Motion, Forces and Energy

PHY101

Dr. Tim RichardsonSemester 1 (Block 1) 2005

What does the Universitymotto mean?

Objectives of PHY101To gain a deep understanding of the behaviour of moving bodiesand the causes of their motion.

Course FormatFormal Lectures (3 per week for 6 weeks).

Tutorials (small group sessions with a tutor aimed at cementing the lecturematerial).

Problems Classes in which you have an opportunity to develop skillsnecessary to tackle challenging physics problems (FRIDAYS 11:10 – 12:00)

Recommended Book & MaterialSee First Year Noticeboard

Various websites advertised throughout the course.

Motion, Forces and Energy

PHY101

Dr. Tim RichardsonSemester 1 (Block 1) 2005

Part 1a: Scale and All ThatPart 1b: Motion in 1D and 2DPart 2: That Guy NewtonPart 3: Circles and ResistancePart 4: Work – A Strange ConceptPart 5: Potential Energy & ConservationPart 6: Snooker (Collisions)Part 7: Rotational KinematicsPart 8: Not-So-Simple Harmonic MotionPart 9: That Sinking Feeling

Part 1a: Scale and All That

Standards of Length, Mass and Time

The metre (m) and the length scaleThe kilogram (kg) – platinum / iridium alloy cylinderTime – Cesium 133 atoms

Units and Dimensions

Scale: From a Quark to a Universe

Units and Related Matters

Seriously, units and dimensions are important. Without units, a value of a particular quantity may be meaningless.

It is truly amazing how often physics students omit units when quoting a value of a quantity. Get into the habit of using units at all times – it is oneof the trademarks of a good and professional physicist!

FORGETTING UNITS IS A PUNISHABLE OFFENCE!

Who’s the scientist in the picture?

Introductory Concepts in Motion

Displacement – the change in position of an object, x (vector quantity)

x = xf – xi

where the subscripts f and i refer to final and initial positions respectively.It is important to understand the difference between displacement and distance(purely scalar):

Who’s the athlete on the right?

Running one lap of a running track givesyou a displacement of 0 m but you havetravelled a distance of 400 m.

Velocity and speed

Average velocity of a particle v is given by:

v = x / t

where t is the time interval. Average velocity is a vector quantity whereas averagespeed is a scalar quantity (distance / t). Again we can use the example of a 400 mathlete who completes one lap; her average speed might be around 11 ms-1 but her average velocity is formally zero.

Instantaneous velocity

The instantaneous velocity vx is the limiting value of the above ratio as Dt approaches zero. In calculus notation, this limit is called the derivative of x with respect to t, written as dx.dt:

vx = lim x/t = dx/dtt~0

~ 10-3 ms-1

~ 104 ms-1

Example:

A particle moves along the x axis such that its displacement x is given byx = -4t + 2t2 .

(a) Find the instantaneous velocity of the particle at t = 2.5 s and t = 3.8 s.

We could plot this function and measure the gradient of the curve at the abovetime values (do this in your own time), but using basic calculus is easier:

v = dx/dt = 4t – 4

Now substituting the values for t yield the velocity:

For t = 2.5 s, v = 6.0 ms-1 and for t = 3.8 s, v = 11.2 ms-1.

See how close your graphical estimate comes the above values.

Acceleration

We can define average acceleration as a = vx / t but more importantlyInstantaneous acceleration, ax as:

ax = lim vx / t = dv/dt t~0

Now we can link displacement to acceleration using calculus:

ax = dv/dt = d (dx/dt) = d2x / dt2

dt

For our example where x = -4t + 2t2 , we can now see that the accelerationis single-valued and equal to + 4 ms-2 (think about the meaning of this).

Problems to try:

1 An athlete runs along part of a track with a displacement given by x = 5t + 4. Find (i) her change in position over the first 3 seconds, (ii) her velocity at t=2s and t=3s and (iii) her acceleration.

2 A particle’s displacement varies with time as x = 4t3 - 3.5t2 + t + 9.Find (i) the change in position of the particle between t=4 s and t=6s,(ii) its instantaneous velocity at t=3.2 s and (iii) its instantaneous acceleration at t=4.1s. (iv) Does the particle ever stop in the time interval 0 < t < 0.5 s?

(NB I sometimes like “trick” questions.)

Part 1b: Motion in 1- and 2-D.

If the acceleration of a particle varies in time, its motion can be complex anddifficult to analyse. However, one-dimensional motion in which the accelerationis a constant value is straightforward to deal with. We saw earlier that we candefine acceleration as a where:

a = v – u (1)

t

where v is the final velocity and u is the initial velocity for the time interval, t.

We can rearrange this expression to yield our first KINEMATIC EQUATION:

v = u + at (KE 1)

This equation allows us to relate velocity to acceleration and time.

Using x for the change in position, we can find our second kinematicequation:

Average velocity = u + v (2)

2

and displacement is average velocity x time, so

x = (u + u + at). t 2

So x = ut + ½ at2 (KE2)

Finally we use equations (2) and (KE1) to yield our 3rd kinematic equation:

x = ( u + v ) ( v – u ) = v2 - u2 2 a 2a

Therefore v2 = u2 + 2ax (KE3)

Deriving the Kinematic Equations from Calculus

We know that a = dv/dt and that v = dx/dt:

So dv = a dt or v = a dt

In the case of constant acceleration, v = at + c, but when t=0, c is just the initial velocity u, so we derive v = u + at .

Also we can write dx = v dt or x = v dt

But from above we know that v = u + at,

so x = (u + at) dt = u dt + at dt

Finally therefore we have x = ut + ½ at2

What was the nickname of the athlete on the right?

Freely Falling Objects

We’ll be talking about gravity at great length later in the course, but for now we justneed to remember that in the absence of air resistance, all bodies fall at the samerate under the influence of the Earth’ gravitational field.

We use +g for objects falling and –g for rising objects decelerating.

vy = v0y + g t

y = v0y t + ½ g t2

vy2 = v0y

2 + 2 g y

What year was the “hammer and feather”experiment performed on the Moon?”

The Coin in the Well Experiment

We use h = v0y t + ½ g t2 to find the depth of the well, counting how long ittakes for the coin to splash into the water.

Since the coin is dropped from rest, weKnow that v0y = 0 so that:

h = ½ g t2

If for example it takes 4.3 s for the coin to fall, then the depth of the well would be90.6 m.

What’s wrong with this diagram?

Reaction Time Game

A further refinement on this theme gives us a “reaction time” testing game. Allyou need is a 30 cm ruler (unless you’re a real dork in which case you might need a 1 metre ruler) and a partner. Ask your partner to drop the ruler through your forefinger and thumb without telling when he/she is going to do it. As soon as yousee the ruler falling, try to catch it. Then use h = ½ g t2 to calculate your reactiontime. Usually students’ reaction times are in the range 0.15 – 0.3 s.

t = 2 h / g

If we assume that the reaction time for pressing a car brakepedal is similar to that found above, then we can work outthe distance travelled at various speeds before we even begin to brake!

eg. t ~ 0.2s, so that:

Speed (ms-1) (mph) Distance (m) 13 30 2.6 22 50 4.4 31 70 6.2 39 90 7.8http://www.chss.montclair.edu/psychology/museum/mrt.html

Projectile Motion

Projectile motion is quite simple to analyse if we make two assumptions:

1 The free-fall acceleration due to gravity g is constant over the range of motion2 The effect of air resistance is negligible.

We’ll start by showing that the projectile path is parabolic. We know thatay = -g and ax = 0:

See Board Work