The Student Room Group

M3: slack strings

how would I do part b? for the string to become slack, the particle must travel back past its natural length. the EPE stored in the string at point b will allow for the particle to travel back up the slope - how do I determine that?

or am I thinking about it wrong?
Reply 1
Original post by Maths&amp
how would I do part b? for the string to become slack, the particle must travel back past its natural length. the EPE stored in the string at point b will allow for the particle to travel back up the slope - how do I determine that?

or am I thinking about it wrong?


I would use conservation of energy.
When particle is at instantaneous rest its speed is zero, and it is about to start travelling back up the plane.

As it travels up the plane it will gain gpe and KE, the sum of which is equal to the loss of EPE.

U know the EPE when particle is at instantaneous rest cus u know its extension, u know how far it must travel up the plane to have no tension in string, u should be able to work out the speed when it goes slack now.

I havent done it myself if there are any problems lmk.
Original post by Shaanv
I would use conservation of energy.
When particle is at instantaneous rest its speed is zero, and it is about to start travelling back up the plane.

As it travels up the plane it will gain gpe and KE, the sum of which is equal to the loss of EPE.

U know the EPE when particle is at instantaneous rest cus u know its extension, u know how far it must travel up the plane to have no tension in string, u should be able to work out the speed when it goes slack now.

I havent done it myself if there are any problems lmk.


what do you mean it will gain GPE? gravity is a constant?

I see how it will gain KE as its moving up the plane and this movement is due to EPE stored in the string at b.

so: [tex ]EPE(B) = KE(A) + GPE(A) + EPE(A)

but still dont see how we gain GPE because I though gravity acts towards (towards B).
(edited 6 years ago)
Reply 3
Original post by Maths&amp
what do you mean it will gain GPE? gravity is a constant?

I see how it will gain KE as its moving up the plane and this movement is due to EPE stored in the string at b.

so: [tex ]EPE(B) = KE(A) + GPE(A) + EPE(A)


but still dont see how we gain GPE because I though gravity acts towards (towards B).

It moves up slope, it gains vertical height, GPE=mgh says that its gpe will increase as it moves up slope.
Original post by Maths&physics
what do you mean it will gain GPE? gravity is a constant?

I see how it will gain KE as its moving up the plane and this movement is due to EPE stored in the string at b.

so: [tex ]EPE(B) = KE(A) + GPE(A) + EPE(A)


but still dont see how we gain GPE because I though gravity acts towards (towards B).

What's A,B?? I can only assume that B is the the starting point which is the point at which the particle stops in part (a), and A is the point where the string becomes slack.

Gravity acts vertically down, by moving up the plane the particle displaces vertically therefore there is change in the GPE.

From what you put, it seems like you are taking the point of reference to be the starting point. Note that at A we have EPE=0 because the string is slack.
(edited 6 years ago)
Original post by Shaanv
It moves up slope, it gains vertical height, GPE=mgh says that its gpe will increase as it moves up slope.


duh! so as long as there is a change in height, there is a change in GPE. in the direction of gravity, its positive and against the direction of gravity, its negative? or it doesn't work like that?
(edited 6 years ago)
Original post by RDKGames
What's A,B?? I can only assume that B is the the starting point which is the point at which the particle stops in part (a), and B is the point where the string becomes slack.

Gravity acts vertically down, by moving up the plane the particle displaces vertically therefore there is change in the GPE.

From what you put, it seems like you are taking the point of reference to be the starting point. Note that at A we have EPE=0 because the string is slack.


yep and yes. :smile:
Reply 7
Original post by Maths&amp
duh! so as long as there is a change in height, there is a change in GPE. in the direction of gravity, its positive and against the direction of gravity, its negative? or it doesn't work like that?


If something moves vertically against gravity it gains GPE as work has to be done to move it against gravity in the first place. If it moves vertically with gravity, gravity is doing the work and so it loses GPE.

Its all to do with conservation of energy
Original post by Shaanv
If something moves vertically against gravity it gains GPE as work has to be done to move it against gravity in the first place. If it moves vertically with gravity, gravity is doing the work and so it loses GPE.

Its all to do with conservation of energy


thanks. its as simple as the alternating value the h: you go up, h increases so GPE increase. you move down, the value of h decreases and you lose GPE.

thanks.

also, does R (reaction force) = mg even on a slope?
(edited 6 years ago)
Original post by Maths&physics
also, does R (reaction force) = mg even on a slope?


No. R is equal to the weight component perpendicular to the slope.
Original post by RDKGames
No. R is equal to the weight component perpendicular to the slope.


how would I approach this question?
Original post by Maths&physics
how would I approach this question?


At this point you may as well make a note and stick it on top of your screen to always post the full question lol...
What's part (a)?? Post all of it.
Original post by RDKGames
At this point you may as well make a note and stick it on top of your screen to always post the full question lol...
What's part (a)?? Post all of it.


sorry :colondollar:
Original post by RDKGames
At this point you may as well make a note and stick it on top of your screen to always post the full question lol...
What's part (a)?? Post all of it.


I should use vertical forces when resolving because theyre in equilibrium?
Original post by Maths&physics
I should use vertical forces when resolving because theyre in equilibrium?


TBH I'm having issues modelling the situation from their wording, so I'll let someone else give it a go.
Original post by RDKGames
TBH I'm having issues modelling the situation from their wording, so I'll let someone else give it a go.


thanks
Original post by Maths&physics
I should use vertical forces when resolving because theyre in equilibrium?


Yes. You have 3 forces acting, the weight, the normal reaction and the friction. Because the car is going at the maximum speed that is can without slipping sideways, Fr = 0.6 R and the friction is acting down the slope as the car is on the point of sliding outwards on the bend (17 year old driver pushing it to the limits?).

As you say, the vertical forces are in equilibrium. Horizontally the sum of the components of the friction and the normal reaction = m a

Here a = v^2/r
(edited 6 years ago)
Original post by tiny hobbit
Yes. You have 3 forces acting, the weight, the normal reaction and the friction. Because the car is going at the maximum speed that is can without slipping sideways, Fr = 0.6 R and the friction is acting down the slope as the car is on the point of sliding outwards on the bend (17 year old driver pushing it to the limits?).

As you say, the vertical forces are in equilibrium. Horizontally the sum of the components of the friction and the normal reaction = m a

Here a = v^2/r


thank you. I managed to work it out though.

however, here (according to the mark scheme), why do we take GPE and KE at the top point (A) but no GPE at the bottom point (P). is it because we must always take the lower point to have GPE = 0? this would make sense if point P was at the Bottom of the circle but not midway. can you explain please????
(edited 6 years ago)
Original post by Maths&physics
thank you. I managed to work it out though.

however, here (according to the mark scheme), why do we take GPE and KE at the top point (A) but no GPE at the bottom point (P). is it because we must always take the lower point to have GPE = 0? this would make sense if point P was at the Bottom of the circle but not midway. can you explain please????


You choose where to take the 0 for GPE. It is easier if you take it at a low fixed place, since then the GPE anywhere else will be positive since it will be ABOVE the zero level.

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