What is the tension as a function of theta

AI Thread Summary
The discussion focuses on determining the tension in a string as a function of the angle theta for a mass m moving with an initial speed V. Participants emphasize the importance of applying Newton's second law and considering centripetal acceleration. Conservation of energy is highlighted as essential, leading to the equation relating initial and final speeds along with gravitational potential energy. The correct expression for tension is derived as T = mg cos(theta) + (mv^2)/L, factoring in the forces acting on the mass. The conversation confirms the need for energy conservation in solving the problem accurately.
amackeytexas
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Need some help with this answer.
A mass m on the end of a string in length L. At the bottom it is given a push to a speed V. What is the tension as a function of theta, where theta is the angle from the intial position?
 
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Start by identifying the forces acting on the mass. Then apply Newton's 2nd law, realizing that the mass is centripetally acceleration. Conservation of energy will come in handy as well.
 
Thanks. How close is this? mv^2/L=Tcos theta
 
amackeytexas said:
Thanks. How close is this? mv^2/L=Tcos theta
Check it for \theta = 0. Does it make sense? (What forces act on the mass at that angle?)
 
forces at that angle are Tsin theta=mg
 
Hi makeytexas,

in order to find the corect answer you must take into account the energy's conservation law too:

\frac{m v_0^2}{2}=\frac{m v^2}{2}+m g L cos \theta

where v_0 is the initial speed.

The static equilibrium condition along the string direction would be

T=mg cos \theta + \frac{m v^2}{L}

...
 
Last edited:
clive said:
in order to find the corect answer you must take into account the energy's conservation law too:

\frac{m v_0^2}{2}=\frac{m v^2}{2}+m g L cos \theta

where v_0 is the initial speed.
That should be:
\frac{m v_0^2}{2}=\frac{m v^2}{2}+m g L (1 - cos \theta)
 
You're right Doc Al !
 
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