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PHYA4 Forces Question watch

1. Hi,

I had an attempt at PHYA4 June 2011, Got 56/75 which I think was just about an A* but I did incredibly poorly on Q1 scoring 1/7.

I was wondering if you could please help me understand how to derive the formula and for part 1a)ii and how to correctly answer the question for 1b) as a mark scheme can only go so far.

page1
page2

thanks a lot.
2. For 1aii the principle is that there needs to be a resultant centripetal force mv2/r on the object in the van if it is to move in a circle. This force will be downwards when the van is at the top of the bridge.
There are two actual forces acting on the parcel, its weight mg downwards (positive) and the reaction force R upwards (negative).
The sum of these two must be equal to the centripetal force.
So mg - R = mv2/r

Rearrange for R

b.
This formula tells you what happens to R as v gets larger.

The question is referring to what happens when R=0
This happens when mg = mv2/r
At this point the parcel just starts to leave the shelf as R becomes zero.
Plug the numbers in to see what happens when v=15m/s

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