laurawatt
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I’m stuck on this question and can’t find anything for it on the internet: (part b)
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Here’s what I’ve got so far:
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I know the scalar product must =0 for them to be perpendicular, I just can’t work out how to get there!
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5hyl33n
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I’m sorry this is not going to be a useful response but your handwriting is beautiful!
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RogerOxon
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(Original post by laurawatt)
I’m stuck on this question and can’t find anything for it on the internet: (part b)
Name:  B40D1436-1DCF-4372-851F-281E4C55CFE5.jpeg
Views: 8
Size:  88.4 KB
Here’s what I’ve got so far:
Name:  7CC1BD1D-705B-4855-B323-EB17207ED5F4.jpeg
Views: 5
Size:  35.2 KB
I know the scalar product must =0 for them to be perpendicular, I just can’t work out how to get there!
What you've written for the scalar product is a vector - it shouldn't be. Sum the two terms that you wrote.
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laurawatt
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(Original post by RogerOxon)
What you've written for the scalar product is a vector - it shouldn't be. Sum the two terms that you wrote.
To get this?
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And then because it =0 that proves it’s perpendicular right?
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simon0
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Hint:  \boldsymbol{v} \centerdot \boldsymbol{a} .

Also what have you been told about the direction of acceleration?
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RogerOxon
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(Original post by laurawatt)
To get this?
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Size:  24.2 KB
And then because it =0 that proves it’s perpendicular right?
If the scalar product is 0, then either the two vectors are perpendicular, or one (or both) are of zero magnitude.

You still haven't got a scalar for the scalar product. The LHS is a vector - it is not the scalar product.

Here's an example. If we have two vectors (1,2) and (2,-1), the scalar product is 1*2 + 2*-1 = 0, so, as both are non-zero vectors, they are perpendicular.
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laurawatt
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(Original post by simon0)
Hint:  \boldsymbol{v} \centerdot \boldsymbol{a} .

Also what have you been told about the direction of acceleration?
The acceleration is towards the centre of the circle (so it’s parallel to the radius)
(Original post by RogerOxon)
If the scalar product is 0, then either the two vectors are perpendicular, or one (or both) are of zero magnitude.

You still haven't got a scalar for the scalar product. The LHS is a vector - it is not the scalar product.

Here's an example. If we have two vectors (1,2) and (2,-1), the scalar product is 1*2 + 2*-1 = 0, so, as both are non-zero vectors, they are perpendicular.
Using v•a, which gives me the same overall result as using r
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(sorry, I must sound thick as mud, I just can’t figure out what to do)
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davros
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(Original post by laurawatt)
The acceleration is towards the centre of the circle (so it’s parallel to the radius)

Using v•a, which gives me the same overall result as using r
Name:  D830333E-ED6A-4A98-ADCF-21184A9BF885.jpeg
Views: 7
Size:  41.0 KB
(sorry, I must sound thick as mud, I just can’t figure out what to do)
You've basically got it

You can use \boldsymbol{v} \centerdot \boldsymbol{a} or \boldsymbol{v} \centerdot \boldsymbol{r} because you know that the acceleration is parallel to the radius.

Your dot product produces 2 terms which are equal but of opposite sign, so they sum to zero. Unless you have the trivial case where one of the vectors is identically zero you have therefore proved that they are perpendicular to each other.
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laurawatt
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(Original post by davros)
You've basically got it

You can use \boldsymbol{v} \centerdot \boldsymbol{a} or \boldsymbol{v} \centerdot \boldsymbol{r} because you know that the acceleration is parallel to the radius.

Your dot product produces 2 terms which are equal but of opposite sign, so they sum to zero. Unless you have the trivial case where one of the vectors is identically zero you have therefore proved that they are perpendicular to each other.
thank you (PRSOM!)
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