Advanced Higher Physics 2009

Hmm... just started at the moment.

Both my prelims I've got low 90's... losing about 6% on Rotational Mechanics both times - if anyone's got a spare moment to help explain it I'd be grateful! Scholar's site hasn't helped at all :frown:

Reply 4

OP

Davey_SMFC

Hmm... just started at the moment.

Both my prelims I've got low 90's... losing about 6% on Rotational Mechanics both times - if anyone's got a spare moment to help explain it I'd be grateful! Scholar's site hasn't helped at all :frown:

Mechanics is my strong point, and I have a spare moment. What specifically would you like me to explain?

Reply 5

Slumpy

16

Davey_SMFC

Hmm... just started at the moment.

Both my prelims I've got low 90's... losing about 6% on Rotational Mechanics both times - if anyone's got a spare moment to help explain it I'd be grateful! Scholar's site hasn't helped at all :frown:

Which bits are troubling you?
I might be able to remember it(though I should be revising :p:

)

Reply 6

Meteorshower

Mechanics is my strong point, and I have a spare moment. What specifically would you like me to explain?

Slumpy

Which bits are troubling you?
I might be able to remember it(though I should be revising :p:

)

Centripetal force
Rotational dynamics

I know how to get equations, how to pull the equations out of the equation sheet and fill them in and get all the easy marks. Centripetal force isn't hard, and when it's Gravitation involved I find it easy, but for Pendulums and "loop-loop" tracks (these were the 2 questions I got) I find it hard to answer the actual questions, with the actual Physics answer. With Rotational Dynamics I can just think of it linearly and adjust it, but I'd rather be able to think of it properly, because it's the one part of the course I don't actually understand (fundamentally!).

I don't quite know what I'm asking for though, because I'm not expecting you to go over everything that the course expects you to know, I just find that I struggle answering the harder bits of the questions.

EDIT: As well as the Pendulum (where I can do the maths, I can't remember where I went wrong, but there was a point where I was just stuck and couldn't go anywhere) and the "loop-the-loop" track (where the ball rolls round it, with but a linear motion and angular motion) there were questions about the Moment of Inertia of a cylinders or other 3D shapes (or even CDs!) that I wasn't getting right.

Reply 7

OP

Davey_SMFC

Centripetal force
Rotational dynamics

I know how to get equations, how to pull the equations out of the equation sheet and fill them in and get all the easy marks. Centripetal force isn't hard, and when it's Gravitation involved I find it easy, but for Pendulums and "loop-loop" tracks (these were the 2 questions I got) I find it hard to answer the actual questions, with the actual Physics answer. With Rotational Dynamics I can just think of it linearly and adjust it, but I'd rather be able to think of it properly, because it's the one part of the course I don't actually understand (fundamentally!).

I don't quite know what I'm asking for though, because I'm not expecting you to go over everything that the course expects you to know, I just find that I struggle answering the harder bits of the questions.

I'm not sure what you're asking either, If you have a particular question find difficult I could try and elaborate on the theory behind that maybe?

Reply 8

Meteorshower

I'm not sure what you're asking either, If you have a particular question find difficult I could try and elaborate on the theory behind that maybe?

Paper 2004: q2,3; 2005: q1,2; 2006: q2 and basically just similar ones. I presume the theory behind them all is the same though. For most of them it's the part d)'s and stuff that I just don't know :confused:

Reply 9

ukdragon37

Davey_SMFC

Centripetal force
Rotational dynamics

I know how to get equations, how to pull the equations out of the equation sheet and fill them in and get all the easy marks. Centripetal force isn't hard, and when it's Gravitation involved I find it easy, but for Pendulums and "loop-loop" tracks (these were the 2 questions I got) I find it hard to answer the actual questions, with the actual Physics answer. With Rotational Dynamics I can just think of it linearly and adjust it, but I'd rather be able to think of it properly, because it's the one part of the course I don't actually understand (fundamentally!).

