Goods
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In this question on part 2 what does in mean by instantaneously rolling?
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davros
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(Original post by Goods)


In this question on part 2 what does in mean by instantaneously rolling?
I think it's just their way of pointing out that the 'axis' AB isn't a fixed line - at any point in time there is a particular line AB about which the sphere is 'instantaneously rolling', but if you look at the motion at a later time, the line AB is in a different place.

(You would probably have made this assumption anyway, I'm just trying to interpret what they meant by using those particular words.)
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(Original post by davros)
I think it's just their way of pointing out that the 'axis' AB isn't a fixed line - at any point in time there is a particular line AB about which the sphere is 'instantaneously rolling', but if you look at the motion at a later time, the line AB is in a different place.

(You would probably have made this assumption anyway, I'm just trying to interpret what they meant by using those particular words.)
So it's still rotating about an axis through it's CoM?
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DFranklin
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(Original post by Goods)
So it's still rotating about an axis through it's CoM?
No. Look at the diagram - does the line AB go through the CofM?
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davros
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(Original post by DFranklin)
No. Look at the diagram - does the line AB go through the CofM?
Out of interest - I haven't done things like this for years! - is this question basically requiring an application of the parallel axes theorem to get a new MI and hence a revised acceleration, or is there some other principle involved here? It looks quite an interesting question but I haven't had a bash at the algebra
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(Original post by DFranklin)
No. Look at the diagram - does the line AB go through the CofM?
Well treating it as rotating around the CoM i got the right solution. If you treat it as rotating around the axis A-B there's no moment due to the friction?


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DFranklin
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(Original post by davros)
Out of interest - I haven't done things like this for years! - is this question basically requiring an application of the parallel axes theorem to get a new MI and hence a revised acceleration, or is there some other principle involved here? It looks quite an interesting question but I haven't had a bash at the algebra
That would be my approach but according to the OP this isn't right...

[I must say that although these questions were completely routine in my FM A-level, I haven't studied it since. There was one STEP question on this (which I don't believe was any harder than the A-level questions used to be) but I understand it was decided my solution was wrong there. So maybe I just don't know what I'm talking about anymore...]
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ghostwalker
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(Original post by Goods)
If you treat it as rotating around the axis A-B there's no moment due to the friction?
True, but there is a moment due to a component of the weight of the ball down the plane/wedge.

I took AB as the centre of rotation and got the correct result.
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(Original post by DFranklin)
That would be my approach but according to the OP this isn't right...

[I must say that although these questions were completely routine in my FM A-level, I haven't studied it since. There was one STEP question on this (which I don't believe was any harder than the A-level questions used to be) but I understand it was decided my solution was wrong there. So maybe I just don't know what I'm talking about anymore...]
I'd be interested in seeing how to do it with parallel axis, i tried it originally but i couldn't get an answer. Looking the examiner report and it says that most people recalled the theorem but couldn't apply it, so its probably faster with parallel axis and what i've done mist take it into account inadvertently.


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ghostwalker
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(Original post by Goods)
I'd be interested in seeing how to do it with parallel axis
Somewhat tersely.

\text{M.of I. about AB}=\frac{2}{5}ma^2+m(a\sin\frac  {\phi}{2})^2

Considering rotation about AB we get the initial equation:

mg\sin\theta\times a\sin\frac{\phi}{2}=ma^2(\frac{2  }{5}+\sin^2\frac{\phi}{2})\times  \frac{\text{acceleration}}{a\sin  \frac{\phi}{2}}

Rearrange and simplify
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(Original post by ghostwalker)
Somewhat tersely.

\text{M.of I. about AB}=\frac{2}{5}ma^2+m(a\sin\frac  {\phi}{2})^2

Considering rotation about AB we get the initial equation:

mg\sin\theta\times a\sin\frac{\phi}{2}=ma^2(\frac{2  }{5}+\sin^2\frac{\phi}{2})\times  \frac{\text{acceleration}}{a\sin  \frac{\phi}{2}}

Rearrange and simplify
Thanks, \frac{\text{acceleration}}{a\sin  \frac{\phi}{2}} am i right in thinking this comes from the non slipping condition? If you consider it rotating around AB and you say at A,B the instantaneous velocity is zero then v-wr=0 but as A is on the axis surely r is zero and that would imply v must also be zero?


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ghostwalker
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(Original post by Goods)
Thanks, \frac{\text{acceleration}}{a\sin  \frac{\phi}{2}} am i right in thinking this comes from the non slipping condition? If you consider it rotating around AB and you say at A,B the instantaneous velocity is zero then v-wr=0 but as A is on the axis surely r is zero and that would imply v must also be zero?


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The acceleration is the linear/tangential acceleration of the centre of mass, which is the distance from the line AB to the centre of mass times the angular accleration about AB; and rearrange.

And yep, rotation about AB, as it's not slipping.
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atsruser
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(Original post by Goods)


In this question on part 2 what does in mean by instantaneously rolling?
At the risk of repeating what has already been said, or what you already know:

1. In this kind of rotational motion question, "instantaneously rolling" is one of those magic phrases you see in maths/physics questions. It is a big hint to you that you need to consider the no-slip condition at the point in question.

2. That's because if it wasn't "rolling", then there would be relative motion between the sphere and the surface i.e. it would be slipping. The word "instantaneously" also stresses the fact that the point that is not slipping is changing all the time, as the sphere rolls. The fact that the sphere is rolling about that point (even if it's not the same point from instant to instant) tells you to use the MI of the sphere about that point (or the axis defined by two such points). This inspires the use of the parallel axis theorem here.

3. You actually need to find the acceleration of the sphere down the slope and usually you would need to solve \Sigma F = ma down the slope. However, the no-slip condition is a nice strong kinematical constraint, that forces the accln down the slope to be proportional to the angular accln of the sphere via:

v = a\omega \Rightarrow \dot{v} = a\alpha

where v is the speed of the the c-o-m of the sphere down the slope. So now you just need to find \alpha via T=I\alpha and that also gives you the required accln.

4. Since the point of no-slip is also the point through which the friction acts, it also makes sense to take the torque about this point since that then removes the need to know anything about the friction.
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