A2 Physics | Post your doubts here

[smzimran]

AOA,
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_41.pdf
Q3 (b) how do we know that acceleration is of constant magnitude
My guess:
because it is 'g' (acceleration due to gravity) ???

 

[XPFMember]

AOA,
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_41.pdf
Q3 (b) how do we know that acceleration is of constant magnitude
My guess:
because it is 'g' (acceleration due to gravity) ???

Waalaikumassalam wr wb!

well my thinking:

Question says: "It then moves down plane LA and rises up plane RA to its original height."


now since it moves the same distance, it indicates, acceleration is constant...



that's what i thought when i did this question and now when i was typing this, i got confused... in shm also the same thing happens

 

[candid24hours]

by first part I mean the answer to the question

I get that we are supposed to consider this formula here: F=mrw^2. In the question it says omega is slowly increased which means its not constant. Keeping this in view, kindly explain: since centripetal force is directly proportional to r, and with increase in r, centripetal force should increase right? meaning that the mud at the far end SHOULDN'T fly off and that it should remain in the circular path! And in marking scheme, they have stated otherwise. Help.

 

[smzimran]

Waalaikumassalam wr wb!

well my thinking:

Question says: "It then moves down plane LA and rises up plane RA to its original height."


now since it moves the same distance, it indicates, acceleration is constant...



that's what i thought when i did this question and now when i was typing this, i got confused... in shm also the same thing happens

Still not clear!

 

[XPFMember]

Still not clear!

forget it...even i dont know what i meant! realized the theory i had in my head...at the time i tried to understand the question...just failed!

 

[smzimran]

forget it...even i dont know what i meant! realized the theory i had in my head...at the time i tried to understand the question...just failed!

Np bro, it will be solved!

 

[XPFMember]

Np bro, it will be solved!

inshaAllah
*waiting*

 

[smzimran]

AOA,
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_41.pdf
Q3 (b) how do we know that acceleration is of constant magnitude
My guess:
because it is 'g' (acceleration due to gravity) ???

 

[waleedsmz]

AOA,
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_41.pdf
Q3 (b) how do we know that acceleration is of constant magnitude
My guess:
because it is 'g' (acceleration due to gravity) ???

It's is the acceleration due to gravity you're right. Anywho, in simple harmonic motion, acceleration is directly proportional to displacement. In this question, the acceleration of the ball is constant since the ball moves by it's constant component of weight. Therefore this is not simple harmonic...

I find this question conceptual. Kindly please help:
Question no.1 part.c

Marking scheme says greater the radius, greater will be centripetal force. In the formula: F=mv^2/r centripetal force is INVERSELY proportional to r. So what's going on here?

Omega does not depend on the radius, that's right. But there's still a centripetal on object right?? And we know, that F= ma, that when there's constant force, there's acceleration of the object. And when you recall the definition of centripetal acceleration, it's the rate of change of angular velocity. Angular velocity, that is omega , is changing at a constant rate. It's changing at a constant rate irrespective of the change of radius. Whether radius changes or not, The angular velocity , omega changes constantly. When omega is constant, there's no centripetal acceleration and with no centripetal acceleration, the object might just not move in a circular motion, but in translatory motion.

Sorry dude but this is not true! When omega is constant there is still centripetal force and thus centripetal acceleration! Centripetal acceleration does not result in a change in speed but it does change the direction of the speed ( i.e. the velocity ). If there is a body moving in a circle and we simply remove the centripetal force ( centripetal acceleration ) then the body wouldn't go in a circle anymore and would travel in a straight line, at this point there won't be an omega because simple there won't be a circular motion! This has to mean that the at constant omega, there is a centripetal force (acceleration). This is actually a concept in the physics syllabus and the mathematics syllabus that we should solve most of the questions with...

Not to mention that the centripetal acceleration is not defined by the rate of change of angular velocity, it's the rate of change of the tangential velocity... It's is the acceleration needed to keep the velocity of the body changing so that it revolves in a circle... Angular velocity is only a property of circular motion...
Here: http://theory.uwinnipeg.ca/physics/circ/node6.html

Now for the paper 4 question...

The answer to the original question is that, as you increase the radius, the force needed to keep the body moving in circular motion must increase. Since according to the rule C.F. = m.w^2.r .However, the centripetal force in this question is the friction force which is a limited force. (1)In other words, the body at a greater distance from the center requires higher force to keep it in the circular motion. (2)Accordingly, as omega increases and by using the same rule previously mentioned, the force required to keep the mud in circular motion gradually increases until it's no longer can be achieved through friction ( only force acting on the object ).

