Equilibrium and tension (2 questions)

  • Thread starter Thread starter nightshade123
  • Start date Start date
  • Tags Tags
    Equilibrium Tension
Click For Summary
SUMMARY

The discussion focuses on two physics problems involving equilibrium and tension in mechanical systems. The first problem involves calculating the time to reach equilibrium for a mass 'm' dropped from a spring with a constant 'k', using the formula T = 2π√(m/k). The second problem addresses finding the tension at the lowest point of a pendulum with length 'L' and mass 'M', where the tension is derived from the equation T = mg + m(v²/r), with v calculated as v = ωr and ω = √(g/L).

PREREQUISITES
  • Understanding of harmonic motion and spring constants
  • Knowledge of pendulum dynamics and tension forces
  • Familiarity with conservation of mechanical energy principles
  • Basic calculus for deriving velocity and angular frequency
NEXT STEPS
  • Study the derivation of the time period for simple harmonic motion using T = 2π√(m/k)
  • Learn about the forces acting on a pendulum at different points in its swing
  • Explore the conservation of energy in mechanical systems, particularly in pendulums
  • Investigate the relationship between angular velocity and linear velocity in circular motion
USEFUL FOR

Students and professionals in physics, mechanical engineering, and anyone interested in understanding the principles of equilibrium and tension in mechanical systems.

nightshade123
Messages
82
Reaction score
0
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. I am 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 doesn't seem right
 
Physics news on Phys.org
anyone?
 
nightshade123 said:
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. I am 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:
 
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*rim not to sure what to do for height
 
Last edited:

Similar threads

  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 17 ·
Replies
17
Views
1K
  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 20 ·
Replies
20
Views
2K
Replies
2
Views
2K
Replies
6
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K