Uniform circular motion and tension of a string

[robbyrandhawa]

Homework Statement

a 0.60 kg sphere rotates around a vertical shaft supported by 2 strings, as shown. if the tension in upper string is 18N calculate.

a) tension in lower string?
b) rotation rate (in rev/min) of the system.

Homework Equations

The Attempt at a Solution

there is image of the picture.

i have part a.
Fnvertical=18Ncos53 = 10.8N
Fn2=mg-Fn1
Fn2 = (0.60kg*9.8m/s^2)-10.8
Fn2 = 4.92

T2= 4.92/cos53 = 8.2N

Now part b is where i am confused. I did this so far but then i get lost.
Fn= 18Nsin53 + 8.2Nsin53
Fn = 21N --> centripetal force

Now what?? :S

 

[voko]

Mass and centripetal force give you centripetal acceleration. Centripetal acceleration and radius give you angular speed.

 

[robbyrandhawa]

so you saying... since Fn= 21N = centripetal force, i then... r=6.371X10^6m
Fr =maR
Fr = m (v^2/r)
21N = 0.60kg (v^2/6.371X10^8m)
V^2 = 2.223X10^8 ---> take tje square root
v = 1.49X10^4 m/s

from this i plugg into

v = 2(3.14)r / T

get T.. and that is the answer?

 

[robbyrandhawa]

OOPS! sorry i got my questions mixed up.. i used the incorrect r... but is the concept behind what i said correct?

 

[nil1996]

in the first answer how you have considered Fnvertical=18Ncos53 = 10.8N ?
isnt fn the vertical component of tension t1

 

[robbyrandhawa]

yes i have that for part a... if yu scroll up you will see it there in my original post...

my trouble i am having is for part B

 

[nil1996]

i mean Isn't Fn the vertical component of tension t1

 

[robbyrandhawa]

yes... not sure I am quite following you.

 

[nil1996]

You have taken the horizontal componet Fnvertical=18Ncos53 = 10.8N
it should be sin instead of cos

 

[robbyrandhawa]

when u look at the image, theta =53 is close to the vertical component, so Fnvertical = a/h = cos 53
to find the vertical component the angle is adjacent to the vertical side that's why it is cos.

 

[nil1996]

Have you figured out the answer?

 

[haruspex]

when u look at the image, theta =53 is close to the vertical component, so Fnvertical = a/h = cos 53
to find the vertical component the angle is adjacent to the vertical side that's why it is cos.

The confusion may be because the diagram shows theta as angle to the vertical but there is also a '53' written as the angle to the horizontal.
Anyway, it is not a good idea to find the numerical value of the angle. It introduces extra rounding errors. You only care about the trig functions of the angle, and you can get those directly from the triangle dimensions.
It would help greatly if you would define your variables. it's a pain having to deduce what they mean from the equations, especially since the equations might be wrong. And which directions are you defining as +ve?

 

[nil1996]

Homework Statement

Fn = 21N --> centripetal force

Now what?? :S

centripetal force =mω²r

m is the mass
ω is the angular velocity
r is the radius

from this you can find ω
from ω you can easily find rev/min

 

[robbyrandhawa]

w = 2pi(r)/T is that the formula you are talking about?

 

[robbyrandhawa]

then. T= 1/f ...

 

[haruspex]

w = 2pi(r)/T is that the formula you are talking about?

I wouldn't think so. What formula is that?
What formulae do you know for centripetal acceleration? (If you don't know any, search the net.)