The Student Room Group

AQA Mechanics 2 Question

Hi, can anyone help me with question 14 please? Am i right in thinking that the particle travels in a horizontal circle below AP but above BP? So that by resolving vertically you get TensionAcosX=mg +TensionBcosY?

Scroll to see replies

Original post by Firefox23
Hi, can anyone help me with question 14 please? Am i right in thinking that the particle travels in a horizontal circle below AP but above BP? So that by resolving vertically you get TensionAcosX=mg +TensionBcosY?


Yes.

Depending on what the angles are that you're refering to.
Reply 2
Original post by ghostwalker
Yes.

Depending on what the angles are that you're refering to.

The angles I am referring to at the ones that the strings make with the vertical, if this is correct then how can I eliminate both tensions? I have found r using Pythagorus but still have too many unknowns
Original post by Firefox23
The angles I am referring to at the ones that the strings make with the vertical, if this is correct then how can I eliminate both tensions? I have found r using Pythagorus but still have too many unknowns


You're going to need the motion in a circle equation as well.

Note: You're not going to get a numerical answer - it's going to be a function of angular velocity.
Hi guys. I'd like to know which are the hardest topics on the m2 exam in your opinions
Reply 5
Original post by ghostwalker
You're going to need the motion in a circle equation as well.

Note: You're not going to get a numerical answer - it's going to be a function of angular velocity.

Is that the f=mv^2/r?
Original post by Firefox23
Is that the f=mv^2/r?


That's one version, though I'd go for mrω2mr\omega^2 since you need omega in the second bit.
Reply 7
Original post by ghostwalker
That's one version, though I'd go for mrω2mr\omega^2 since you need omega in the second bit.


Ok so I'm goin to get T1sinx +T2siny=2.4ma^2w^2 (2.4a is what I found the radius to be)
Original post by Firefox23
Ok so I'm goin to get T1sinx +T2siny=2.4ma^2w^2 (2.4a is what I found the radius to be)


OK. But it should be "a" on the RHS, not a^2.

So, put your two equations together and solve for T1 and T2. You should be able to put in values for the sines and cosines.

For T2, I got:

Spoiler

(edited 9 years ago)
Reply 9
Original post by ghostwalker
OK. But it should be "a" on the RHS, not a^2.

So, put your two equations together and solve for T1 and T2. You should be able to put in values for the sines and cosines.

For T2, I got:

Spoiler



Ok I have found the two tensions and my T2 matches yours. how can i now prove w^2>5g/16a? If i sub my values for T1 and T2 back into the equation w^2 will just cancel out
Original post by Firefox23
Ok I have found the two tensions and my T2 matches yours. how can i now prove w^2>5g/16a? If i sub my values for T1 and T2 back into the equation w^2 will just cancel out


Consider the object you're modelling. What must be true of T2? (and T1 for that matter).
Reply 11
Original post by ghostwalker
Consider the object you're modelling. What must be true of T2? (and T1 for that matter).


T1 and T2 must be greater than 0 for the object to keep moving in a circle I think
Reply 12
Original post by ghostwalker
Consider the object you're modelling. What must be true of T2? (and T1 for that matter).


Making T2 greater than 0 gives the correct value for w, but making T1 greater than 0 does not
Original post by Firefox23
T1 and T2 must be greater than 0 for the object to keep moving in a circle I think


So, apply that restriction to T2 and see what you get.

Note that T1 will always be >0.
Reply 14
Original post by ghostwalker
So, apply that restriction to T2 and see what you get.

Note that T1 will always be >0.


So i can't use T1 because the question asks for w^2 to be greater than or equal to the given value. I get the correct value for w but it's negative
Original post by Firefox23
So i can't use T1 because the question asks for w^2 to be greater than or equal to the given value. I get the correct value for w but it's negative


You can't use T1 as it doesn't give any restriction on w. T1 > 0 even if w=0. A negative value of w is meaningless here, as we haven't given it a direction, clockwise or anticlockwise.

Think about it. The mass is moving in the shape of a conical pendulum. It's radius depends on how fast it is going. AP will always be taut, but BP will only be taut once the angular velocity exceeds a certain value, and the radius of the circular motion is large enough.
Reply 16
Original post by ghostwalker
You can't use T1 as it doesn't give any restriction on w. T1 > 0 even if w=0. A negative value of w is meaningless here, as we haven't given it a direction, clockwise or anticlockwise.

Think about it. The mass is moving in the shape of a conical pendulum. It's radius depends on how fast it is going. AP will always be taut, but BP will only be taut once the angular velocity exceeds a certain value, and the radius of the circular motion is large enough.


Oh ok i understand be because a smaller angular speed will result in a smaller radius so the sting is slack but once the speed picks u and the radius increases the string can become taut. So I need to find this minimum value of the radius?
Original post by Firefox23
Oh ok i understand be because a smaller angular speed will result in a smaller radius so the sting is slack but once the speed picks u and the radius increases the string can become taut. So I need to find this minimum value of the radius?


You know the radius.

You need to know the value of omega that gives that radius, which is the same as the one that makes T2 >=0. Hence the calculations you've already done.
Reply 18
Original post by ghostwalker
You know the radius.

You need to know the value of omega that gives that radius, which is the same as the one that makes T2 >=0. Hence the calculations you've already done.


But subbing values for T1 and T2 into the calculations i have done eliminates omega, i don't know how I can make omega the subject apart from doing T2 greater than or equal to 0 which gives me a minus value for omega
Original post by Firefox23
But subbing values for T1 and T2 into the calculations i have done eliminates omega, i don't know how I can make omega the subject apart from doing T2 greater than or equal to 0 which gives me a minus value for omega


Why are you mentioning T1 again?

You just need T2 >=0. Post working if it's not coming out.
(edited 9 years ago)

Quick Reply

Latest