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Edexcel M3 Circular Motion on a Rotating Disc

Textbook's Question and Solutions: https://gyazo.com/64f164a4f144e8a66f808d6355e6b1cc
My questions: For answer b, why did they take the friction as negative? I get why they'd take it as positive since it acts in the same direction as the centripetal force.
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Zacken
22
Original post by Nerrad
Textbook's Question and Solutions: https://gyazo.com/64f164a4f144e8a66f808d6355e6b1cc
My questions: For answer b, why did they take the friction as negative? I get why they'd take it as positive since it acts in the same direction as the centripetal force.


I'm not sure I understand your question? For one case (either maximum or minimum) friction acts with tension and resolving radially gives it as positive or negative . For the other case (either maximum or minimum) friction acts in the opposite direction against tension and resolving radially gives it as negative or positive, i.e it switches sign from the previous case. These give you a lower and upper bound for ω2\omega^2.
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Nerrad
OP
7
Original post by Zacken
I'm not sure I understand your question? For one case (either maximum or minimum) friction acts with tension and resolving radially gives it as positive or negative . For the other case (either maximum or minimum) friction acts in the opposite direction against tension and resolving radially gives it as negative or positive, i.e it switches sign from the previous case. These give you a lower and upper bound for ω2\omega^2.

I don't know what you mean by "resolving radially". Also doesn't friction acts to the left (towards the centre of the circle), therefore you resolve (friction+Tension= Centripetal force), so why when you get the lower bound of Omega they resolve (-friction + tension= centripetal force) wouldn't this mean that friction is now acting to the right instead of to the left(towards the centre of the circle)?

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