How Is Tension Calculated in a Pendulum with Varying Angles?

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The discussion focuses on calculating the tension (T) in a pendulum system with a mass (m) hanging from a cord of length (L) that swings at varying angles (alpha) up to a maximum angle (alpha max). The user derives an expression for tension using principles of circular motion and energy conservation, ultimately arriving at T = 3mgcos(alpha) - 2mgcos(alpha max). There is uncertainty about the necessity of including length (L) in the final expression for tension. The conversation highlights the relationship between tension, gravitational force, and angular displacement, emphasizing the importance of understanding the system dynamics. The user seeks clarification on their calculations and the role of L in the expression.
devanlevin
how can i express tension in this case, using alpha, alpha max,g, m, and L ??

a case of a mass of m hanging from a ceiling on a cord with a length of L swinging at a maximum angle of "alpha max" on either side, and at any given moment and an angle of alpha(changes all the time)

how can i express T, the strings tension, using the parameters m,g,alpha,alpha max,and L??

what i did was say:
according to circular movement, looking at the radial acceleration ar
F=ma
T-mg*cos(alpha)=mar=m*v^2/L
T=m(v^2/L+g*cos(alpha))

now how can i express v using the parameters, what i think i need to do is look at the maximum angle, and say V=0 so mar=0 so T(point of max)=mgcos(alpha max) but i don't see how that cann further me at all,
again, in the end i need an expression T=? using only m,g,L,alph, alpha max

***no friction at all, mass will reach alpha max every time, no loss of energy
 
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T-mg*cos(alpha)=(m/L)*v^2

V^2=(L/m)(T-mgcos(alpha))

from here using energy,
E(max point)=mgh=const
my point of reference for potential energy being the ceiling

h=cos(alpha)*L

E=0.5mv^2+mgh=-mgcos(alpha max)*L
v^2=(L/m)(T-mgcos(alpha))

0.5(m)(L/m)(T-mgcos(alpha))-mgcos(alpha)*L=-mgcos(alpha)*L

0.5(T-mgcos(alpha))-mgcos(alpha)=-mgcos(alpha)
T-mgcos(alpha)-2mgcos(alpha)=-2mgcos(alpha)

T=3mgcos(alpha)-2mgcos(alpha max)

is this correct??
the question asked me to use L in the expression, do i need it??
 
I would start by determining the period of the pendulum which will be a function of length and gravity. You then have your alpha max which will determine your displacement. Period and displacement you can then equate to velocity.

EDIT: Bit of a time lapse with post #2, wasn't able to read it before this post. What happened to this L?

0.5(m)(L/m)(T-mgcos(alpha))-mgcos(alpha)*L=-mgcos(alpha)*L
 
i divided the whole equation by L
 
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