I don't quite know what I'm asking for though, because I'm not expecting you to go over everything that the course expects you to know, I just find that I struggle answering the harder bits of the questions.

EDIT: As well as the Pendulum (where I can do the maths, I can't remember where I went wrong, but there was a point where I was just stuck and couldn't go anywhere) and the "loop-the-loop" track (where the ball rolls round it, with but a linear motion and angular motion) there were questions about the Moment of Inertia of a cylinders or other 3D shapes (or even CDs!) that I wasn't getting right.

Rotational dynamics is analogous to linear except that you must take into account the distance from the mass and force from the axis of rotation, since the larger the distance the "heavier" the object is (i.e. harder to move rotationally). Check that you are using the correct Moment of Intertia equations. A dimensional extension of one sort of object will have the same moment of inertia equation as the original object as long as the masses remain the same distance from the axis of rotation. To see what I mean:

A point mass has I=mr^2
Its 2D equivalent, the ring, also has I=mr^2 since the masses are the same distance from the axis.
This means that the 3D equivalent (hollow cylinder) will also have I=mr^2

A disc has I=0.5mr^2 (found using integration blah blah)
Its 3D equivalent the rod, will have I=0.5mr^2 when rotating about the axis.
When the rod rotates about other axes, such as about end or about centre I changes since the distance of mass to the axis changes.

Reply 10

Thanks. Never actually thought about the dimensional extension bit, even though I knew the moment of inertia equations.

Does I = kMR² where k depends only on the distribution of mass around the rotational axis - nothing else?

Other examples of the questions I find harder: Cars going round corners (banked or flat!) and then the Flywheel questions I never fully understood why I was doing what I done.

Reply 11

OP

Davey_SMFC

Paper 2004: q2,3; 2005: q1,2; 2006: q2 and basically just similar ones. I presume the theory behind them all is the same though. For most of them it's the part d)'s and stuff that I just don't know :confused:

Ok, I'll go for d 2004 because I assume you can do the calculations if you can get 90% in the prelim!

For anything to be moving in a circle, it must have a force acting on it perpendicular to it's velocity. For a car going round a bend, that can be friction, or a component due to gravity if the bend is banked. This force is the centripetal force, and it's directly proportional to mass and velocity and inversely proportional to the radius of the circle the object being acted upon is travelling in. This means that the faster an object is travelling, the greater the force required to induce circular motion assuming radius remains constant. Same with mass. And with constant velocity, the greater the radius of the circle the less force needed to have circular motion.

Applying this to 2004 question 2 part d -

The question states that at high angular velocities the friction pads move away from the axle. The reason they are there in the first place is that tension stored in the spring exerts a force on them. As the angular velocity increases of the yo-yo increases, the tangential velocity of the friction pads increases and the central force needed to keep them going in a circle increases. At a certain velocity, there will be an equilibrium where the central force necessary is exactly 5N and therefore the friction pads will be exerting no force upon the axle. At velocities higher than this, the central force necessary will be greater than 5N. In order for the friction pads to keep travelling circularly, the radius of the circle must decrease to bring the required central force down to 5N. As the friction pads move away to do this, the springs further compress aswell resulting in more tension that can be converted to central force.

I hope that helps a bit, I can explain the other ones later if you need.

Reply 12

ukdragon37

Davey_SMFC

Thanks. Never actually thought about the dimensional extension bit, even though I knew the moment of inertia equations.

Does I = kMR² where k depends only on the distribution of mass around the rotational axis - nothing else?

Other examples of the questions I find harder: Cars going round corners (banked or flat!) and then the Flywheel questions I never fully understood why I was doing what I done.

For AH, yes. An example is that if you squish a tube down to a ring and then bring in the circumference of the ring into a single point while keeping the distance the same you will still have the same moment of inertia.

Reply 13

Meteorshower

For a car going round a bend, that can be friction, or a component due to gravity if the bend is banked.

In order for the friction pads to keep travelling circularly, the radius of the circle must decrease to bring the required central force down to 5N.

