- #1
Tibriel
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Homework Statement
You spin a 1.20 kg mass at the end of a 1.10 m string. The tangential velocity is 13 m/s. What is the tension in the string?
Homework Equations
x = angle string makes to the vertical
L = string length
r = L*sin(x)
T*cos(x) = mg (because the sum of the vertical forces has to be zero)
T*sin(x) = Fc = mv^2/r
The Attempt at a Solution
Below is the solution I've come up with. My real question is if there is an easier way of solving this. or something I'm missing. In class we ended up putting the 2 equations below into desmos.com and picking the first intersection before 90 degrees to see what the solution should be.
After setting up the free body diagram I get the equations above. Substituting L*sin(x) in for r and I get
T = mg/cos(x)
T = mv^2/[L*sin^2(x)]
we can set these equal to each other and eliminate the mass and I get
sin^2(x)/cos(x) = v^2/(Lg)
substitute: sin^2(x) = 1-cos^2(x)
since the whole right side is made up of known values let's just call all that 'b'
and while we're at it let's set cos(x) = a to make things look simpler
that leaves us with
(1 - a^2)/a = b
1 - a^2 = ab
a^2 + ab = 1
a^2 + ab + (b^2)/4 = 1 + (b^2)/4 <--- complete the square
(a + b/2)^2 = 1 + (b^2)/4
a + b/2 = root [1 + (b^2)/4]
a = -b/2 + root [1 + (b^2)/4]
so now we can say the angle, x = Cos^-1 (a)
and then substitute that into here: T = mg/cos(x), to get the Tension.
T = mg/a