HSC Tips - Mechanics (2 Viewers)

[McLake]

Tip page 6? 7? Oh well, one of them anyway ...

Tips for Mechanics:
- DRAW a diagram of the situation.
- Try not to let the physics you have learn't for the HSC influence you, this is maths, they do things differently.


I can't think of any more tips (pathetic really) so please someone else post some more ...

 

[JizZ]

Hmm... this looks kinda retarted when typed up, but anyways, this is what I had written on a summary sheet i found. I dunno if its helpful, but thought I might post it up.

If a = f(t), use dv/dt or (d^2)x/d(t^2)

If a = f(x), use d(0.5v^2)/dx

If a = f(v), use dv/dt if initial conditions are on (t,v)
use v.dv/dx if initial conditions are on (x,v)

- When working with resisted motion and object changes direction, remember to split working out into two if the resisting force is now actually assisting the motion (eg. vertical projectile)

- Know how to work out escape velocity using inverse square law for gravity

- Know Circular motion and banked track formulas

- Know conversions of kg wt

- Know how to work out..

'A car of mass 1 tonne passes over a bridge formed by the arc of a circle of radius 10 metres,
a) Find the force exerted by the car on the road at the top of the bridge if the car travels 8m/s
b) What speed would cause the car to be on the point of leaving the bridge at its highest point

 

[ezzy85]

know how to derive equaitons of motion and projectile motion etc...

 

[Affinity]

It's not that different in maths and physics, just remember to derive your formulae always, leaning physics gives you an edge in this.

basics, results that can be used:
1.)change in momentum with respect to time is force:
dp/dt = F where p = mv
and when m is constant(almost always in HSC), m(dv/dt)=F or F=ma

2.)action - reaction

3.) a = d^2x/dt^2 = v*(dv/dx) = (1/2)*[d(v^2)/dx]

The concept of 'energy' may be handy and can be introduced with some simple equations:

m[gx + (v^2)/2] is conserved for uniform accelerated motion.

m[(v^2)/2 - (k/x)] is conserved for acceleration that varies with distance squared.

If you are stuck, list all equations and and see which ones you can solve.

 

[wogboy]

I think an important point for a conical pendulum is that a centripetal force is NOT a "causal" force. It is not a force that just exists due to nature, like gravity & tension do. Rather the centripetal force is caused by gravitation (weight or mg) and the tension in the string, both of which are "causal" forces.

By the word "causal", I mean something that causes something else to happen (e.g. hunger is causal, since it causes you to eat )

In a conical pendulum (which isn't moving up or down, and which is contantly spinning), the resultant force on the bob is EQUAL to the centripetal force acting on the bob. The centripetal force is the end result.

So what does this all mean in terms of maths? It simply means that:
W + T = C,

where:
W = weight of the bob = mg

T = sum of ALL tension/tensions in the string/strings (there may be more than one string)

C = centripetal force acting on the bob = m(v^2)/r = m(w^2)r

(also v = linear speed of the bob, w = angular speed of the bob, r = radius of the circle travelled by the bob, m = mass of bob)

so remember: Weight + Tension(s) = Centripetal force

When I first started mechanics, I used to think that W + C = T. I used to think that the tension in the string was the resultant force, and it was caused by the weight mg & the centripetal force. This is obviously WRONG, get this idea out of your head.(unless of course it is already out of your head)

Finally, note the directions of the forces as this is important so you know when to put a minus sign. I personally prefer to define forces pointing down to be negative (so subtract weight from forces that go up, like tension in an upper string, don't add them) and forces going to the left to be negative, however this is just personal preference. Whatever way you choose, stick to it

That's all the advice on conical pendulums I have folks

 

[Affinity]

positive and negative are assigned to directions, so if acceleration is considered negative in one direction so is velocity, displacement, forces, etc etc.

following that, sometimes you have resistance which is proportional to v^2, and that will complicate things a little when you assign the signs for expressions.

sometimes it saves writing if you refer to all three axes (x,y,z) in your answer. but worry about basics first.

 

[JizZ]

Affinity are you doing the HSC this year?

 

[Affinity]

yah unfortunately

 

[JizZ]

man I wish I knew this stuff as well as you do..

 

[Dumbarse]

lake they were some cruddy tips for mechanics dude

 

[underthesun]

I can at least see a person that posts in this thread that will get 99 on both 4unit and 3unit (know who? )

 

[McLake]

Originally posted by Dumbarse
lake they were some cruddy tips for mechanics dude

Sorry, I just couldn't think of anything.

And thankyou underthesun, but I didn't get 99 in 3 or 4U

 

[underthesun]

Well, someone else, could, for example affinity looks like a good candidate.

 

[ezzy85]

in the patel book theres chapter 7 which is about tension and forces on cars and strings. i didnt see this type of stuff in cambridge so is that also considered as "resisted motion"?

 

[Newbie]

ezzy here is also a good candidate for 100 in both maths

in fact, ezzy > affinity ! :evilwhip:

anyway whats this tension business, never learnt it

 

[ezzy85]

noones as good as you, newbs. since you guys havent learnt it, i guess thats less work for trials.

 

[OLDMAN]

Use integral limits in solving mechanics rather than the more cumbersome 2-unit "integrate, add constant and check boundary conditions" way. See example in "trivial mechanics" thread.

 

[TheStar]

That helps alot, however I find that I have problems if the initial point is NOT the origin - eg:

2. A particle of mass m is released from a height H, where H is much smaller than the radius of the earth. If it experiences an air resistance which is numerically equal to its momentum (=mv) find an expression that relates its instantaneous velocity and its distance from the ground.

What I have to do in this case, is to integrate dx from _0_ to x, rather than from H to x - and at the moment I can't quite figure out why - any thoughts?

 

[OLDMAN]

__________________________________________________
What I have to do in this case, is to integrate dx from _0_ to x, rather than from H to x - and at the moment I can't quite figure out why - any thoughts?
___________________________________________________
Good students labour this point.

Suppose the problem is : find H if it hits the ground at 8 m/s, assuming g=10. Note this is tremendous air resistance, giving terminal velocity of only 10 m/s, why?.

Two ways of measuring x - (1) from the ground up, (2)or from H down.

(1) from the ground up motion eqn.
vdv/dx=-g-v , why minus g? because g is acting to diminish x. why minus v? because the particle falling means v is negative, minus it again to become positive since air resistance is acting to "preserve" x.

thus I{0--->-8}(v/(10+v)dv =I{H--->0}(-1)dx

(2) from H down, meaning x=0 means particle at H, positive down.

vdv/dx=g-v , why plus g? because g is acting to increase x. why minus v? because the particle falling means v is positive, minus it since air resistance is acting to decrease x.

thus I{0--->8}(v/(10-v)dv =I{0--->H}(1)dx

Both approaches should give the same H value.

Einstein famously said that he learnt all his Physics by age 3. Einstein never bragged, I believe what he was saying was he knew which way was up, which way was down, which way negative, which positive.

Note notation: I{a--->b}f(x)dx integral upper limit b, lower limit a wrt x.

 

[OLDMAN]

Parallel to Affinity's acceleration triad below:

3.) a = d^2x/dt^2 = v*(dv/dx) = (1/2)*[d(v^2)/dx]

is the angular acceleration triad,

d^2@/dt^2=w*(dw/d@)=(1/2)*[d(w^2)/d@

where w (omega) = d@/dt.