#1
Hi, i'm doing some past papers for mechanics and i can't get this question to work, i keep ending up doing the inverse sin of a number > 1.

In a simple model of a 'rotating swing', a particle of mass 30kg is attached to one end of a light inextensible rope of length 2 metres. The other end is attached to a fixed point O. The particle moves in a horizontal circle at a constant angular speed of 3 radians per second. The rope is inclined at a constant angle theta degrees to the verticle. Taking g to be 9.8ms^-2, calculate theta.

If you gave it a go, thanks

The answer should come out as 57.0 (to 3sf)
2 ducks and a rabbi walk into a bar.
The rabbi enjoys a nice drink and the ducks are shooed out so that health services aren't called in.
The day proceeds as normal.
#2
arrrrgh - id dig my a-level papers out for you mate - if i hadnt burnt them in disgust after killing myself working all year on Physics crap.

good luck, youll need it - i feel for you.
#3
thanks
2 ducks and a rabbi walk into a bar.
The rabbi enjoys a nice drink and the ducks are shooed out so that health services aren't called in.
The day proceeds as normal.
#4
centrifugal force = m * angular speed ^2 * radius (which you can get from theta and the length of the string)

and then you have the regular gravitational force which goes down, and again, from theta you can work out the some kind of connection between these 2

maybe some kind of drawring would illustrate it better
#5
of course! i took the length of the string as the radius!!
thank you!
2 ducks and a rabbi walk into a bar.
The rabbi enjoys a nice drink and the ducks are shooed out so that health services aren't called in.
The day proceeds as normal.
#6
note, a = theta

resolving horizontally gives t sin a=30w^2 r
vertically gives t cos a = 30g

divide 1st by second to give tan a = w^2 r/g

tan a = w^2 L sin a

cos a = g/2 w^2

put the values in and there you go
#7
Fcf = m * r * sinθ * ω^2
Fg = m*g

tan θ = m * r * sinθ * ω^2/(m*g) = r * sinθ * ω^2 / g

tan θ = r * sinθ * ω^2 / g
sin θ / cos θ = r * sinθ * ω^2 / g
1 / cos θ = r *ω^2 / g
cos θ = g/(r *ω^2)

θ = arccos(g/(r *ω^2) )

http://www.google.si/search?hl=sl&client=firefox-a&rls=org.mozilla%3Aen-GB%3Aofficial&hs=Igd&q=arccos%289.81m%2Fs%5E2%2F%282m+*%283%2Fs%29%5E2%29+%29+in+degrees&btnG=Iskanje&meta=
#8
m=30kg
g=9.81ms-1
w (rotational velocity)=3rads-1
L=2.0m

using @ for theta

cos@=g/(L(w^2))
cos@=9.81/2.0*3^2
cos@=0.545
@=acos (or cos^-1) 0.545
@=56.9 degrees

just like magic... no - not magic - physics!!!

hope this helps

richard - (final year civil engineering student) (and chronic UG'er)
#10
haha yeah i know, i was bored writing up all my physic experiments for my project so i decided to give it a go. I forgot how easy it was XD
#11
Quote by l3vity
haha yeah i know, i was bored writing up all my physic experiments for my project so i decided to give it a go. I forgot how easy it was XD

yeah, i was almost hoping for something you needed a differential equation to solve (the masochist that I am. our TA really loved pendulums :P)