Uniform Circular Motion Question

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GradeKing99
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Could someone explain this to me?

I would have thought that R = Centripetal force + mg, so the measured weight would be 'more than' the actual weight.

However, the answer is less than.

Why is this so? A relevant explanation/equation and/or diagram would be much appreciated
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Pangol
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(Original post by GradeKing99)
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Could someone explain this to me?

I would have thought that R = Centripetal force + mg, so the measured weight would be 'more than' the actual weight.

However, the answer is less than.

Why is this so? A relevant explanation/equation and/or diagram would be much appreciated
Your statement that R = Centripetal force + mg is equivalent to saying that Centripetal force = R - mg. You know the direction that the centripetal force must be acting in, so is R - mg correct?
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GradeKing99
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(Original post by Pangol)
Your statement that R = Centripetal force + mg is equivalent to saying that Centripetal force = R - mg. You know the direction that the centripetal force must be acting in, so is R - mg correct?
They are not looking for an expression. They are asking if the weight measured on a balance will be less than or greater than their actual weight. Does that not mean we are trying to find either R > mg or R < mg?
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Ethanity007
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The centripetal force is the net force.
mg-R = ma
R=M(g-a)

R is equivalent to the scale reading. Thus less than weight
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itslitbro
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I thought it was more about the positioning of the person and Newton's 3rd Law. The equator is the furthest point away from the centre of gravity on the surface of earth. So (N3L) that means a smaller weight (because they feel a smaller 'pulling' force), and so a smaller reading (reaction force) on the scale.
Last edited by itslitbro; 1 month ago
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itslitbro
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Stonebridge please help OP!
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Stonebridge
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The student would have to use a spring balance to detect a difference and not a scale with weights. (You might like to think why.)

1. As he is on the equator and moving round in a circle he will need a small resultant centripetal force acting towards the centre of the Earth
Let's call this resultant force C, acting downwards towards the centre of the Earth.
2. Let's have a look at the only two actual physical forces acting on him that can give him this force.
a) his weight due to gravity acting downwards (W)
b) the scale pushing up on his feet (R) (acting upwards)
3. These two act in opposite directions, but the resultant must give this small downwards centripetal force. See point 6
4. From this argument alone, it's clear that his weight downwards must be greater than the force of the scales pushing upwards.
5. As the measurement on the scale is caused by the force of the scale on the person (and person on scale), this measurement will be less than his weight.

6. In maths
C = W - R
Means
R = W - C
That is, the reaction force of the scale (which measures his weight) is less than W, the 728N that the earth is pulling him down with.
Last edited by Stonebridge; 1 month ago
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