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lots of m2 help needed! (ocr)

I'm really panicking about this exam. I know all the formulae and stuff but the bloody exams seem to want to test actual knowledge of what's going on! If anybody could possibly help me, I need it on the june 2002 paper question 6:

A ball of mass 0.08kg is attatched by 2 strings to a fixed vertical post. The strings have lengths 2.5m for the upper string and 2.4m for lower string, which is horizontal. They are both taut. The ball moves in a horizontal circle of radius 2.4 m at constant speed vm/s.

i. Find the tension in each string when v=10.5

ii. Find the least value of v for which the lower string is taut.

I managed part i with a little help from the mark scheme but i don't understand how to do part ii. Is the tension in the upper string constant? I don't get motion in a circle at all- i just manage to put in values to get other things! please help!!! and any random tips to remember for the exam would be good too...i thought p3 was a hard module but this is worse! x

Reply 1

Squishy

Circular motion on the Edexcel syllabus is in M3, which I haven't done for a while, so forgive me if I'm a bit rusty.

I get the centripetal force to be 3.675N, and hence the tension in the upper (longer) string is 2.8N, and the tension in the lower string is 0.987N.

If the lower string is about to become slack, then there is no tension in it. Hence the centripetal force causing circular motion is provided entirely by the upper string (= 2.8N)

F = (mv^2)/r
2.8 = (0.08v^2)/2.4
v^2 = 84
so v is 9.17 m/s to 2DP

Reply 2

Jonny W

Let S be the tension in the lower string, and T be the tension in the higher string.

The higher string makes an angle theta = arccos(24/25) ~= 0.283 with the horizontal. Observe that 25^2 - 24^2 = 7^2 (it's a nice triangle), so sin(theta) = 7/25.

Resolve vertically at the particle:

0.08g = T sin(theta) = 7T/25,
T = 2.8.

The acceleration of the particle is v^2/2.4 inwards. Resolve horizontally at the particle:

S + T cos(theta) = 0.08 v^2/2.4,
S = 0.08 v^2/2.4 - 2.8*(24/25) = v^2/30 - 336/125

(i) When v = 10.5, S = (10.5)^2/30 - 336/125 = 0.987.

(ii) If you decrease v from 10.5 then S will go down. The lower string will be on the point of going slack when S = 0, ie, when

v^2/30 - 336/125 = 0,
v^2 = 2016/25,
v = 8.980.

Reply 3

Jonny W

Squishy

If the lower string is about to become slack, then there is no tension in it. Hence the centripetal force causing circular motion is provided entirely by the upper string (= 2.8N)

The upper string isn't horizontal. You have to multiply 2.8 by the cosine of the angle that the upper string makes with the horizontal.

Reply 4

Squishy

Jonny W

The upper string isn't horizontal. You have to multiply 2.8 by the cosine of the angle that the upper string makes with the horizontal.

Yeah, sorry, your answer's right...I just spotted my mistake. Chalk it up to severe sleep deprivation. :smile:

Reply 5

picmar

It says find the value of V for which the lower string remains taut. the string is not taut unless the tension is as it is in the first part of the question? so to solve it doint you take the longer string to have zero tension and the lower one to be as it is in the first part?