is distributed about the axis of rotation.

Consider a mass m at a distance r from the axis of rotation. The moment of inertia can be calculated using:

r

In data booklet

(additional relationships)

m

m is the mass of the object, measured in kg

r is the distance from the axis of rotation, measured in m

I is the moment of inertia measured in kgm²

Moment of inertia : simple situation

All the mass can be considered to be at the same distance from the axis.

Masses on a very light rod

r

Wheel with heavy rim and very light spokes

r

In these cases I=mr² where m is the total mass.

Moment of inertia: rods

rod about centre

With:

ltotal length of the rod

m total mass of the rod

rod about end

The moment of inertia for a rodrotating about end is 4 times bigger

than the moment of inertia for a rod rotating about centre as it is

harder to do so.

This is because there are now more particles at a greater distance

from the axis of rotation.

Moment of inertia: solid disc about centre

In data booklet

(additional relationships)

r

Where m is the total mass of the disc

r is the radius of the circle

Moment of inertia: Sphere about centre

Where m is the total mass of the sphere

r is the radius of the sphere

(all the moment of inertia formulas can be found

in the data booklet in Additional Relationships)

Newton 2nd Law

An unbalanced Torque will produce an angular acceleration.

In data booklet

With I, the moment of inertia in kgm²

α, the angular acceleration in radsˉ²

T, the Torque in Nm

Angular momentum

The angular momentum is defined as the moment of the linear momentum.

r

m

For this particle of mass m:

The linear momentum p = mv

v

w

The angular momentum = the moment of p = mvr = mr²w, since v=rw.

In data booklet

With L angular momentum measured in kgm²sˉ¹.

Angular momentum of a rigid body

A rigid body is an object in which all the individual parts have the

same angular velocity w.

The angular momentum of this body is the total of the angular

momenta of its particles:

w is constant as all particles must be rotating at the same rate.

In data booklet

Conservation of angular momentum

In the absence of external Torque, the total angular momentum

before impact equal the total angular momentum after impact.

Not in data booklet

Example: lump of mud stuck to a bike wheel

Before:

After:

Total angular momentum before = total angular momentum after

Iwheelw0 = w (Iwheel + Imud)

The moment of inertia of the wheel and the mud after impact is larger

than the moment of inertia of the wheel before impact.

Therefore the angular velocity of the wheel is smaller after impact.

Example: pupil spinning on a chair

After:

Before:

Pupil draws arms in

Pupil pushes arms out

Total angular momentum before = total angular momentum after

Ioutw1 = Iinw2

Iin is smaller than Iout because the particles are closer to the axis

of rotation.

Therefore w2 is larger than w1.

Example: mass dropped on a turntable

Before:

After:

Axis of rotation

Total angular momentum before = total angular momentum after

Idiscw1 = w2(Idisc+Imass)

The moment of inertia of the disc and the mass after impact is larger

than the moment of inertia of the disc before impact.

Therefore the angular velocity of the disc is smaller after impact.

Rotational Kinetic energy

In data booklet

With I moment of inertia, measured in kgm²

w angular velocity, measured in radsˉ¹

Erot rotational kinetic energy, measured in J

Work done

Not in data booklet

With T, Torque measured in Nm

θ, angular displacement in rad

Ew, work done in J

Objects rolling down an inclined plane

Ep = mgh

Erot = ½Iw²

h

w

Ek = ½mv²

v

Potential energy at top = total of linear and angular kinetic energy

mgh = ½mv² + ½Iw²

Comparison of linear and angular motion

The equations of angular motion are similar to those of linear motion.

linear motion

angular motion

10 kg

6m

- Find:
- Total I
- Unbalanced T
- Driving T
- Driving F
- Final w
- Deceleration when driving T removed
- Time taken to move to rest.

Frictional torque = 1000 Nm

time = 5s

r

- Find:
- Driving Torque
- Moment of inertia of solid disc
- Angular acceleration
- Angular displacement
- Angular velocity

F

Solid disc mass = 20 kg

Radius = 3m

Axle radius = 1cm

Force applied to axle = 4000 N

Frictional Torque = 30 Nm

Cord length = 50 cm

- A turntable of mass 5kg and radius 25 cm is rotating at 10 radsˉ¹.
- A metal ring of mass 2 kg and radius 10 cm is dropped over the centre of the turntable.
- Find the new angular velocity of the system.
- Using rotational energy determine whether this is an elastic or an inelastic situation.

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