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Circular motion

Can somebody help explain the second part of this question for me please?

I got as far as:
Minimum speed -> tension in lower string = 0
If

\theta

is the angle that the upper string makes with the vertical, resolving vertically:

T\cos{\theta} = 2g

Horizontally:

T\sin{\theta} = \frac{2v^2}{r}

The mark scheme seems to have found r = 1.2 but I can't see where they get that from. It's probably just me having a moment and it's really obvious :colondollar:

Thank you!

Reply 1

ghostwalker

Original post by felecita

Can somebody help explain the second part of this question for me please?

I got as far as:
Minimum speed -> tension in lower string = 0
If

\theta

is the angle that the upper string makes with the vertical, resolving vertically:

T\cos{\theta} = 2g

Horizontally:

T\sin{\theta} = \frac{2v^2}{r}

The mark scheme seems to have found r = 1.2 but I can't see where they get that from. It's probably just me having a moment and it's really obvious :colondollar:

Thank you!

Diagram if you haven't already.

A horizontal through P creates two triangles, whose sides are in ratio 3:4:5. Hence radius =1.2.

Or use Pythagoras.

Reply 2

Zacken

Original post by felecita

Can somebody help explain the second part of this question for me please?

You're going to have to exercise your 3D visualtion skills here. The horiztal circle means that the there's a radial length and then a vertical downwards length from the centre of the circle downwards to the particle A/B with length 3.2/2 and a hypotenuse with length of the string, 2. I can't explain it very well, so hopefully the crappy diagrams below will help:

Side view:

3D-ish view:

You'll have to kind of imagine the green circle bulging out with the centre being the midpoint of the red line, where the red line passes through the middle of the centre. Kind of like a hula hoop thrown over a pole where the red line is the pole and the green circle is the hula hoop.

The blue lines are a crappily drawn triangle that's showcased in the side view above.

Reply 3

Zacken

Grrr at ghostwalkers ninja. :shakecane:

Reply 4

exo97

Original post by felecita

Can somebody help explain the second part of this question for me please?

I got as far as:
Minimum speed -> tension in lower string = 0
If

\theta

is the angle that the upper string makes with the vertical, resolving vertically:

T\cos{\theta} = 2g

Horizontally:

T\sin{\theta} = \frac{2v^2}{r}

The mark scheme seems to have found r = 1.2 but I can't see where they get that from. It's probably just me having a moment and it's really obvious :colondollar:

Thank you!

Yeah you are having one of those moments lol. Shame this isn't a moments question :biggrin:

Anyways, I think you are right with what you did. r = 1.2 comes from the string A or B which are both 2m and when you do 2sin

\theta

, (where

\theta

= 36.869... degrees), you get 1.2m for the horizontal distance.

I got 2.97ms^-1 for part ii. Is that the answer for that question?

Reply 5

Zacken

Original post by exo97

I got 2.97ms^-1 for part ii. Is that the answer for that question?

Yes.

Reply 6

felecita

Original post by ghostwalker

Original post by Zacken

Original post by exo97

Sorry for the delayed reply, thank you so much everybody! So the diagram just stays the same as the first part of the question? For some reason I thought that if the tension was 0 in the lower string, the whole thing would move in a bit and it wouldn't be 1.6 on the left. I don't think that made any sense whatsoever but yeah :lol: