Uniform Circular Motion/Newton's 2nd Law?

  • #1
Cryptologica
22
0

Homework Statement



Question: A 3.45 kg object is attached to a vertical rod by two strings as in Figure P6.11. The object rotates in a horizontal circle at constant speed 5.65 m/s. What are the tensions of the two strings, individually?

Variables:
m = 3.45kg
length of both strings (individually) = 2m
vertical distance between two strings = 3m


Homework Equations



Sum(F) = m(v^2/r)
Sum(F) = m(a)
Basic trig.

The Attempt at a Solution



I assumed that the ball was revolving around the middle of the rod, thus we have a triangle with hypotenuse of 2m and side 1.5m and by solving we find the radius:
r = sqrt(1.75) = 1.32m.
We should be able to find the sum of the forces by using m(v^2/r):
(3.45kg)((5.65m/s)^2/1.32m) = 83.25N.
This is the centripetal force, right?
a_c = v^2/r, thus a_c = (5.65m/s)^2/(√1.75m)=33.429m/s^2
I then used Sum(F) = ma to find the x-component of T (tension of string):
(3.45kg)(33.429m/s^2) = 115.33N
We have a triangle with legs 1.5m and √1.75m and hypotenuse 2m, thus the angle made by T and r is θ = sin^-1(1.5m/2m) = 48.59°.
Therefore, T = 115.33N/cos(48.59°) = 174.36N
Both strings share the force equally and both have this value, correct?
I just want to make sure I am doing this right...I have already lost points at wrong attempts already (online homework). Could someone look this over and make sure I did it correctly and notify me if there's anything I might have misunderstood? Thanks!
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
9
Hi Cryptologica, welcome to PF.
In the equilibrium condition, weight of the object acts in the downward direction. In your attempt there in no mg term. Obviously the tensions in the two strings are not equal. Now try it again.
 
  • #3
Cryptologica
22
0
So which one will have more tension? How do I account for mg?
 
  • #4
rl.bhat
Homework Helper
4,433
9
Go through the thread " Newton's 2nd Law for a Particle in Uniform Circular motion" below.
 
  • #5
Cryptologica
22
0
So would the y-comp(of T) = Tsin(θ) - mg
and x = Tcosθ? I don't have to factor that into x because mg is a vertical force? Is that going in the right direction?
 
  • #6
Cryptologica
22
0
Ah, sorry...I am trying to follow that other thread, but am having a hard time. I have been working on this for too long probably, but I think I am almost there? Any more hints you can give, I am still kind of confused?
 
  • #7
rl.bhat
Homework Helper
4,433
9
Σ Fy = T1cosθ - (T2cosθ + mg) = 0
ΣFx = (T1 + T2)sinθ - Fc = 0
Here cos θ = 3/4
Now solve for T1 and T2.
 
  • #8
Cryptologica
22
0
θ = cos^-1(3/4) = 41.41
t1cos(41.41)-(t2cos(41.41) + (3.45kg)(9.8m/s^2)) = 0
(3/4)t1 - (3/4)t2 - 33.81N = 0
t1 = 48.08N+t2
substitute:
(48.08N+2(t2))sin(41.41) = fc
fc/2sin(41.41) + 24.04N = t2

Did I do something wrong, or do I know fc? Then I could plug and solve for t2 and t1? fc = centripetal force, which is?
 
  • #9
Cryptologica
22
0
Oh, is fc = m(v^2/r)?
fc = (3.45kg)[(5.65m/s)^2/sqrt(1.75)] = 83.25N
Then use that to solve, right?
 
  • #10
rl.bhat
Homework Helper
4,433
9
t1 = 48.08N+t2
check this step.
fc/2sin(41.41) + 24.04N = t2
Check this step.
 

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