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Equilibrium and tension (2 questions)

  • #1
1)A mass 'm' is dropped from a spring with constant 'k'. find the time it takes to reach equilibrium.

im pretty sure i can use this eqn
T = 2*PI*sqrt(m/k)






2)find the tension at the lowest point of the pendulum, with length L and mass M.

there will be zero work done by tension at the bot, it is at a r. angle. im thinking you have to use k + U = k_0 + U_0 for conservation of mechanical energy, but i was also thinking of using the y component of the problem.

T - m*g = a * m

a = v^2 / r

T = m*g + m*(v^2/r)

v = omega*r and omega = sqrt (g/L)

T = m*g +(m*g) / L


this dosnt seem right
 

Answers and Replies

  • #2
anyone?
 
  • #3
Doc Al
Mentor
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1)A mass 'm' is dropped from a spring with constant 'k'. find the time it takes to reach equilibrium.

im pretty sure i can use this eqn
T = 2*PI*sqrt(m/k)
You will certainly need that result.

2)find the tension at the lowest point of the pendulum, with length L and mass M.

there will be zero work done by tension at the bot, it is at a r. angle. im thinking you have to use k + U = k_0 + U_0 for conservation of mechanical energy, but i was also thinking of using the y component of the problem.

T - m*g = a * m

a = v^2 / r

T = m*g + m*(v^2/r)
Looks good. How will you find the speed? From what height was it released?

v = omega*r and omega = sqrt (g/L)

T = m*g +(m*g) / L
:confused:
 
  • #4
wouldnt v or h be given in the question? these are just general questions i have right now not h/w or anything.

but
for velocity

(mv^2)/r = m*g

v^2 = g*r


im not to sure what to do for height
 
Last edited:

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