As the friction pads move away to do this, the springs further compress aswell resulting in more tension that can be converted to central force.

First point, is it a component due to gravity, or the horizontal component of the reaction force (enbankment on the car)? I'd have thought it was the second one.. I presume the vertical component of the reaction force would equal the weight of the car (there's no vertical motion!), but the actual centripetal force would be due to horizontal component of R. (Btw I'm not critising you, just making sure I've got it sorted in my head!)

Secondly, I assume you mean the radius must increase? The required centripetal force varies inversely with the radius (perpendicular distance of the friction pads to the axle?)

And finally, just to make sure I've got it, the required central force was greater than the actual centripetal force (5N), so to "compensate" the friction pads move from the axle, increasing the radius, and hence decreasing the required centripetal force so that the actual centripetal force (5N) is greater than or equal to the required centripetal force?

Reply 14

abstraction98

*subscribing.

What a great idea starting a thread Meteorshower :tongue:

Reply 15

OP

Davey_SMFC

First point, is it a component due to gravity, or the horizontal component of the reaction force (enbankment on the car)? I'd have thought it was the second one.. I presume the vertical component of the reaction force would equal the weight of the car (there's no vertical motion!), but the actual centripetal force would be due to horizontal component of R. (Btw I'm not critising you, just making sure I've got it sorted in my head!)

Secondly, I assume you mean the radius must increase? The required centripetal force varies inversely with the radius (perpendicular distance of the friction pads to the axle?)

And finally, just to make sure I've got it, the required central force was greater than the actual centripetal force (5N), so to "compensate" the friction pads move from the axle, increasing the radius, and hence decreasing the required centripetal force so that the actual centripetal force (5N) is greater than or equal to the required centripetal force?

If you have a car on a banked curve where x is the angle of embankment Rsinx would be the force that provides the centripetal force assuming no friction. Since R is equivalent to mg in such a situation the centripetal force can also be expressed as mgsinx, so the two are equivalent really and we are both right :smile:

As far as I can tell at midnight anyway :p:

If it helps, draw a force diagram - might make things clearer

For your second point, yes that's what I do mean. F = mv^2/r so the bigger r is the smaller required centripetal force. At first glance this may appear to contradict the alternate equation for centripetal force, F=mrw^2 , but w is inversely proportionate to r this is not the case.

For your last paragraph yes, although it wouldn't compensate to a position where the available force from tension in the springs was more than the required centripetal force, because then there would be an unbalanced force and it would just move in again.

Reply 16

OP

abstraction98

*subscribing.

What a great idea starting a thread Meteorshower :tongue:

I'm glad I had it :p:

(Abstraction told me to!) :woo:

Reply 17

A serving table of mass 2.5 kg and length 80 cm is hinged to a wall, and is supported by a chain which makes an angle of 50° with the horizontal table top. The chain is attached to the edge of the table furthest from the wall (80 cm). The table is uniform, so its centre-of-mass is 40 cm from the wall. A full serving dish of mass 1.0 kg is placed on the table 60 cm from the wall. Calculate the tension in the chain.

Right see a question like this... even if I got the answer I don't understand how the Physics of this works at all!

Reply 18

OP

Davey_SMFC

A serving table of mass 2.5 kg and length 80 cm is hinged to a wall, and is supported by a chain which makes an angle of 50° with the horizontal table top. The chain is attached to the edge of the table furthest from the wall (80 cm). The table is uniform, so its centre-of-mass is 40 cm from the wall. A full serving dish of mass 1.0 kg is placed on the table 60 cm from the wall. Calculate the tension in the chain.

Right see a question like this... even if I got the answer I don't understand how the Physics of this works at all!

I'm afraid you'll have to wait til tomorrow before I at least have a proper look at that, I've just realised how tired I am (bloody maths) and there's too much crap in that question to visualise without a diagram :p:

Off to bed for me!

Reply 19

ukdragon37