m3 circular motion

https://160a4a0fd863e9dcf6f7a7eb2ecff88758e1bb02.googledrive.com/host/0B1ZiqBksUHNYOXlTZ2NiQWdoRm8/Edexcel-Set-2/Ch.4%20Motion%20in%20a%20Circle.pdf

Question 5
Why when resolving towards the centre would it be t+tsin theta=mrw^2?
thats what it sais in the mark scheme

There are multiple Q5's in there.

the first question 5 lol
sorry bout that (techinaccly the 2nd question)

Okay, so there's two things acting radially. One is the tension in the string BR which is of magnitude T and acts in the direction of the centre, so when you resolve you get a $T$ term. The other term comes from resolving the tension in the string AR. That has tension $T \sin \theta$ radially and $T \cos \theta$ vertically. So when you resolve radially you get $T + T \sin \theta = m\omega^2 r$ which is the centripetal force.

oh yeah thanks bro
so two tensions in a string would always have equal tension ?

Well no, my wording was rather sneaky because when I said "two strings" I really meant "two different parts of the same string" and it's an inextensible string so the tension at any point in the string is the same. i.e: the two different parts of the string have the same tension.

ah thanks