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# Circular Motion Question watch

1. Hi all need a bit of help please

A small fluffy toy hangs on a string in the rear window of a car. The car rounds a bend at 30ms-1 and the string hangs 20 degrees from the vertical.

There is a diagram just of the string hanging 20 degrees from vertical.

a) mark the forces acting on the toy

I have no idea so any help would be good.

Thanks
2. (Original post by TheStrokess)
Hi all need a bit of help please

A small fluffy toy hangs on a string in the rear window of a car. The car rounds a bend at 30ms-1 and the string hangs 20 degrees from the vertical.

There is a diagram just of the string hanging 20 degrees from vertical.

a) mark the forces acting on the toy

I have no idea so any help would be good.

Thanks
For a) i got the tenson in the string, weight acting down on the toy, and centripetal force towards the center.

for b) Well i"m pretty weak at this topic myself, but I got 252.1m, seems a lot. I did it by using tan(20)=30^2/9.81xR R=252.1m. ?

Anyone confirm this or am i totally wrong?
3. I'm poor at it also. I always thought the car was experiencing the centripetal force and the bodies inside the car kept 'trying to move in a straight line' according to newtons law. Thanks for the help so far.
4. (Original post by Im_a_cyborg)
For a) i got the tenson in the string, weight acting down on the toy, and centripetal force towards the center.

for b) Well i"m pretty weak at this topic myself, but I got 252.1m, seems a lot. I did it by using tan(20)=30^2/9.81xR R=252.1m. ?

Anyone confirm this or am i totally wrong?
I just checked the MS and 252 is correct. Can you explain what you've done here in more detail? I would appreciate it.

Thanks
5. Yes, no problem, from the diagram of the toy, sin(20) gives the centripetal force (towards the centre) so sin(20)=mv^2/R and
cos(20)= mg (the weight)

You cannot use sin(20)=mv^2/R since you do not have the mass of the car, so you combine mv^2/R with mg. 'm" cancels leaving v^2/gr which is equal to tan(20) and get 'r' from that

Its hard to explain but if you remember;
sin(angle)=mv^2/R
cos(angle)= mg
tan(angle)=v^2/gr

You can't go wrong.
Hope that helps.
6. (Original post by Im_a_cyborg)
Yes, no problem, from the diagram of the toy, sin(20) gives the centripetal force (towards the centre) so sin(20)=mv^2/R and
cos(20)= mg (the weight)

You cannot use sin(20)=mv^2/R since you do not have the mass of the car, so you combine mv^2/R with mg. 'm" cancels leaving v^2/gr which is equal to tan(20) and get 'r' from that

Its hard to explain but if you remember;
sin(angle)=mv^2/R
cos(angle)= mg
tan(angle)=v^2/gr

You can't go wrong.
Hope that helps.
Yeah i know sin/cos is tan. Brilliant! Just didnt see where the dividing by g came from but of course now i do.
Thanks. much appreciated.
So for the first part is it safe to assume for these q's anything in the car experiences the same centripetal force as the car itself?
7. (Original post by TheStrokess)
Yeah i know sin/cos is tan. Brilliant! Just didnt see where the dividing by g came from but of course now i do.
Thanks. much appreciated.
So for the first part is it safe to assume for these q's anything in the car experiences the same centripetal force as the car itself?
Yes I think so, as everything in the car is traveling at the same speed and direction.

No problem.
8. I got 205 for this answer, can anyone tell me what i've done wrong, i think the mark scheme may be incorrect

v^2/g x tan(20)

30^2 = 900
g=9.81
tan(20) = 2.24

r = 900 x 2.24 / 9.81 = 205.2m

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