Using the last two sentences ( 1 & 2 ), we reach that the mud at a greater distance ( near the edge ) will leave the plate first.

Hope that helped. : )

 

[waleedsmz]

Still not clear!

forget it...even i dont know what i meant! realized the theory i had in my head...at the time i tried to understand the question...just failed!

Since you guys didn't know the answer as well. Check mine. ; )

 

[candid24hours]

It's is the acceleration due to gravity you're right. Anywho, in simple harmonic motion, acceleration is directly proportional to displacement. In this question, the acceleration of the ball is constant since the ball moves by it's constant component of weight. Therefore this is not simple harmonic...





Sorry dude but this is not true! When omega is constant there is still centripetal force and thus centripetal acceleration! Centripetal acceleration does not result in a change in speed but it does change the direction of the speed ( i.e. the velocity ). If there is a body moving in a circle and we simply remove the centripetal force ( centripetal acceleration ) then the body wouldn't go in a circle anymore and would travel in a straight line, at this point there won't be an omega because simple there won't be a circular motion! This has to mean that the at constant omega, there is a centripetal force (acceleration). This is actually a concept in the physics syllabus and the mathematics syllabus that we should solve most of the questions with...

Not to mention that the centripetal acceleration is not defined by the rate of change of angular velocity, it's the rate of change of the tangential velocity... It's is the acceleration needed to keep the velocity of the body changing so that it revolves in a circle... Angular velocity is only a property of circular motion...
Here: http://theory.uwinnipeg.ca/physics/circ/node6.html

Now for the paper 4 question...

The answer to the original question is that, as you increase the radius, the force needed to keep the body moving in circular motion must increase. Since according to the rule C.F. = m.w^2.r .However, the centripetal force in this question is the friction force which is a limited force. (1)In other words, the body at a greater distance from the center requires higher force to keep it in the circular motion. (2)Accordingly, as omega increases and by using the same rule previously mentioned, the force required to keep the mud in circular motion gradually increases until it's no longer can be achieved through friction ( only force acting on the object ).

Using the last two sentences ( 1 & 2 ), we reach that the mud at a greater distance ( near the edge ) will leave the plate first.

Hope that helped. : )

Okay, so you are saying that the centripetal force is increasing with increase in omega. Until eventually it becomes greater than friction which is providing the centripetal force. Which is why the object no longer stays intact and flies off. Correct me if I have stated something wrong.
And one more thing, is there any opposite force to friction which may be acting on the mud? Every force has an opposite reaction force, right?

 

[waleedsmz]

Okay, so you are saying that the centripetal force is increasing with increase in omega. Until eventually it becomes greater than friction which is providing the centripetal force. Which is why the object no longer stays intact and flies off. Correct me if I have stated something wrong.
And one more thing, is there any opposite force to friction which may be acting on the mud? Every force has an opposite reaction force, right?

This is exactly what I am saying.

Friction in itself is a reaction force. It is caused by both the weight of the body ( The reaction of the weight to be specific ) and the force causing it's motion ( The moving plate ). Keep in mind however that there is something called the centrifugal force which is not really included in our syllabus but it takes place with the centripetal force but in opposite direction as far as I think. I don't know any more details unfortunately.

 

[Mustehssun Iqbal]

Sorry dude but this is not true! When omega is constant there is still centripetal force and thus centripetal acceleration! Centripetal acceleration does not result in a change in speed but it does change the direction of the speed ( i.e. the velocity ). If there is a body moving in a circle and we simply remove the centripetal force ( centripetal acceleration ) then the body wouldn't go in a circle anymore and would travel in a straight line, at this point there won't be an omega because simple there won't be a circular motion! This has to mean that the at constant omega, there is a centripetal force (acceleration). This is actually a concept in the physics syllabus and the mathematics syllabus that we should solve most of the questions with...

