Uniform circular motion and tension of a string

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Homework Help Overview

The discussion revolves around a problem involving uniform circular motion and the tension in strings supporting a rotating sphere. The original poster presents a scenario where a 0.60 kg sphere rotates around a vertical shaft, with specific tensions in the strings and seeks to calculate the tension in the lower string and the rotation rate of the system.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between tension and centripetal force, with some attempting calculations for the lower string's tension and the centripetal force involved in the motion.
  • Questions arise regarding the correct application of trigonometric functions to resolve components of tension, particularly whether to use sine or cosine.
  • There is confusion about the radius used in calculations, with participants discussing the implications of using incorrect values.

Discussion Status

The conversation is ongoing, with participants providing guidance on how to relate centripetal force to angular velocity and discussing the importance of defining variables clearly. There is no explicit consensus on the correct approach to the problem, as different interpretations of the tension components and radius are being explored.

Contextual Notes

Participants note the potential for confusion due to the diagram's angles and the need for clarity in variable definitions. The original poster expresses uncertainty about the next steps in their calculations, indicating a lack of complete information on how to proceed.

robbyrandhawa
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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
 

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Mass and centripetal force give you centripetal acceleration. Centripetal acceleration and radius give you angular speed.
 
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?
 
OOPS! sorry i got my questions mixed up.. i used the incorrect r... but is the concept behind what i said correct?
 
in the first answer how you have considered Fnvertical=18Ncos53 = 10.8N ?
isnt fn the vertical component of tension t1
 
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
 
i mean Isn't Fn the vertical component of tension t1
 
yes... not sure I am quite following you.
 
You have taken the horizontal componet Fnvertical=18Ncos53 = 10.8N
it should be sin instead of cos
 
  • #10
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.
 
  • #11
Have you figured out the answer?
 
  • #12
robbyrandhawa said:
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?
 
  • #13
robbyrandhawa said:

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
 
  • #14
w = 2pi(r)/T is that the formula you are talking about?
 
  • #15
then. T= 1/f ...
 
  • #16
robbyrandhawa said:
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.)
 

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