Not to mention that the centripetal acceleration is not defined by the rate of change of angular velocity, it's the rate of change of the tangential velocity... It's is the acceleration needed to keep the velocity of the body changing so that it revolves in a circle... Angular velocity is only a property of circular motion...
Here: http://theory.uwinnipeg.ca/physics/circ/node6.html

Hope that helped. : )

"This has to mean that the at constant omega, there is a centripetal force (
Let's start from basic, F = ma. When the velocity is constant, there's no acceleration. When there's no acceleration, there's no resultant force on the object.
Similarly, In circular motion, the effect of the centripetal force is to produce a centripetal acceleration. And the effect of centripetal acceleration is to cause a rate of change of angular velocity. It changes the angular velocity, when you say it changes direction of speed, or it changes the angular velocity of the object, both are more or less the same thing. So when angular velocity is constant, centripetal acceleration is zero. When acceleration is zero, centripetal force is zero.
I hope you get the point. And pls correct me if I'm wrong...

 

[Mustehssun Iqbal]

And check my post again. It wasn't the direct answer to the question. The answer was given by Unique280. I just mentioned a correction in her answer, other than that, her answer was quite correct.

 

[waleedsmz]

"This has to mean that the at constant omega, there is a centripetal force (
Let's start from basic, F = ma. When the velocity is constant, there's no acceleration. When there's no acceleration, there's no resultant force on the object.
Similarly, In circular motion, the effect of the centripetal force is to produce a centripetal acceleration. And the effect of centripetal acceleration is to cause a rate of change of angular velocity. It changes the angular velocity, when you say it changes direction of speed, or it changes the angular velocity of the object, both are more or less the same thing. So when angular velocity is constant, centripetal acceleration is zero. When acceleration is zero, centripetal force is zero.
I hope you get the point. And pls correct me if I'm wrong...

Sorry dude this is still wrong. On a more general look, you're basically saying that if an object is moving in a circle with a constant angular velocity it requires no force to keep it moving that way, which is wrong on all concepts.

Furthermore, you said " When the velocity is constant there's no acceleration " ... Who said that the velocity is constant? Velocity is a vector quantity that has both a direction and a magnitude. When moving in a circle the direction of the velocity is constantly changing, which means that there is a change in velocity and thus acceleration that affects only the direction ( Centripetal acceleration ). This acceleration requires a force, which is the centripetal force. I'll explain more later since I have to go now...

 

[zenyatales]

Question about the electromagnetic forces between parallel conductors ...
Can anyone explain the questions in the figure with the Fleming's Rule ?
Also, in a long straight current carrying wire, the magnetic field is circular right ? But what is the direction of that magnetic field when trying to
put it into Fleming's Left Hand Rule ( as the first finger- the external magnetic field ?

 

[Mustehssun Iqbal]

Sorry dude this is still wrong. On a more general look, you're basically saying that if an object is moving in a circle with a constant angular velocity it requires no force to keep it moving that way, which is wrong on all concepts.

Furthermore, you said " When the velocity is constant there's no acceleration " ... Who said that the velocity is constant? Velocity is a vector quantity that has both a direction and a magnitude. When moving in a circle the direction of the velocity is constantly changing, which means that there is a change in velocity and thus acceleration that affects only the direction ( Centripetal acceleration ). This acceleration requires a force, which is the centripetal force. I'll explain more later since I have to go now...

I didn't say that object moving in a circle with a constant angular velocity requires no force to keep it moving in that way.What I was trying to tell was An object can't move in a uniform circle with a constant angular velocity.

 

[Mustehssun Iqbal]

Sorry dude this is still wrong. On a more general look, you're basically saying that if an object is moving in a circle with a constant angular velocity it requires no force to keep it moving that way, which is wrong on all concepts.

Furthermore, you said " When the velocity is constant there's no acceleration " ... Who said that the velocity is constant? Velocity is a vector quantity that has both a direction and a magnitude. When moving in a circle the direction of the velocity is constantly changing, which means that there is a change in velocity and thus acceleration that affects only the direction ( Centripetal acceleration ). This acceleration requires a force, which is the centripetal force. I'll explain more later since I have to go now...

You didn't get my post right, I guess
And no , when I wrote when the velocity is constant, there's no acceleration, I wasn't referring to rotational motion, but I was trying to refer to translatory motion, and later analyse both the motions in order to get to the right point. Read the post again slowly.

 

[candid24hours]

This is exactly what I am saying.

Friction in itself is a reaction force. It is caused by both the weight of the body ( The reaction of the weight to be specific ) and the force causing it's motion ( The moving plate ). Keep in mind however that there is something called the centrifugal force which is not really included in our syllabus but it takes place with the centripetal force but in opposite direction as far as I think. I don't know any more details unfortunately.

Thank you very much. It was very helpful. =>

 

[doc.ahmed]

can somebody please explain me Q6c m/j 09